IRISAN KERUCUT

 A. Pendahuluan

Irisan Kerucut yang dimasud di sini adalah bangun-bangun geometri yang diperoleh dengan cara mengiris sebuah kerucut tegak berselimut ganda dengan sebuah bidang datar dengan arah pengirisan tertentu. Garis potong antara bidang datar dan kerucut tegak tersebut akan menyebabkan beberapa kemungkinan, di antaranya:

• lingkaran, jika kerucut dipotong oleh bidang datar tegak lurus dengan sumbu kerucut dan tidak melalui titik puncak kerucut atau dapat juga dengan kondisi dipotong oleh bidang datar dan sejajar dengan bidang alas
• elips, jika kerucut dipotong pada bidang miringdari garis pelukis sampai garis pelukislainnya, atau dapat juga dikatakan dengan kondisi di mana kerucut dipotong oleh bidang datar membentuk sudut lancip terhadap sumbu dan tidak melalui puncak kerucut
• parabola, jika bidang datar membentuk sejajar dengan garis pelukis kerucut dan tidak melalui puncak kerucut
• hiperbola, jika bidang datar sejajar dengan sumbu kerucut dan tidak melalui titik nol

Kerucut tegak ganda

Kerucut dipotong oleh bidang datar sejajar alas kerucut, penampang irisannya berupa lingaran

penampang irisannya berupa elips

Penampang irisannya berupa parabola

Penampang irisannya berupa hiperbola

$\begin{aligned}&\textrm{Eksentrisitas}\\ &e=\displaystyle \frac{PF}{PL}\\ &\begin{array}{|c|c|c|c|}\hline \textrm{Lingkaran}&\textrm{Elips}&\textrm{Parabola}&\textrm{Hiperbola}\\ (\textrm{Circle})&(\textrm{Ellips})&(\textrm{Parabola})&(\textrm{Hyperbola})\\\hline e=0&e<1&e=1&e>1\\\hline \end{array} \end{aligned}$

DAFTAR PUSTAKA

1. Kurnia, N, dkk. 2017. Jelajah Matematika SMA Kelas XI Peminatan MIPA. Jakarta: YUDHISTIRA

CONTOH SOAL TRANSFORMASI FUNGSI BAGIAN 5

 $\begin{array}{ll}\\ 21.&\textrm{Jika}\: \: T_{1}=\begin{pmatrix} 1 & 2\\ 1 & 1 \end{pmatrix}\: \: \textrm{dan}\: \: T_{2}=\begin{pmatrix} -2 & 5\\ -1 & 3 \end{pmatrix}\\ &\textrm{maka bayangan garis}\: \: x+y+1=0\\ &\textrm{oleh}\: \: T_{2}\circ T_{1}\: \: \textrm{adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad x-2y-1=0&&\\ \textrm{b}.\quad x+2y-1=0&\\ \textrm{c}.\quad x+2y+1=0&\\ \textrm{d}.\quad x-2y+1=0\\ \textrm{e}.\quad x+y-1=0 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui bahwa}:\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=T_{2}\circ T_{1}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} -2 & 5\\ -1 & 3 \end{pmatrix}\begin{pmatrix} 1 & 2\\ 1 & 1 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} -2+5 & -4+5\\ -1+3 & -2+3 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} 3 & 1\\ 2 & 1 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} 3x+y\\ 2x+y \end{pmatrix}\\ \textrm{Dipe}&\textrm{roleh}\\ &\begin{array}{lllllllll}\\ \quad x'&=3x+y\\ \quad y'&=2x+y\qquad\quad-\\\hline x'-y'&=x\\ \Leftrightarrow \quad x&=x'-y'\qquad ....(1)\\ \textrm{maka}\\ \qquad y&=x'-3x\\ &=x'-3(x'-y')\\ &=3y'-2x'\quad ....(2) \end{array}\\ \textrm{Sehin}&\textrm{gga}\\ x+y&+1=0\\ x'-&y'+3y'-2x'+1=0\\ -x'+&2y'+1=0\\ x'-&2y'-1=0\\ \textrm{maka}&\: \textrm{bayangan garisnya}\\ x-2&y-1=0 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 22.&\textrm{Diketahui}\: \: M\: \: \textrm{adalah pencerminan terhadap}\\ &\textrm{garis}\: \: y=-x\: \: \textrm{dan}\: \: T\: \: \textrm{adalah transformasi} \\ &\textrm{yang dinyatakan oleh matriks}\: \: \begin{pmatrix} 2 & 3\\ 0 & -1 \end{pmatrix}\\ &\textrm{Koordinat bayangan titik}\: \: A(2,-8)\: \: \textrm{oleh}\\ &\textrm{transformasi}\: \: M\: \: \textrm{dilanjutkan oleh}\: \: T\: \: \textrm{adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad (-10,2)&&\\ \textrm{b}.\quad (-2,-10)&\\ \textrm{c}.\quad (10,2)&\\ \textrm{d}.\quad (-10,-2)\\ \textrm{e}.\quad (2,10) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui bahwa}:\\ \begin{pmatrix} x'\\ y' \end{pmatrix} &=T\circ M\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} 2 & 3\\ 0 & -1 \end{pmatrix}\begin{pmatrix} 0 & -1\\ -1 & 0 \end{pmatrix}\begin{pmatrix} 2\\ -8 \end{pmatrix}\\ &=\begin{pmatrix} 0-3 & -2+0\\ 0+1 & 0+0 \end{pmatrix}\begin{pmatrix} 2\\ -8 \end{pmatrix}\\ &=\begin{pmatrix} -3 & -2\\ 1 & 0 \end{pmatrix}\begin{pmatrix} 2\\ -8 \end{pmatrix}\\ &=\begin{pmatrix} -6+16\\ 2+0 \end{pmatrix}\\ &=\begin{pmatrix} 10\\ 2 \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 23.&\textrm{Jika}\: \: W\: \: \textrm{adalah transformasi oleh}\\ &\textrm{matriks}\: \: \begin{pmatrix} 1 & 0\\ 3 & 1 \end{pmatrix},\: \: \textrm{maka titik mula}\\ &\textrm{dari}\: \: W'(-2,5)\: \: \textrm{adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad (-11,-2)&&\\ \textrm{b}.\quad (11,-2)&\\ \textrm{c}.\quad (-2,11)&\\ \textrm{d}.\quad (2,11)\\ \textrm{e}.\quad (12,11) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\textrm{Dimi}&\textrm{salkan}:\\ A&=\begin{pmatrix} -2\\ 5 \end{pmatrix},\: \: \textrm{dan}\\ W&=\begin{pmatrix} 1 & 0\\ 3 & 1 \end{pmatrix},\: \: \textrm{serta}\: \: X=\begin{pmatrix} x\\ y \end{pmatrix}\\ \textrm{mak}&\textrm{a}\\ &\begin{array}{|c|}\hline \begin{aligned}A&=BX\\ B^{-1}A&=B^{-1}BX\\ B^{-1}A&=I.X\\ B^{-1}A&=X\\ X&=B^{-1}A \end{aligned}\\\hline \end{array}\\ \begin{pmatrix} x\\ y \end{pmatrix}&=\displaystyle \frac{1}{\begin{vmatrix} 1 &0 \\ 3 & 1 \end{vmatrix}}\begin{pmatrix} 1 & 0\\ -3 & 1 \end{pmatrix}\begin{pmatrix} -2\\ 5 \end{pmatrix}\\ &=1.\begin{pmatrix} -2+0\\ 6+5 \end{pmatrix}\\ &=\begin{pmatrix} -2\\ 11 \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 24.&\textrm{Jika setiap titik pada grafik dengan}\\ &\textrm{dengan persamaan}\: \: y=\sqrt{x}\: \: \textrm{dicerminkan} \\ &\textrm{terhadap garis}\: \: y=x\: ,\: \textrm{maka persamaan}\\ &\textrm{grafik yang dihasilkan adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad y=x^{2}\: ,\: x\geq 0&&\\ \textrm{b}.\quad y=-\sqrt{x}\: ,\: x\geq 0&\\ \textrm{c}.\quad y=-x^{2}\: ,\: x\leq 0&\\ \textrm{d}.\quad y=\sqrt{-x}\: ,\: x\leq 0\\ \textrm{e}.\quad y=-\sqrt{-x}\: ,\: x\leq 0 \end{array}\\\\ &\quad\quad\qquad \textbf{UMB Tahun 2011 Kode 152}\\\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui bahwa}:\\ y&=\sqrt{x},\: \: \textrm{atau}\: \: y^{2}=x\\ \textbf{Alt}&\textbf{ernatif 1}\\ \textrm{mak}&\textrm{a}\: \: \textrm{saat dicerminkan terhadap}\\ \textrm{gari}&\textrm{s}\: \: y=x,\: \textrm{adalah}\: \: x^{2}=y\\ \textrm{atau}&\: \: y=x^{2}.\\ \textbf{Alt}&\textbf{ernatif 2}\\ \textrm{Jika}\: &\textrm{ingin dikerjakan dengan rumus}\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=M_{x=y}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} y\\ x \end{pmatrix}\\ \textrm{Sela}&\textrm{njutnya hasilnya disubstitusikan}\\ \textrm{ke p}&\textrm{ersamaan}\: \: y=\sqrt{x}\Rightarrow x'=\sqrt{y'}\\ \sqrt{y'} &=x'\: \: \: \textrm{maka}\\ y'&=\left ( x' \right )^{2}\: \: \: \textrm{selanjutnya}\\ y&=x^{2} \end{aligned} \end{array}$.

Sebelum dicerminkan terhadap garis y=x

Gambar kurva/grafik setelah cerminkan terhadap garis y=x

$\begin{array}{ll}\\ 25.&\textrm{Transformasi}\: \: T\: \: \textrm{adalah pencerminan}\\ &\textrm{terhadap garis}\: \: y=\displaystyle \frac{x}{3}\: \: \textrm{dilanjutkan oleh} \\ &\textrm{pencerminan terhadap garis}\: \: y=-3x.\\ &\textrm{Matriks yang bersesuian dengan}\\ &\textrm{transformasi}\: \: T\: \: \textrm{adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad \begin{pmatrix} -1 & 0\\ 0 & 1 \end{pmatrix}&&\\ \textrm{b}.\quad \begin{pmatrix} -1 & 0\\ 0 & -1 \end{pmatrix}&\\ \textrm{c}.\quad \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}&\\ \textrm{d}.\quad \begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix}\\ \textrm{e}.\quad \begin{pmatrix} 0 & -1\\ -1 & 0 \end{pmatrix} \end{array}\\\\ &\quad\quad \textbf{SBMPTN Tahun 2013 Kode 433}\\\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui bahwa}:\\ \textrm{sebu}&\textrm{ah persamaan garis lurus}\\ \textrm{dapa}&\textrm{t dituliskan dengan}:\: y=mx\\ \textrm{Dike}&\textrm{tahui pula bahwa ada 2 garis}:\\ y_{1}&=\displaystyle \frac{1}{3}x\quad \textrm{dan}\: \: \: y_{2}=-3x\\ \textrm{seba}&\textrm{gai representasi transformasi}\: \: T.\\ \textrm{Kare}&\textrm{na}\: \: m_{1}\times m_{2}=\left ( \displaystyle \frac{1}{3} \right )(-3)=-1\\ \textrm{bera}&\textrm{rti 2 garis di atas saling tegak}\\ \textrm{luru}&\textrm{s dan hal ini seperti rotasi 2}\\ \textrm{kali}\: \: &90^{\circ}\: \: \textrm{atau}\: \: 180^{\circ}\\ \textrm{Jadi},&\: T=\begin{pmatrix} \cos 180^{\circ} & -\sin 180^{\circ}\\ \sin 180^{\circ} & \cos 180^{\circ} \end{pmatrix}\\ \Leftrightarrow &\: T=\begin{pmatrix} -1 & 0\\ 0 & -1 \end{pmatrix} \end{aligned} \end{array}$.

DAFTAR PUSTAKA

1. Johanes, Kastolan, Sulasim, 2006. Kompetensi Matematika 3A SMA Kelas XII Program IPA Semester Pertama. Jakarta: YUDHISTIRA.
2. Nugroho, P. A. Gunarto, D. 2013. Big Bank Soal-Bahas MAtematika SMA/MA. Jakarta: WAHYUMEDIA.
3. Santoso, N.A., Aksin, N. 2024. PR Matematika untuk SMA/MA/SMK/MAK Kelas 12. Yogyakarta: INTAN PARIWARA EDUKASI
4. Sharma,S.N., dkk. 2017. Jelajah Matematika SMA Kelas XI Program Wajib. Jakarta: YUDHISTIRA.

CONTOH SOAL TRANSFORMASI FUNGSI BAGIAN 4

 $\begin{array}{ll}\\ 16.&\textrm{Garis}\: \: 2x+y+4=0\: \: \textrm{ditranslasikan}\\ &\textrm{oleh}\: \: \begin{pmatrix} -2\\ 5 \end{pmatrix}\: \: \textrm{dilanjutkan transformasi} \\ &\textrm{oleh}\: \: \begin{pmatrix} 1 & 2\\ 0 & 1 \end{pmatrix}\: \: \textrm{persamaan bayangannya}\\ &\textrm{adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad 2x+y+3=0&&\\ \textrm{b}.\quad 2x-3y+3=0&\\ \textrm{c}.\quad 2x+3y+3=0&\\ \textrm{d}.\quad 3x+2y+3=0\\ \textrm{e}.\quad 3x-2y+3=0 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{Diket}&\textrm{ahui bahwa}:\\ \begin{pmatrix} x'\\ y' \end{pmatrix}\: &=\begin{pmatrix} x\\ y \end{pmatrix}+\begin{pmatrix} -2\\ 5 \end{pmatrix}=\begin{pmatrix} x-2\\ y+5 \end{pmatrix}\\ \begin{pmatrix} x''\\ y'' \end{pmatrix}&=\begin{pmatrix} 1 & 2\\ 0 & 1 \end{pmatrix}\begin{pmatrix} x'\\ y' \end{pmatrix}\\ &=\begin{pmatrix} 1 & 2\\ 0 & 1 \end{pmatrix}\begin{pmatrix} x-2\\ y+5 \end{pmatrix}\\ &=\begin{pmatrix} x-2+2y+10\\ y+5 \end{pmatrix}\\ &=\begin{pmatrix} x+2y+8\\ y+5 \end{pmatrix}\\ \textrm{Diper}&\textrm{oleh}\\ &\begin{array}{lllllllll}\\ \quad x''&=x+2y+8\\ \: \: \: 2y''&=2y+10\qquad\quad-\\\hline x''-2y''&=x-2\\ \Leftrightarrow \quad x&=x''-2y''+2\: ....(1)\\ \textrm{maka}\\ \qquad y&=y''-5\quad \qquad....(2) \end{array}\\ \textrm{sehin}&\textrm{gga}\\ 2x+&y+4=0\\ 2(x''&-2y''+2)+(y''-5)+4=0\\ 2x''-&3y''+3=0\\ \textrm{maka}&\: \textrm{bayangan garisnya}\\ 2x-&3y+3=0 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 17.&\textrm{Titik A(4,-4) dicerminkan terhadap}\\ &\textrm{garis}\: \: y=x\tan 15^{\circ}\: \: \textrm{menghasilkan}\\ &\textrm{bayangan}\: \: A'(a,b)\: \: \textrm{adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad \sqrt{3}&&\textrm{d}.\quad 4\sqrt{3}\\ \textrm{b}.\quad 2\sqrt{3}&\textrm{c}.\quad 3\sqrt{3}&\textrm{e}.\quad 6\sqrt{3} \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\begin{pmatrix} a\\ b \end{pmatrix}&=\begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} \cos 2.15^{\circ}& \sin 2.15^{\circ}\\ \sin 2.15^{\circ} & -\cos 2.15^{\circ} \end{pmatrix}\begin{pmatrix} 4\\ -4 \end{pmatrix}\\ &=\begin{pmatrix} \cos 30^{\circ}&\sin 30^{\circ}\\ \sin 30^{\circ}&-\cos 30^{\circ} \end{pmatrix}\begin{pmatrix} 4\\ -4 \end{pmatrix}\\ &=\begin{pmatrix} \displaystyle \frac{1}{2}\sqrt{3} & \displaystyle \frac{1}{2}\\ \displaystyle \frac{1}{2} & -\displaystyle \frac{1}{2}\sqrt{3} \end{pmatrix}\begin{pmatrix} 4\\ -4 \end{pmatrix}\\ &=\begin{pmatrix} 2\sqrt{3}-2\\ 2+2\sqrt{3} \end{pmatrix}\\ &\begin{cases} a &=2\sqrt{3}-2 \\ b &=2+2\sqrt{3} \end{cases}\\ \textrm{mak}&\textrm{a nilai dari}\\ a+b&=\left ( 2\sqrt{3}-2+2+2\sqrt{3} \right )\\ &=4\sqrt{3} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 18.&\textrm{Lingkaran}\: \: x^{2}+y^{2}-5x+8y+7=0\\ & \textrm{ditranslasikan oleh}\: \: T=\begin{pmatrix} m\\ n \end{pmatrix}\: \: \textrm{menghasilkan}\\ &\textrm{bayangan}\: \: x^{2}+y^{2}-9x+2y+6=0.\\ & \textrm{Nilai}\: \: m+n=\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad 2&&\textrm{d}.\quad 5\\ \textrm{b}.\quad 3&\qquad\textrm{c}.\quad 4\qquad&\textrm{e}.\quad 6 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\textrm{Dik}&\textrm{etahui sebuah lingkaran dengan persamaan}:\\ & x^{2}+y^{2}-5x+8y+7=0\\ \textrm{kar}&\textrm{ena akibat translasi, maka}\\ &\begin{cases} x & =x'-m \\ y & =y'-n \end{cases}\\ &x^{2}+y^{2}-5x+8y+7=0\\ \textrm{seh}&\textrm{ingga}\\ &\Leftrightarrow (x'-m)^{2}+(y'-n)^{2}-5(x'-m)+8(y'-n)+7=0\\ &\Leftrightarrow x'^{2}+y'^{2}-2mx'-2ny'+m^{2}+n^{2}-5x'+5m+8y'-8n+7=0\\ &\Leftrightarrow x'^{2}+y'^{2}-(2m+5)x'+(8-2n)y'+m^{2}+n^{2}+5m-8n+7=0\\ &\qquad \equiv \: x'^{2}+y'^{2}-9x'+2y'+6=0\qquad (\textbf{akhir bayangan})\\ &\begin{cases} 9 &=2m+5 \Rightarrow m=2\\ 2 & =8-2n \: \Rightarrow \, \: n=3 \end{cases}\\ \textrm{Jad}&\textrm{i , nilai}\: \: m+n=2+3=5\end{aligned} \end{array}$.

$\begin{array}{ll}\\ 19.&\textrm{Jika titik A(-2,1) dicerminkan terhadap garis}\\ & y=-\displaystyle \frac{1}{3}x\sqrt{3}\: ,\: \textrm{maka bayangan dari}\\ &\textrm{titik A tersebut adalah}\, ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad A'\left ( 1-\displaystyle \frac{1}{2}\sqrt{3},-\displaystyle \frac{1}{2}+\sqrt{3} \right )&&\\ \textrm{b}.\quad A'\left ( -1-\displaystyle \frac{1}{2}\sqrt{3},-\displaystyle \frac{1}{2}+\sqrt{3} \right )&\\ \textrm{c}.\quad A'\left (-1-\displaystyle \frac{1}{2}\sqrt{3},\displaystyle \frac{1}{2}-\sqrt{3} \right )&\\ \textrm{d}.\quad A'\left ( 1-\displaystyle \frac{1}{2}\sqrt{3},\displaystyle \frac{1}{2}-\sqrt{3} \right )\\ \textrm{e}.\quad A'\left ( -1+\displaystyle \frac{1}{2}\sqrt{3},-\displaystyle \frac{1}{2}+\sqrt{3} \right ) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{Diketahui}&\: \textrm{bahwa}:\\ y&=-\displaystyle \frac{1}{3}x\sqrt{3}=\left ( -\displaystyle \frac{1}{3}\sqrt{3} \right )x\\ &=\left (-\tan 30^{\circ} \right )x=\tan \left ( 180^{\circ}-30^{\circ} \right )x\\ &=\tan 150^{\circ}.x\\ \textrm{maka}\: \: \theta &=150^{\circ}\quad \Rightarrow \quad 2\theta =300^{\circ}\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} \cos 300^{\circ} & \sin 300^{\circ} \\ \sin 300^{\circ} & -\cos 300^{\circ} \end{pmatrix}\begin{pmatrix} -2\\ 1 \end{pmatrix}\\ &=\begin{pmatrix} \displaystyle \frac{1}{2} & -\displaystyle \frac{1}{2}\sqrt{3}\\ -\displaystyle \frac{1}{2}\sqrt{3} & -\displaystyle \frac{1}{2} \end{pmatrix}\begin{pmatrix} -2\\ 1 \end{pmatrix}\\ &=\begin{pmatrix} -1-\displaystyle \frac{1}{2}\sqrt{3}\\ \sqrt{3}-\displaystyle \frac{1}{2} \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 20.&\textrm{Bayangan titik A(2,4) dicerminkan }\\ &\textrm{terhadap garis}\: \: y-x=0\: \: \textrm{dilanjutkan}\\ &\textrm{ke garis}\: \: x\sqrt{3}-3y=0\: \: \textrm{adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad A'\left ( 2+\sqrt{3},-1+2\sqrt{3} \right )&&\\ \textrm{b}.\quad A'\left ( 2+\sqrt{3},1-2\sqrt{3} \right )&\\ \textrm{c}.\quad A'\left ( 1-\sqrt{3},-2+\sqrt{3} \right )&\\ \textrm{d}.\quad A'\left ( -2+\sqrt{3},1+2\sqrt{3} \right )\\ \textrm{e}.\quad A'\left ( 2-\sqrt{3},1-2\sqrt{3} \right ) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui bahwa}:\\ &\begin{cases} x\sqrt{3}-3y=0 & \Leftrightarrow y=\displaystyle \frac{1}{3}\sqrt{3}x\\ &\Leftrightarrow y=\tan 30^{\circ}.x\\\\ x-y=0 & \Leftrightarrow y=x \end{cases}\\\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} \cos 2.30^{\circ} & \sin 2.30^{\circ} \\ \sin 2.30^{\circ} & -\cos 2.30^{\circ} \end{pmatrix}\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} \displaystyle \frac{1}{2} & \displaystyle \frac{1}{2}\sqrt{3}\\ \displaystyle \frac{1}{2}\sqrt{3} & -\displaystyle \frac{1}{2} \end{pmatrix}\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} 2\\ 4 \end{pmatrix}\\ &=\begin{pmatrix} \sqrt{3}+2\\ -1+2\sqrt{3} \end{pmatrix} \end{aligned} \end{array}$.

CONTOH SOAL TRANSFORMASI FUNGSI BAGIAN 3

$\begin{array}{ll}\\ 11.&\textrm{Titik A(1,-2) dirotasikan sejauh}\: \: 15^{\circ}\\ & \textrm{kemudian dilanjutkan}\: \: 75^{\circ}\: \: \textrm{dengan pusat }\\ &O(0,0)\: \: \textrm{maka bayangan akhir titik A adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad (-2,1)&&\textrm{d}.\quad (2,1)\\ \textrm{b}.\quad (-1,2)&\textrm{c}.\quad (1,2)&\textrm{e}.\quad (-2,-1) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} \cos (\theta _{1}+\theta _{2}) & -\sin (\theta _{1}+\theta _{2})\\ \sin (\theta _{1}+\theta _{2}) & \cos (\theta _{1}+\theta _{2}) \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} \cos (75^{\circ}+15^{\circ})& -\sin (75^{\circ}+15^{\circ})\\ \sin (75^{\circ}+15^{\circ}) & \cos (75^{\circ}+15^{\circ}) \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} \cos 90^{\circ}&-\sin 90^{\circ}\\ \sin 90^{\circ}&\cos 90^{\circ} \end{pmatrix}\begin{pmatrix} 1\\ -2 \end{pmatrix}\\ &=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1\\ -2 \end{pmatrix}\\ &=\begin{pmatrix} 2\\ 1 \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 12.&\textrm{Sebuah lingkaran yang berpusat di (3,4) }\\ &\textrm{dan menyinggung sumbu-X dicerminkan}\\ &\textrm{terhadap garis}\: \: y=x\: \textrm{, maka persamaan }\\ &\textrm{akhir lingkaran yang terjadi adalah}\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad x^{2}+y^{2}-8x-6y+9=0&&\\ \textrm{b}.\quad x^{2}+y^{2}+8x+6y+9=0&\\ \textrm{c}.\quad x^{2}+y^{2}+6x+8y+9=0&\\ \textrm{d}.\quad x^{2}+y^{2}-8x-6y+16=0\\ \textrm{e}.\quad x^{2}+y^{2}+8x+6y+16=0\end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\textrm{Refleksi l}&\textrm{ingkaran yang berpusat di (3,4) }\\ \textrm{dan men}&\textrm{yinggung sumbu-X, }\\ \textrm{dengan}\: \: r&=(y)=4,\\ \textrm{maka}\: \textrm{per}&\textrm{samaan lingkarannya adalah}:\\ (x-3)^{2}+&(y-4)^{2}=4^{2}.\: \textrm{Karena}\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} y\\ x \end{pmatrix}\\ &\begin{cases} x & =y' \\ y & =x' \end{cases}\\ \textrm{selanjutn}&\textrm{ya untuk persamaan bayangan }\\ \textrm{lingkaran} &\textrm{nya adalah}:\\ &(y'-3)^{2}+(x'-4)^{2}=4^{2},\\ & \textbf{menjadi}\\ &(y-3)^{2}+(x-4)^{2}=4^{2},\quad \textrm{atau}:\\ &x^{2}+y^{2}-8x-6y+9=0 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 13.&\textrm{Jika}\: \: M_{x}\: \: \textrm{adalah pencerminan terhadap sumbu-X }\\ &\textrm{dan}\: \: M_{y=x}\: \: \textrm{adalah pencerminan terhadap garis}\\ & y=x\: ,\: \textrm{maka matriks transformasi tunggal }\\ &\textrm{yang mewakili}\: \: M_{x}\circ M_{y=x}=\, ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \begin{pmatrix} 0 &-1 \\ 1 & 0 \end{pmatrix}&&\textrm{d}.\quad \begin{pmatrix} -1 & 0\\ 0 & -1 \end{pmatrix}\\ \textrm{b}.\quad \begin{pmatrix} 0 &1 \\ -1 & 0 \end{pmatrix}&&\textrm{e}.\quad \begin{pmatrix} -1 & 0\\ 0 & 1 \end{pmatrix}\\ \textrm{c}.\quad \begin{pmatrix} 0 & -1\\ -1 & 0 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{Diketahui}\: &\textrm{bahwa}:\\ &\begin{cases} M_{x} & = \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}\\ M_{y=x} & =\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} \end{cases}\\ M_{x}\circ M_{y=x}&=\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\\ &=\begin{pmatrix} 0+0 & 1+0\\ 0-1 & 0+0 \end{pmatrix}\\ &=\begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 14.&\textrm{Diketahui vektor}\: \: \vec{x}\: \: \textrm{dirotasikan terhadap titik asal}\\ & O\: \: \textrm{sebesar}\: \: \theta >0\: \: \textrm{searah jarum jam}.\\ &\textrm{Kemudian hasilnya dicerminkan terhadap garis}\: \: y=0\\ & \textrm{menghasilkan vektor}\: \: \vec{y}.\\ &\textrm{Jika}\: \: \vec{y}=A.\vec{x}\: ,\: \textrm{maka matriks}\: \: A-\textrm{nya adalah}\, ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}&&\\ \textrm{b}.\quad \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}&&\\ \textrm{c}.\quad \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}&&\\ \textrm{d}.\quad \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}\\ \textrm{e}.\quad \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\begin{aligned}&\textrm{Diketahui bah}\textrm{wa}:\\ &\begin{cases} M_{x} & = \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}\\ R_{-\theta } & =\begin{pmatrix} \cos (-\theta ) & -\sin (-\theta )\\ \sin (-\theta ) & \cos (-\theta ) \end{pmatrix}=\begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \end{cases}\\ &A=M_{x}\circ R_{-\theta }=\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}\begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 15.&\textrm{Jika garis}\: \: 3x-2y+5=0\: \: \textrm{dicerminkan }\\ &\textrm{terhadap garis}\: \: y=-x\: \: \textrm{kemudian}\\ &\textrm{didilatasikan dengan pusat (1,-2) }\\ &\textrm{dengan faktor skala 2, maka persamaan}\\ & \textrm{bayangannya adalah}\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad x-2y-10=0&&\\ \textrm{b}.\quad x+2y-10=0&\\ \textrm{c}.\quad x-6y+5=0&\\ \textrm{d}.\quad x+2y-12=0\\ \textrm{e}.\quad 2x-3y+18=0 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{e}\\ &\begin{aligned}\textrm{Proses}&\: \textrm{untuk refleksinya}\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} 0&-1\\ -1&0 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} -y\\ -x \end{pmatrix}\\ \textrm{proses}&\: \textrm{dilatasinya}\\ \begin{pmatrix} x''\\ y'' \end{pmatrix}&=\begin{pmatrix} 2&0\\ 0&2 \end{pmatrix}\begin{pmatrix} x'-1\\ y'+2 \end{pmatrix}+\begin{pmatrix} 1\\ -2 \end{pmatrix}\\ &=\begin{pmatrix} 2x'-2\\ 2y'+4 \end{pmatrix}+\begin{pmatrix} 1\\ -2 \end{pmatrix}\\ &=\begin{pmatrix} 2x'-1\\ 2y'+2 \end{pmatrix}\\ &=\begin{pmatrix} 2(-y)-1\\ 2(-x)+2 \end{pmatrix}\\ &\begin{cases} x &=-\displaystyle \frac{1}{2}(y''-2) \\ y &=-\displaystyle \frac{1}{2}(x''+1) \end{cases} \end{aligned}\\ &\begin{aligned}\textrm{Sehingga persam}&\textrm{aan bayangan}\\ \textrm{garisnya adalah}:&\\ 3x&-2y+5=0\\ 3\left ( -\displaystyle \frac{1}{2}(y''-2) \right )&-2\left ( -\displaystyle \frac{1}{2}(x''+1) \right )+5=0\\ -\displaystyle \frac{3}{2}y''+3 &+(x''+1)+5=0\\ 2x&-3y+6+2+10=0\\ 2x&-3y+18=0 \end{aligned} \end{array}$.

CONTOH SOAL TRANSFORMASI FUNGSI BAGIAN 2

$\begin{array}{ll}\\ 6.&\textrm{Sebuah transformasi yang didefiniskan oleh}\\ & \begin{cases} x' & =2x+3y \\ y' & =3x+2y \end{cases}\\ &\textrm{Maka bayangan titik M}(2,-1)\: \: \textrm{adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad (7,10)&&\textrm{d}.\quad (1,10)\\ \textrm{b}.\quad (10,7)&\textrm{c}.\quad (1,4)&\textrm{e}.\quad (4,1) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\textrm{Diketahui}&\: \textrm{bahwa}:\\ &\begin{cases} x' & =2x+3y \\ y' & =3x+2y \end{cases}\\ \left.\begin{matrix} x=2\\ y=-1 \end{matrix}\right\}&\Rightarrow \begin{cases} x' & =2(2)+3(-1)=4-3=1 \\ y' & =3(2)+2(-1)=6-2=4 \end{cases} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Bayangan untuk titik A(1,3) oleh rotasi }\\ &\textrm{dengan pusat}\: \: \textit{O}(0,0)\textrm{sejauh}\: \: 90^{\circ}\: \: \textrm{adalah}\, ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad (-1,3)&&\textrm{d}.\quad (1,-3)\\ \textrm{b}.\quad (-1,-3)&\textrm{c}.\quad (-3,1)&\textrm{e}.\quad (3,1) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\textrm{Karena rotasi d}&\textrm{engan pusat O sebesar}\: \: 90^{\circ},\\ \textrm{maka}\qquad\qquad\: \: &\\ R\left ( O(0,0),90^{\circ} \right )&=\begin{pmatrix} \cos 90^{\circ} & -\sin 90^{\circ}\\ \sin 90^{\circ} & \cos 90^{\circ} \end{pmatrix}\\ &=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\\ \textrm{sehingga}\quad\qquad&\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1\\ 3 \end{pmatrix}\\ &=\begin{pmatrix} -3\\ 1 \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 8.&\textrm{Suatu lingkaran dengan jari-jari 4 }\\ &\textrm{dengan pusat di O(0,0) dtranslasikan}\\ &\textrm{oleh}\: \: \textrm{T}=\begin{pmatrix} 2\\ -3 \end{pmatrix},\: \textrm{maka luas }\\ &\textrm{bayangan lingkaran tersebut adalah}\\ & ....\: \textrm{satuan luas}\\ &\begin{array}{lll}\\ \textrm{a}.\quad \pi &&\textrm{d}.\quad 8\pi \\ \textrm{b}.\quad 2\pi &\textrm{c}.\quad 4\pi &\textrm{e}.\quad 16\pi \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{e}\\ &\begin{aligned}&\textrm{Diketahui persamaan lingkaran berpusat}\\ &\textrm{di O dengan}\: \: r=4.\: \textrm{Karena translasi adalah}\\ &\textrm{termasuk transformasi isometri(kongruen)}\\ &\textrm{maka jari-jari lingkaran bayangannya }\\ &\textrm{akan sama dengan bendanya. Sehingga}\\ &\textrm{ luas bayangan lingkarannya}\\ &=\pi r^{2}=\pi \times 4^{2}=16\pi \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 9.&\textrm{Sebuah transformasi yang didefiniskan oleh}\\ & \begin{cases} x' & =4-3x \\ y' & =2x-y-4 \end{cases}\\ &\textrm{Yang merupakan titik invarian (tidak berubah) }\\ &\textrm{adalah}\: ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad (0,0)&&\textrm{d}.\quad (0,-1)\\ \textrm{b}.\quad (1,-1)&\textrm{c}.\quad (1,0)&\textrm{e}.\quad (1,1) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{D}&\textrm{iketahui bahwa}:\\ &\begin{cases} x' & =4-3x \\ y' & =2x-y-4 \end{cases}\\ &\begin{array}{|c|c|c|c|}\hline \textrm{NO}&\textrm{Titik}&\begin{aligned}&\textrm{Disubstitusikan ke}\\ & \begin{cases} x' & =4-3x \\ y' & =2x-y-4 \end{cases} \end{aligned}&\begin{aligned}&\textrm{Keterangan}\\ &\quad\textrm{Titik} \end{aligned}\\\hline \textrm{a}.&(0,0)&\begin{cases} x' & =4-3(0)=4 \\ y' & =2(0)-(0)-4=-4 \end{cases}&\textrm{Varian}\\\hline \textrm{b}&(1,-1)&\begin{cases} x' & =4-3(1)=1 \\ y' & =2(1)-(-1)-4=-1 \end{cases}&\textbf{Invarian}\\\hline \textrm{c}&(1,0)&\begin{cases} x' & =4-3(1)=1 \\ y' & =2(1)-(0)-4=-2 \end{cases}&\textrm{Varian}\\\hline \textrm{d}&(0,-1)&\begin{cases} x' & =4-3(0)=4 \\ y' & =2(0)-(-1)-4=-3 \end{cases}&\textrm{Varian}\\\hline \textrm{e}&(1,1)&\begin{cases} x' & =4-3(1)=1 \\ y' & =2(1)-(1)-4=-3 \end{cases}&\textrm{Varian}\\\hline \end{array} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 10.&\textrm{Bayangan kurva}\: \: xy=6\: \: \textrm{oleh rotasi sebesar}\\ & \displaystyle \frac{\pi }{2}\: \: \textrm{dengan pusat}\: \: O(0,0)\: \: \textrm{adalah}\, ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad xy=-6&&\textrm{d}.\quad x(y-x)=6\\ \textrm{b}.\quad xy=6&&\textrm{e}.\quad x(x+y)=-6\\ \textrm{c}.\quad x(x-y)=6&\end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\textrm{Karena rotasi d}&\textrm{engan pusat O sebesar}\\ \displaystyle \frac{\pi }{2}=90^{\circ},\: \: \textrm{maka}&\\ R\left ( O(0,0),90^{\circ} \right )&=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\\ \textrm{sehingga bayan}&\textrm{gan semua titik yang }\\ \textrm{terletak pada k}& \textrm{urva adalah}:\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} -y\\ x \end{pmatrix}\\ &\begin{cases} x & =y' \\ y & =-x' \end{cases}\\ \textrm{Selanjunya}\: \: \: \textrm{unt}&\textrm{uk bayangan kurvanya }\\ \textrm{adalah}:\qquad\quad&\\ xy&=6\\ y'.(-x')&=6\\ x'y'&=-6\\ \textrm{Jadi , persamaa}&\textrm{n kurva bayangannya}\\ \textrm{adalah}\: &\: xy=-6 \end{aligned} \end{array}$.

CONTOH SOAL TRANSFORMASI FUNGSI BAGIAN 1

$\begin{array}{ll}\\ 1.&\textrm{Sebuah fungsi}\quad y=3x^{\displaystyle 2}-4\quad \textrm{ditranslasikan}\\ &\textrm{oleh}\quad \left( \begin{matrix} 0 \\ 3 \end{matrix} \right),\quad \textrm{maka hasilnya adalah}\, ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle y=3x^{\displaystyle 2}-1\\ \textrm{b}.&\displaystyle y=3x^{\displaystyle 2}+2\\ \textrm{c}.&\displaystyle y=3x^{\displaystyle 2}-18x+23\\ \textrm{d}.&\displaystyle y=4x^{\displaystyle 2}-15x+23\\ \textrm{e}.&\displaystyle y=6x^{\displaystyle 2}+12x+26 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\begin{aligned}&\textrm{Perhatikan bahwa untuk translasi tersebut}\\ &\textrm{di atas, berlaku}\\ &\left( \begin{matrix} x' \\ y' \end{matrix} \right)=\left( \begin{matrix} 0 \\ 3 \end{matrix} \right)+\left( \begin{matrix} x \\ y \end{matrix} \right).\quad \textrm{Selanjutnya}\\ & \left( \begin{matrix} x \\ y \end{matrix} \right)=\left( \begin{matrix} x' \\ y' \end{matrix} \right)-\left( \begin{matrix} 0 \\ 3 \end{matrix} \right)=\left( \begin{matrix} x' \\ y'-3 \end{matrix} \right)\\ &\textrm{Sehingga bayangan garis}\quad y=3x^{\displaystyle 2}-4\\ &\textrm{adalah}:\\ &y=3x^{\displaystyle 2}-4\qquad (\textrm{mula-mula})\\ &(y'-3)=3(x')^{\displaystyle 2}-4\Leftrightarrow y'=3(x')^{2}-1\\ &\textrm{Jadi, bayangan garis}\quad y=3x^{\displaystyle 2}-4\\ &\textrm{adalah}\quad y=3x^{\displaystyle 2}-1 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Diketahui garis}\quad 3x+4y+12=0\quad \textrm{direfleksikan}\\ &\textrm{terhadap sumbu-Y. Hasil dari refleksi garis}\\ &\textrm{tersebut adalah}\, ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 4x-3y-12=0\\ \textrm{b}.&\displaystyle 4x-3y+12=0\\ \textrm{c}.&\displaystyle 3x+4y-12=0\\ \textrm{d}.&\displaystyle 3x-4y+12=0\\ \textrm{e}.&\displaystyle 3x-4y-12=0 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{e}\\ &\begin{aligned}&\textrm{Perhatikan bahwa untuk refleksi tersebut}\\ &\textrm{dengan simulasi bayangan sebuah titik}\\ &(x,y)\overset{\displaystyle M_{\textrm{sumbu}-Y}}{\Longrightarrow }(-x,y)=(x',y')\\&\textrm{Selanjutnya}\\ & \left( \begin{matrix} x \\ y \end{matrix} \right)=\left( \begin{matrix} -x' \\ y' \end{matrix} \right)\\ &\textrm{Sehingga bayangan garis}\quad 3x+4y+12=0\\ &\textrm{adalah}:\\ &3x+4y+12=0\qquad (\textrm{mula-mula})\\ &3(-x')+4y'+12=0\Leftrightarrow -3x'+4y'+12=0\\ &\textrm{Jadi, bayangan garis}\quad 3x+4y+12=0\\ &\textrm{adalah}\quad -3y+4y+12=0\quad \textrm{atau}\\ &3x-4y-12=0\quad (\textrm{dikali dengan} -1) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Bayangan kurva}\quad y=2^{\displaystyle x+1}-4\quad \textrm{jika}\\ &\textrm{direfleksikan terhadap sumbu-X adalah}\, ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle y=2^{\displaystyle x+1}+4\\ \textrm{b}.&\displaystyle y=-2^{\displaystyle x+1}-4\\ \textrm{c}.&\displaystyle y=-2^{\displaystyle x+1}+4\\ \textrm{d}.&\displaystyle y=-2^{\displaystyle -x+1}+4\\ \textrm{e}.&\displaystyle y=-2^{\displaystyle -x+1}-4 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\begin{aligned}&\textrm{Perhatikan bahwa untuk refleksi tersebut}\\ &\textrm{dengan simulasi bayangan sebuah titik}\\ &(x,y)\overset{\displaystyle M_{\textrm{sumbu}-X}}{\Longrightarrow }(x,-y)=(x',y')\\&\textrm{Selanjutnya}\\ & \left( \begin{matrix} x \\ y \end{matrix} \right)=\left( \begin{matrix} x' \\ -y' \end{matrix} \right)\\ &\textrm{Sehingga bayangan kurva}\quad y=2^{\displaystyle x+1}-4\\ &\textrm{adalah}:\\ &y=2^{\displaystyle x+1}-4\qquad (\textrm{mula-mula})\\ &(-y')=2^{\displaystyle x'+1}-4\Leftrightarrow y'=-2^{\displaystyle x+1}+4\\ &\textrm{Jadi, bayangan kurva}\quad y=2^{\displaystyle x+1}-4\\ &\textrm{adalah}\quad y=-2^{\displaystyle x+1}+4 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Bayangan kurva}\quad y=x^{\displaystyle 3}-6x^{\displaystyle 2}+4x+12\quad \textrm{jika}\\ &\textrm{dirotasikan sebesar}\quad 90^{0}\quad \textrm{searah jarum jam}\\ &\textrm{dengan pusat di}\quad (0,0)\quad \textrm{adalah}\, ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle x=y^{\displaystyle 3}+6y^{\displaystyle 2}+4y+12\\ \textrm{b}.&\displaystyle x=y^{\displaystyle 3}-6y^{\displaystyle 2}-4y+12\\ \textrm{c}.&\displaystyle x=y^{\displaystyle 3}-6y^{\displaystyle 2}-4y-12\\ \textrm{d}.&\displaystyle x=-y^{\displaystyle 3}-6y^{\displaystyle 2}-4y+12\\ \textrm{e}.&\displaystyle x=-y^{\displaystyle 3}+6y^{\displaystyle 2}-4y-12 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{e}\\ &\begin{aligned}&\textrm{Diketahui bahwa untuk rotasi tersebut}\\ &\textrm{dengan simulasi bayangan sebuah titik}\\ &(x,y)\overset{\displaystyle R_{\left[ O,90^{0} \right]}}{\Longrightarrow }(x',y')\\&\textrm{dengan matriks rotasinya adalah}:\\ &\begin{pmatrix} cos\: 90^{0} & -sin\: 90^{0} \\ sin\: 90^{0} & cos\: 90^{0} \end{pmatrix}=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\\ &\textrm{Bayangan proses hasilnya adalah}:\\ &\left( \begin{matrix} x' \\ y' \end{matrix} \right)=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\left( \begin{matrix} x \\ y \end{matrix} \right)=\left( \begin{matrix} -y \\ x \end{matrix} \right)\\ &\textrm{Selanjutnya}\\ & \left( \begin{matrix} x \\ y \end{matrix} \right)=\left( \begin{matrix} \displaystyle y' \\ -x' \end{matrix} \right)\\ &\textrm{Sehingga bayangan kurva}\\ &y=x^{\displaystyle 3}-6x^{\displaystyle 2}+4x+12\\ &\textrm{adalah}:\\ &y=x^{\displaystyle 3}-6x^{\displaystyle 2}+4x+12\qquad (\textrm{mula-mula})\\ &-x'=(y')^{\displaystyle 3}-6(y')^{\displaystyle 2}+4(y')+12\\ &\Leftrightarrow x'=-(y')^{\displaystyle 3}+6(y')^{\displaystyle 2}-4(y')-12\\ &\textrm{Jadi, bayangan kurva}\quad y=x^{\displaystyle 3}-6x^{\displaystyle 2}+4x+12\\ &\textrm{adalah}\quad x=-y^{\displaystyle 3}+6y^{\displaystyle 2}-4y-12 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Bayangan kurva}\quad y=6^{\displaystyle x}+9\quad \textrm{jika}\\ &\textrm{didilatasikan sejajar sumbu-X dengan}\\ &\textrm{skala}\quad \displaystyle \frac{1}{3}\quad \textrm{adalah}\, ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle y=6^{\displaystyle \frac{1}{3}x}+3\\ \textrm{b}.&\displaystyle y=6^{\displaystyle \frac{1}{3}x}+9\\ \textrm{c}.&\displaystyle y=6^{\displaystyle 3x}+3\\ \textrm{d}.&\displaystyle y=3^{\displaystyle \frac{1}{3}x}+3\\ \textrm{e}.&\displaystyle y=3^{\displaystyle 3x}+3 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\begin{aligned}&\textrm{Perhatikan bahwa untuk dilatasi tersebut}\\ &\textrm{dengan simulasi bayangan sebuah titik}\\ &(x,y)\overset{\displaystyle D\left[ O,\frac{1}{3} \right]}{\Longrightarrow }(x',y')=(\frac{1}{3}x,\frac{1}{3}y)\\ &\textbf{Catatan}:\textrm{agak beda dengan translasi, refleksi}\\ &\textrm{dan rotasi dalam pengambilan bayangannya}\\\\&\textrm{Karena yang diminta hanya sejajar}\\ &\textrm{dengan sumbu-X, maka hasilnya}\\ &\textrm{adalah}:\quad \left( x',y' \right)=\left( \displaystyle \frac{1}{3}x,y \right)\\ &\textrm{Selanjutnya}\\ & \left( \begin{matrix} x' \\ y' \end{matrix} \right)=\left( \begin{matrix} \displaystyle \frac{1}{3}x \\ y \end{matrix} \right)\\ &\textrm{Sehingga bayangan kurva}\quad y=6^{\displaystyle x}+9\\ &\textrm{adalah}:\\ &y=6^{\displaystyle x}+9\qquad (\textrm{mula-mula})\\ &y'=6^{\displaystyle x'}+9\Leftrightarrow y=6^{\displaystyle \frac{1}{3}x}+9\\ &\textrm{Jadi, bayangan kurva}\quad y=6^{\displaystyle x}+9\\ &\textrm{adalah}\quad y=6^{\displaystyle \frac{1}{3}x}+9 \end{aligned} \end{array}$.

TRANSFORMASI FUNGSI

 Materi lama di sini

A. Pengetian Trnasformasi Fungsi

Transformasi adalah suatu proses atau operasi yang mengubah suatu objek menjadi objek baru menurut aturan tertentu, sedangkan untuk transfomasi geometri adalah cabang transformasi dalam matematika yang mempelajari perubahan posisi, orientasi, ukuran, atau bentuk suau titik, garis, kurva, atau bangun pada bidang atau ruang berdasarkan aturan tertentu. Selanjutnya terkait bahasan ini, yaitu transformasi fungsi adalah perubahan bentuk atau letak grafik suatu fungsi tanpa mengubah sifat dasar fungsi tersebut.

B. Jenis-Jenis Transformasi fungsi

Berikut empat transformasi geometri dasar

• Translasi (geseran), memindahkan titik sejauh dan searah suatu vektor tertentu dengan bentuk dan ukuran bangun tetap
• Refleksi (pencerminan), memantulkan bangun terhadap suatu garis atau bidang cermin dengan bentuk dan ukuran tetap tetapi orientasi berubah
• Rotasi (perputaran), memutar bangun terhadap suatu titik pusat dengan sudut tertentu dengan hasil bentuk dan ukuran tetap
• Dilatasi (pembesaran/pengecilan), mengubah ukuran bangun dengan faktor skala tertentu dengan bentuk tetap, tetapi ukuran berubah, kecuali fooaktor skalanya 1.

C. Matriks transformasi

Misalkan suatu transfomasi T memetakan sebuah titik A(x,y) ke A'(x',y') 

selanjutnya perhatikan ilustrasi berikut:

$\boxed{\begin{aligned}A(x,y)&\xrightarrow[.]{Transformasi\, =\: T}A'(x',y')=A'\left ( ax+by,cx+dy \right )\\\\ \Rightarrow &\begin{pmatrix} x'\\ y' \end{pmatrix}=\underset{\underset{transformasi}{Matriks}}{\underbrace{\begin{pmatrix} a & b\\ c & d \end{pmatrix}}}\begin{pmatrix} x\\ y \end{pmatrix} \end{aligned}}$.

D. Jenis-Jenis Transformasi dengan matriks yang sesuaian

1. Translasi (Geseran)

$\begin{array}{|l|c|c|}\hline \begin{aligned}&\textrm{Jenis}\\ &\textrm{Transformasi} \end{aligned}&\textrm{Rumus}&\textrm{Matriks}\\\hline \textrm{Translasi}&(x,y)\xrightarrow[.]{\begin{pmatrix} a\\ b \end{pmatrix}}(x+a,y+b)&\begin{pmatrix} a\\ b \end{pmatrix}\\\hline \end{array}$.

2. Rotasi (Perputaran)

$\begin{aligned}&\begin{array}{|l|c|c|}\hline \begin{aligned}&\textrm{Jenis}\\ &\textrm{Transformasi} \end{aligned}&\textrm{Rumus}&\textrm{Matriks}\\\hline \textrm{Rotasi}&&\\\hline \begin{aligned}&\textrm{Pusat rotasi}\\ & \left [ O,\alpha \right ] \end{aligned}&\begin{aligned}&\begin{cases} x' =... \\ y' = ... \end{cases}\\ &\begin{aligned}&\colorbox{yellow}{Lihat}\\ &\colorbox{yellow}{di bawah}\\ &\colorbox{yellow}{tulisan}\\ &\colorbox{yellow}{warna}\\ &\colorbox{yellow}{biru} \end{aligned} \end{aligned}&\begin{pmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix}\\\hline \begin{aligned}&\textrm{Pusat}\\ & (a,b)\: \textrm{sudut}\: \alpha \end{aligned}&\begin{pmatrix} x'-a\\ y'-b \end{pmatrix}=&\begin{aligned}&\colorbox{yellow}{Lihat}\\ &\colorbox{yellow}{di bawah}\\ &\colorbox{yellow}{tulisan}\\ &\colorbox{yellow}{warna}\\ &\colorbox{yellow}{merah} \end{aligned}\\\hline \end{array}\\ &\color{blue}\begin{cases} x' =x\cos \alpha -y\sin \alpha \\ y' = x\sin \alpha +y\cos \alpha \end{cases}\\ &\color{red}\triangleright \triangleright \triangleright \triangleright \begin{pmatrix} x'-a\\ y'-b \end{pmatrix}=\begin{pmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix}.\begin{pmatrix} x-a\\ y-b \end{pmatrix} \end{aligned}$.

3. Refleksi (Pencerminan)

$\begin{array}{|l|c|c|}\hline \textrm{Refleksi}&&\\\hline \textrm{terhadap sumbu}-\textrm{X}&(x,y)\rightarrow (x,-y)&\begin{pmatrix} 1 &0 \\ 0 & -1 \end{pmatrix}\\\hline \textrm{terhadap sumbu}-\textrm{Y}&(x,y)\rightarrow (-x,y)&\begin{pmatrix} -1 & 0\\ 0 & 1 \end{pmatrix}\\\hline \textrm{terhadap garis y = x}&(x,y)\rightarrow (y,x)&\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\\\hline \textrm{terhadap garis y = -x}&(x,y)\rightarrow (-y,-x)&\begin{pmatrix} 0 & -1\\ -1 & 0 \end{pmatrix}\\\hline \textrm{terhadap garis x = h}&(x,y)\rightarrow (2h-x,y)&\begin{pmatrix} -1 & 0\\ 0 & 1 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}+\begin{pmatrix} 2h\\ 0 \end{pmatrix}\\\hline \textrm{terhadap garis y = x}&(x,y)\rightarrow (y,x)&\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\\\hline \textrm{terhadap garis y = -x}&(x,y)\rightarrow (-y,-x)&\begin{pmatrix} 0 & -1\\ -1 & 0 \end{pmatrix}\\\hline \textrm{terhadap garis x = h}&(x,y)\rightarrow (2h-x,y)&\begin{pmatrix} -1 & 0\\ 0 & 1 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}+\begin{pmatrix} 2h\\ 0 \end{pmatrix}\\\hline \textrm{terhadap garis y = k}&(x,y)\rightarrow (x,2k-y)&\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}+\begin{pmatrix} 0\\ 2k \end{pmatrix}\\\hline \textrm{pusat}\: (0,0)\begin{cases} y=mx \\ m=\tan \alpha \end{cases}&&\begin{pmatrix} \cos 2\alpha & \sin 2\alpha \\ \sin 2\alpha & -\cos 2\alpha \end{pmatrix}\\\hline \end{array}$.

4. Dilatasi (Perkalian)

$\begin{aligned}&\begin{array}{|l|c|c|}\hline \begin{aligned}&\textrm{Jenis}\\ &\textrm{Transformasi} \end{aligned}&\textrm{Rumus}&\textrm{Matriks}\\\hline \textrm{Dilatasi}&&\\\hline \textrm{Pusat}\: \left [ O,k \right ]&(x,y)\rightarrow (kx,ky)&\begin{pmatrix} k & 0\\ 0 & k \end{pmatrix}\\\hline \begin{aligned}&\textrm{Pusat}\: (a,b)\\ & \textrm{faktor skala}\: k \end{aligned}&\begin{pmatrix} x'-a\\ y'-b \end{pmatrix}&\begin{aligned}&\colorbox{yellow}{Lihat}\\ &\colorbox{yellow}{di bawah}\\ &\colorbox{yellow}{tulisan}\\ &\colorbox{yellow}{warna}\\ &\colorbox{yellow}{merah} \end{aligned}\\\hline \begin{aligned}&\textrm{Luas bangun}\\ &\textrm{ datar} \end{aligned}&\textrm{Misal bangun A}&\textrm{T}=\begin{pmatrix} a & b\\ c & d \end{pmatrix}\\\hline &\textbf{Bangun A}'&= \textrm{det T}\times \textrm{A}\\\hline \end{array}\\ &\color{red}\triangleright \triangleright \triangleright \triangleright \triangleright \triangleright \begin{pmatrix} x'-a\\ y'-b \end{pmatrix}=\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}.\begin{pmatrix} x-a\\ y-b \end{pmatrix} \end{aligned}$.

Tambahan rumus

$\begin{array}{|l|c|l|}\hline \begin{aligned}&\textrm{Jenis}\\ &\textrm{Dilatasi} \end{aligned}&\textrm{Rumus}&\textrm{Keterangan}\\\hline \textrm{Vertikal}&\begin{aligned}&(x,y)\Longrightarrow (x,ky)\\ &y=f(x)\Longrightarrow y=kf(x)\end{aligned}&\begin{aligned}&\textrm{Jenis dilatasi}\\ &\textrm{yang menghasilkan}\\ &\textrm{perubahan tinggi}\\ &\textrm{suatu objek}\end{aligned}\\\hline  \textrm{Horizontal}&\begin{aligned}&(x,y)\Longrightarrow (kx,y)\\ &y=f(x)\Longrightarrow y=f(kx)\end{aligned}&\begin{aligned}&\textrm{Jenis dilatasi}\\ &\textrm{yang menghasilkan}\\ &\textrm{perubahan lebar}\\ &\textrm{suatu objek}\end{aligned}\\\hline  \end{array}$.

Catatan:

Translasi, refleksi, dan rotasi suatu objek adalah bagian dari transformasi yang hanya mengubah posisi objek saja, sehingga jenis transformasi-transformasi ini juga disebut dengan transformasi isometri

E. Bayangan Kurva dan Komposisi Transformasi

$\begin{array}{|l|l|}\hline \qquad \textrm{Bayangan Kurva}\quad y=f(x)&\qquad\qquad\qquad \textrm{Komposisi Transformasi}\\\hline \begin{aligned}\textrm{Lan}&\textrm{gkah-langkah}:\\ 1.\quad&\textrm{Tentukan bayangan titiknya}\\ &(x,y)\rightarrow \left ( x',y' \right )\\ 2.\quad&\textrm{Salanjutnya tentukan}\: \: x\: \: \textrm{dan}\: \: y\:\\ &\textrm{dalam}\: \: x'\: \: \textrm{dan}\: \: y'\\ 3.\quad&\textrm{Substitusikan}\: \: x\: \: \textrm{dan}\: \: y\\ &\textrm{ke}\: \: \: y=f(x) \end{aligned}&\begin{aligned}\textrm{Lan}&\textrm{gkah-langkah}:\\ 1.\quad&\textrm{Selesaikan sesuai urutan transformasi}\\ &(x,y)\xrightarrow[\qquad.]{T_{1}}(x',y')\xrightarrow[\qquad.]{T_{2}}(x'',y'')\\ 2.\quad&\textrm{Jika dapat disederhanakan kedua transformasi}\\ &\textrm{tersebut di atas, maka cukup dengan}\\ &(x,y)\xrightarrow[\qquad.]{T_{2}\circ T_{1}}(x'',y'') \end{aligned}\\\hline \end{array}$.

$\LARGE{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Tentukanlah bayangan dari segitiga PQR dengan}\\\ & P(0,4),\: Q(-1,1),\: \textrm{dan}\: \: R(3,6).\\ &\textrm{oleh translasi}\: \: \: T=\begin{pmatrix} 5\\ -2 \end{pmatrix}\\\\ &\textbf{Jawab}\\ &\begin{cases} \begin{pmatrix} x_{P}^{'}\\ y_{P}^{'} \end{pmatrix} &=T+\begin{pmatrix} x_{P}\\ y_{P} \end{pmatrix}=\begin{pmatrix} 5\\ -2 \end{pmatrix}+\begin{pmatrix} 0\\ 4 \end{pmatrix}=\begin{pmatrix} 5+0\\ -2+4 \end{pmatrix}=\begin{pmatrix} 5\\ 2 \end{pmatrix} \\ \begin{pmatrix} x_{Q}^{'}\\ y_{Q}^{'} \end{pmatrix} & =\cdots\qquad \textrm{isilah sendiri} \\ \begin{pmatrix} x_{R}^{'}\\ y_{R}^{'} \end{pmatrix} &= \cdots\qquad \textrm{isilah sendiri} \end{cases} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Tentukanlah bayangan dari garis}\: \: y=2x+4\\ & \textrm{oleh translasi}\: \: T=\begin{pmatrix} -1\\ 2 \end{pmatrix}.\\\\ &\textbf{Jawab}\\ &\begin{array}{|c|c|}\hline \textbf{Bayangan Titik-titik}&\textbf{Bayangan Garis}\\\hline \begin{aligned}\begin{pmatrix} x'\\ y' \end{pmatrix}&=T+\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} -1\\ 2 \end{pmatrix}+\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} -1+x\\ 2+y \end{pmatrix}\\ &\begin{cases} x' & =-1+x\Leftrightarrow x=x'+1 \\ y' & =2+y\quad\Leftrightarrow y=y'-2 \end{cases} \end{aligned}&\begin{aligned}y&=2x+4\\ y'-2&=2(x'+1)+4\\ y'&=2x+2+4+2\\ &=2x+8\\ \textrm{Jadi}\, ,&\: \textbf{bayangan garisnya}\\ \textrm{adala}&\textrm{h}:\\ y&=2x+8\\ & \end{aligned}\\\hline \end{array} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Tentukanlah bayangan titik A(4,6) oleh rotasi yang berpusat }\\ &\textrm{di titik P(3,-2) dengan sudut putar sebesar}\: \: 90^{\circ} \\\\ &\textbf{Jawab}\\ &\begin{aligned}\textrm{Untuk Ro}&\textrm{tasi yang berpusat di}\: \: (a,b)\: \: \textrm{dengan sudut}\: \: \alpha \: \: \textrm{adalah}:\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix}\begin{pmatrix} x-a\\ y-b \end{pmatrix}+\begin{pmatrix} a\\ b \end{pmatrix}\\ &=\begin{pmatrix} \cos 90^{\circ} & -\sin 90^{\circ}\\ \sin 90^{\circ} & \cos 90^{\circ} \end{pmatrix}\begin{pmatrix} 4-3\\ 6-(-2) \end{pmatrix}+\begin{pmatrix} 3\\ -2 \end{pmatrix}\\ &=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1\\ 8 \end{pmatrix}+\begin{pmatrix} 3\\ -2 \end{pmatrix}\\ &=\begin{pmatrix} -8\\ 1 \end{pmatrix}+\begin{pmatrix} 3\\ -2 \end{pmatrix}\\ &=\begin{pmatrix} -5\\ -1 \end{pmatrix}\\ \textrm{Jadi}\, ,\: &\textrm{bayangan titik A adalah}\: \: \textrm{A}'(-5,-1) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Tentukanlah bayangan titik A(4,6) }\\ &\textrm{oleh dilatasi yang berpusat di titik P(3,-2)}\\ &\textrm{dengan faktor skala}\: \: k=2 \\\\ &\textbf{Jawab}\\ &\begin{aligned}\textrm{Bayangan}&\: \textrm{titik A-nya adalah}:\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} k & 0\\ 0 & k \end{pmatrix}\begin{pmatrix} x-a\\ y-b \end{pmatrix}+\begin{pmatrix} a\\ b \end{pmatrix}\\ &=\begin{pmatrix} 2 & 0\\ 0 & 2 \end{pmatrix}\begin{pmatrix} 4-3\\ 6-(-2) \end{pmatrix}+\begin{pmatrix} 3\\ -2 \end{pmatrix}\\ &=\begin{pmatrix} 2 & 0\\ 0 & 2 \end{pmatrix}\begin{pmatrix} 1\\ 8 \end{pmatrix}+\begin{pmatrix} 3\\ -2 \end{pmatrix}\\ &=\begin{pmatrix} 2\\ 16 \end{pmatrix}+\begin{pmatrix} 3\\ -2 \end{pmatrix}\\ &=\begin{pmatrix} 5\\ 14 \end{pmatrix}\\ \textrm{Jadi}\: ,\: &\textrm{bayangan titik A-nya adalah}\: \: \textrm{A}'(5,14) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Tentukanlah bayangan titik A(4,6) }\\ &\textrm{oleh translasi}\: \: t\: \: \textrm{dilanjutkan}\: \: s\: \: \textrm{dengan}\\ &\textrm{matriks transformasi berturut-turut }\\ &\textrm{adalah}\: \: T=\begin{pmatrix} 1 & 1\\ 1 & 2 \end{pmatrix}\: \: \textrm{dan}\: \: S= \begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\textrm{Bayangan}&\: \textrm{titik A-nya adalah}:\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=S\times T\times \begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 1\\ 1 & 2 \end{pmatrix}\begin{pmatrix} 4\\ 6 \end{pmatrix}\\ &=\begin{pmatrix} 2 & 3\\ 1 & 2 \end{pmatrix}\begin{pmatrix} 4\\ 6 \end{pmatrix}\\ &=\begin{pmatrix} 26\\ 16 \end{pmatrix}\\ \textrm{Jadi}\: ,\: &\textrm{bayangan titik A-nya adalah}\: \: \textrm{A}'(26,16) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 6.&\textrm{Suatu kurva}\: \: y=\, ^{3}\log (2x-2)\: \: \textrm{memiliki bayangan}\\ & y=\, ^{3}\log \left ( \displaystyle \frac{2x+3}{3} \right )\: \: \textrm{oleh translasi}\\ & T=\begin{pmatrix} a\\ b \end{pmatrix}.\: \textrm{Tentukanlah nilai}\: \: a+b\\\\ &\textbf{Jawab}\\ &\begin{aligned}\textrm{Diketahui}&\: \textrm{bahwa}\\ y&=\, ^{3}\log (2x-2)\quad \Leftrightarrow\quad 3^{y}=2x-2\: (\textrm{benda})\\ y&=\, ^{3}\log \left ( \displaystyle \frac{2x+3}{3} \right )\\ & \Leftrightarrow\quad 3^{y}=\left ( \displaystyle \frac{2x+3}{3} \right )\quad (\textbf{bayangan})\\ \textrm{sehingga}&\: \textrm{untuk bayangan}\\ 3^{y'-b}&=2(x'-a)-2\quad \Leftrightarrow \quad 3^{y'}.3^{-b}=2(x'-a)-2\\ & \Leftrightarrow\quad 3^{y'}=\displaystyle \frac{2(x'-a)-2}{3^{-b}}=\displaystyle \frac{2x'+3}{3}\\ \textrm{Jadi}\, ,\: &\begin{cases} a &=\displaystyle \frac{5}{2} \\ b &=-1 \end{cases}\\ \textrm{Sehingga}&\: a+b=\displaystyle \frac{5}{2}+(-1)=\displaystyle \frac{3}{2} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Tentukanlah bayangan garis}\: \: ax+by+c=0\: \: \textrm{oleh transformasi}\\ &\textrm{yang bersesuaian dengan matriks}\: \: \: \begin{pmatrix} 1&-2\\ 3&-4 \end{pmatrix}\\\\ &\textbf{Jawab}\\ &\begin{array}{|c|c|}\hline \textbf{Proses Awal}&\textbf{Penentuan Bayangan}\\\hline \begin{aligned}\begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} 1 & -2\\ 3 & -4 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ \begin{pmatrix} x\\ y \end{pmatrix}&=\begin{pmatrix} 1 & -2\\ 3 & -4 \end{pmatrix}^{-1}\begin{pmatrix} x'\\ y' \end{pmatrix}\\ &=\displaystyle \frac{1}{\begin{vmatrix} 1 & -2\\ 3 & -4 \end{vmatrix}}\begin{pmatrix} -4 & 2\\ -3 & 1 \end{pmatrix}\begin{pmatrix} x'\\ y' \end{pmatrix}\\ &=\displaystyle \frac{1}{-4+6}\begin{pmatrix} -4x'+2y'\\ -3x'+y' \end{pmatrix}\\ &=\displaystyle \frac{1}{2}\begin{pmatrix} -4x'+2y'\\ -3x'+y' \end{pmatrix}\\ &\begin{cases} x &=-2x'+y' \\ y &=-\displaystyle \frac{3}{2}x'+\displaystyle \frac{1}{2}y' \end{cases} \end{aligned}&\begin{aligned}ax+by+c&=0\\ a\left ( -2x'+y' \right )+b\left ( -\displaystyle \frac{3}{2}x'+\frac{1}{2}y' \right )+c&=0\\ -2ax'-\displaystyle \frac{3}{2}bx'+ay'+\displaystyle \frac{1}{2}by'+c&=0\\ (-4a-3b)x'+(2a+b)y'+2c&=0\\ &\\ \textbf{Jadi, bayangan garisnya adalah}:&\\ &\\ (-4a-3b)x+(2a+b)y+2c&=0\\ &\\ &\\ &\\ & \end{aligned} \\\hline \end{array} \end{array}$.

$\begin{array}{ll}\\ 8.&\textrm{Diketahui kurva}\: \: y=4x^{2}-9\: \: \textrm{dicerminkan terhadap sumbu-X kemudian}\\ &\textrm{ditranslasikan dengan}\: \: \begin{pmatrix} -1\\ 2 \end{pmatrix}.\: \textrm{Ordinat titik potong terhadap sumbu-Y adalah}....\\\\ &\textbf{Jawab}\\ &\begin{array}{|c|c|}\hline \begin{aligned}\begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} -1\\ 2 \end{pmatrix}+\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} -1\\ 2 \end{pmatrix}+\begin{pmatrix} x\\ -y \end{pmatrix}\\ &=\begin{pmatrix} -1+x\\ 2-y \end{pmatrix}\\ &\begin{cases} x &= x'-1\\ y &= 2-y' \end{cases} \end{aligned}&\begin{aligned}y&=4x^{2}-9\\ (2-y')&=4(x'-1)^{2}-9\\ -y'&=4(x'^{2}-2x'+1)-9-2\\ -y'&=4x'^{2}-8x'+4-11\\ y'&=-4x'^{2}+8x'+7\\ &\\ \textbf{Maka}\, ,&\, \textbf{persamaan kurva bayangannya}:\\ y&=-4x^{2}+8x+7 \end{aligned} \\\hline \end{array}\\ &\begin{aligned}\textrm{Sehingga}&\: \textrm{ordinat dari titik potong terhadap sumbu-Y-nya adalah}:\\ y&=-4x^{2}+8x+7,\qquad \textbf{atau}\\ f(x)&=-4x^{2}+8x+7\\ f(0)&=-4(0)^{2}+8(0)+7\qquad\quad \textrm{saat}\: \: x=0\: (\textrm{karena memotong sumbu-Y})\\ &=7\\ \textrm{Jadi}&\: \textrm{ordinatnya adalah}\: \: y=f(0)=7 \end{aligned} \end{array}$.

DAFTAR PUSTAKA

1. Johanes, Kastolan, Sulasim, 2006. Kompetensi Matematika 3A SMA Kelas XII Program IPA Semester Pertama. Jakarta: YUDHISTIRA.
2. Mastd, A. dkk. 2021. Matematika Tingkat Lanjut untuk SMA Kelas XI (Kurikulum Merdeka). Jakarta: Pusat Perbukuan Kemendikbudristek.
3. Nugroho, P. A. Gunarto, D. 2013. Big Bank Soal-Bahas MAtematika SMA/MA. Jakarta: WAHYUMEDIA.
4. Santoso, N.A., Aksin, N. 2024. PR Matematika untuk SMA/MA/SMK/MAK Kelas 12. Yogyakarta: INTAN PARIWARA EDUKASI.

EKSPONEN-ANEKA CONTOH SOAL LANJUTAN (BAGIAN 12)

$\begin{array}{ll}\\ 56.&\textrm{Tentukan jumlah nilai riil}\quad x \quad \textrm{dari persamaan}\\ & (2+\sqrt{3})^{\displaystyle x^{\displaystyle 2}-2x+1}+(2-\sqrt{3})^{\displaystyle x^{\displaystyle 2}-2x-1}=\displaystyle \frac{4}{2-\sqrt{3}}\\ &\textrm{adalah}\:....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 2\\ \textrm{b}.&\displaystyle 3\\ \textrm{c}.&\displaystyle 4\\ \textrm{d}.&\displaystyle 5\\ \textrm{e}.&\displaystyle 6 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&(2+\sqrt{3})^{\displaystyle x^{\displaystyle 2}-2x+1}+(2-\sqrt{3})^{\displaystyle x^{\displaystyle 2}-2x-1}=\displaystyle \frac{4}{2-\sqrt{3}}\\ &\displaystyle \frac{(2+\sqrt{3})^{\displaystyle x^{\displaystyle 2}-2x+1}}{\displaystyle \frac{4}{2-\sqrt{3}}}+\displaystyle \frac{(2-\sqrt{3})^{\displaystyle x^{\displaystyle 2}-2x-1}}{\displaystyle \frac{4}{2-\sqrt{3}}}=1\\ &\displaystyle \frac{(2+\sqrt{3})^{\displaystyle x^{\displaystyle 2}-2x}}{4}+\displaystyle \frac{(2-\sqrt{3})^{\displaystyle x^{\displaystyle 2}-2x}}{4}=1\\ &\displaystyle \frac{(2+\sqrt{3})^{\displaystyle x^{\displaystyle 2}-2x}}{4}+\displaystyle \frac{(2+\sqrt{3})^{-\left(\displaystyle  x^{\displaystyle 2}-2x \right)}}{4}=1\\ &\textrm{dan kondisi di atas terpenuhi saat}\quad x^{\displaystyle 2}-2x=\pm 1\\ & \textrm{Sehingga solusinya adalah:}\\ &x^{\displaystyle 2}-2x=1 \quad \textrm{atau}\quad x^{\displaystyle 2}-2x=-1.\quad \textrm{Selanjutnya}\\ &\textrm{untuk}\\ &\bullet \quad x^{\displaystyle 2}-2x-1=0\Rightarrow x_{1,2}=1\pm \sqrt{2}\\ &\qquad \textrm{Persamaan kuadrat dan untuk solusinya gunakan}\\ &\qquad \textrm{rumus ABC dan nantinya akan didapatkan dua solusi}\\ &\bullet \quad x^{\displaystyle 2}-2x+1=0\Rightarrow x_{3}=1\\ &\qquad \textrm{mirip caranya dengan di atas dan akan didapatkan}\\ &\qquad \textrm{satu solusi saja}\\ &\textrm{Jadi, jumlah semua nilainya adalah}:\\ &x_{1}+x_{2}+x_{3}=\left( 1+\sqrt{2} \right)+\left( 1-\sqrt{2} \right)+1=3 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 57.&\textrm{Bentuk sederhana dari}\\ & \sqrt[\displaystyle 81^{\displaystyle 3^{\displaystyle x}}]{\left( \sqrt[\displaystyle 3]{512^{\displaystyle 3^{\displaystyle 3^{\displaystyle x+1}}}} \right)^{\displaystyle 3^{\displaystyle 3^{\displaystyle x}}}}\qquad=\:....\\\\ &\textrm{Solusi}:\quad \\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&\sqrt[\displaystyle 81^{\displaystyle 3^{\displaystyle x}}]{\left( \sqrt[\displaystyle 3]{512^{\displaystyle 3^{\displaystyle 3^{\displaystyle x+1}}}} \right)^{\displaystyle 3^{\displaystyle 3^{\displaystyle x}}}}\\ &=\sqrt[\left( \displaystyle 3^{\displaystyle 3^{\displaystyle x}} \right)^{\displaystyle 4}]{\left( \sqrt[\displaystyle 3]{512^{\left( \displaystyle 3^{\displaystyle 3^{\displaystyle x}} \right)^{\displaystyle 3}}} \right)^{\displaystyle 3^{\displaystyle 3^{\displaystyle x}}}}\\ &\qquad\textrm{misalkan}\quad 3^{\displaystyle 3^{\displaystyle x}}=m\\ &=\sqrt[\displaystyle m^{\displaystyle 4}]{\left( \sqrt[\displaystyle 3]{512^{\displaystyle m^{\displaystyle 3}}} \right)^{\displaystyle m}}\quad =\sqrt[\displaystyle 3m^{\displaystyle 4}]{512^{\displaystyle m^{\displaystyle 4}}}\\ &=\sqrt[\displaystyle 3]{512}=\sqrt[\displaystyle 3]{8^{\displaystyle 3}}=8\\ &\textrm{Jadi, nilai}\: \:  \sqrt[\displaystyle 81^{\displaystyle 3^{\displaystyle x}}]{\left( \sqrt[\displaystyle 3]{512^{\displaystyle 3^{\displaystyle 3^{\displaystyle x+1}}}} \right)^{\displaystyle 3^{\displaystyle 3^{\displaystyle x}}}}\quad =8 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 58.&\textrm{Bentuk sederhana dari}\\ &\left[ \sqrt[\displaystyle n^{\displaystyle n+1}]{\left( n^{\displaystyle n^{\displaystyle n^{\displaystyle 5n}}}\qquad \right)^{\displaystyle n\:\qquad}}\qquad \right]^{\displaystyle n^{\displaystyle n-\left( \displaystyle n^{n} \right)^{\displaystyle 5}}}\\ &\textrm{adalah}\:....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 1\\ \textrm{b}.&\displaystyle n\\ \textrm{c}.&\displaystyle n^{\displaystyle n}\\ \textrm{d}.&\displaystyle \sqrt[\displaystyle n]{n}\\ \textrm{e}.&\displaystyle \sqrt[\displaystyle n^{n}]{n} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&\left[ \sqrt[\displaystyle n^{\displaystyle n+1}]{\left( n^{\displaystyle n^{\displaystyle n^{\displaystyle 5n}}}\qquad \right)^{\displaystyle n\:\qquad}}\qquad \right]^{\displaystyle n^{\displaystyle n-\left( \displaystyle n^{n} \right)^{\displaystyle 5}}}\\ &=\left( n^{\displaystyle n^{\displaystyle n^{\displaystyle 5n}}} \right)^{\left( \displaystyle \frac{n}{n^{\displaystyle n+1}} \right).\displaystyle n^{\displaystyle n-\left( n^{\displaystyle n} \right)^{\displaystyle 5}}}\\ &=\left( n^{\displaystyle n^{\displaystyle n^{\displaystyle 5n}}} \right)^{\left( \displaystyle \frac{n}{n^{\displaystyle n}.n} \right).\displaystyle n^{\displaystyle n-\left( n^{\displaystyle n} \right)^{\displaystyle 5}}}\\ &=n^{\displaystyle n^{\displaystyle n^{\displaystyle 5n}}.n^{\displaystyle -n}.n^{\displaystyle n}.n^{\displaystyle -n^{\displaystyle 5n}}}=n^{\displaystyle n^{\displaystyle 0}}=n^{\displaystyle 1}=n\\ \end{aligned} \end{array}$.

EKSPONEN-ANEKA CONTOH SOAL LANJUTAN (BAGIAN 11)

$\begin{array}{ll}\\ 51.&\textrm{Penyelesaian dari}\quad 2^{\displaystyle 4^{\displaystyle x}}=4^{\displaystyle 2^{\displaystyle x}}\quad\textrm{adalah}\:....\\\\ &\textrm{Solusi}:\quad \\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&2^{\displaystyle 4^{\displaystyle x}}=4^{\displaystyle 2^{\displaystyle x}}\Leftrightarrow 2^{\displaystyle 2^{\displaystyle 2x}}=2^{\displaystyle 2.2^{\displaystyle x}}\\ &\Leftrightarrow 2^{\displaystyle 2^{\displaystyle 2x}}=2^{\displaystyle 2^{\displaystyle 1+x}}\\ &\Leftrightarrow 2x=1+x\Leftrightarrow x=1 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 52.&\textrm{Penyelesaian dari}\\ &\left(\displaystyle  \frac{1}{x^{\displaystyle 3}} \right)^{\left( \displaystyle \frac{1}{x^{\displaystyle 4}} \right)}=\left(\displaystyle  \frac{1}{\sqrt[\displaystyle 3x]{x}\:} \right)^{\left( \displaystyle \frac{1}{x} \right)}\\ &\textrm{untuk}\quad  x>1\quad\textrm{adalah}\:....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \displaystyle \frac{4}{3}\\ \textrm{b}.&\displaystyle 2\\ \textrm{c}.&\displaystyle \frac{3}{2}\\ \textrm{d}.&\displaystyle 5\\ \textrm{e}.&\displaystyle 3 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&\left(\displaystyle  \frac{1}{x^{\displaystyle 3}} \right)^{\left( \displaystyle \frac{1}{x^{\displaystyle 4}} \right)}=\left(\displaystyle  \frac{1}{\sqrt[\displaystyle 3x]{x}\:} \right)^{\left( \displaystyle \frac{1}{x} \right)}\:\textrm{dengan}\quad  x>1\\ &\Leftrightarrow \left(\displaystyle  \frac{1}{x^{\displaystyle 3}} \right)^{\displaystyle x}=\left(\displaystyle \frac{1}{x^{\displaystyle \frac{1}{3x}}} \right)^{\displaystyle x^{\displaystyle 4}}\\ &\Leftrightarrow \left(\displaystyle  x^{\displaystyle -3} \right)^{\displaystyle x}=\left(x^{\displaystyle -\frac{1}{3x}} \right)^{\displaystyle x^{\displaystyle 4}}\\ &\Leftrightarrow x^{\displaystyle -3x}=x^{\displaystyle -\frac{x^{\displaystyle 4}}{3x}}\\ &\Leftrightarrow -3x=\displaystyle -\frac{x^{\displaystyle 4}}{3x}\Leftrightarrow 9x^{\displaystyle 2}=x^{\displaystyle 4}\\ &\Leftrightarrow x^{\displaystyle 2}=9\Leftrightarrow x=\left| 3 \right|=\pm 3,\quad\textrm{karena}\quad  x>1\\ &\textrm{maka nilai}\quad x=3 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 53.&\textrm{Penyelesaian dari}\quad x^{\displaystyle x^{\displaystyle 4}}=\displaystyle 4\quad\textrm{adalah}\:....\\\\ &\textrm{Solusi}:\quad \\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&x^{\displaystyle x^{\displaystyle 4}}=\displaystyle 4\\ &\textrm{masing-masing ruas dipangkatkan}\quad 4,\\ &\textrm{selanjutnya}\\ &\left( x^{\displaystyle x^{\displaystyle 4}} \right)^{\displaystyle 4}=4^{\displaystyle 4}\Leftrightarrow \left( x^{\displaystyle 4} \right)^{\left( \displaystyle x^{\displaystyle 4} \right)}=4^{\displaystyle 4}\\ &\Leftrightarrow x^{\displaystyle 4}=4\Leftrightarrow \left| x \right|=\sqrt[\displaystyle 4]{4}=\sqrt[\displaystyle 4]{2^{\displaystyle 2}}\\ &\Leftrightarrow \left| x \right|=\sqrt{2}\\ &\textrm{Jadi, nilai}\quad x=-\sqrt{2}\quad \textrm{atau}\quad x=\sqrt{2} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 54.&\textrm{Jika diketahui}\quad x^{\displaystyle 2x^{\displaystyle 6}}=\displaystyle 3\quad\textrm{dengan}\quad x>1,\\ &\textrm{maka nilai}\quad \left( x^{\displaystyle x^{\displaystyle x^{\displaystyle 6}}} \right)^{\displaystyle \sqrt{3}}=\:....\\\\ &\textrm{Solusi}:\quad \\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&x^{\displaystyle 2x^{\displaystyle 6}}=\displaystyle 3\\ &\textrm{masing-masing ruas dipangkatkan}\quad 3,\\ &\textrm{selanjutnya}\\ &x^{\displaystyle 3.2x^{\displaystyle 6}}=\displaystyle 3^{\displaystyle 3}\Leftrightarrow \left( x^{\displaystyle 6} \right)^{\left( \displaystyle x^{\displaystyle 6} \right)}=3^{\displaystyle 3}\\ &\Leftrightarrow x^{\displaystyle 6}=3\Leftrightarrow x^{\displaystyle 2.3}=3\\ &\Leftrightarrow x^{\displaystyle 3}=3^{\displaystyle \frac{1}{2}}\Leftrightarrow x^{\displaystyle 3}=\sqrt{3}.\quad\textrm{dengan}\quad x>1\\ &\textrm{Sehingga}\\ &\left( x^{\displaystyle x^{\displaystyle x^{\displaystyle 6}}} \right)^{\displaystyle \sqrt{3}}=\left( x^{\displaystyle x^{\displaystyle 3}} \right)^{\sqrt{3}}=\left( x^{\displaystyle \sqrt{3}} \right)^{\displaystyle \sqrt{3}}\\ &=x^{\displaystyle 3}=\sqrt{3}\\ &\textrm{Jadi, nilai}\quad \left( x^{\displaystyle x^{\displaystyle x^{\displaystyle 6}}} \right)^{\displaystyle \sqrt{3}}=\sqrt{3} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 55.&\textrm{Penyelesaian dari}\quad x^{\displaystyle x}=3^{\displaystyle x+3^{\displaystyle 2}}\quad\textrm{adalah}\:....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 3\\ \textrm{b}.&\displaystyle 3^{\displaystyle \sqrt{3}}\\ \textrm{c}.&\displaystyle 6\\ \textrm{d}.&\displaystyle 9\\ \textrm{e}.&\displaystyle 3^{\displaystyle 3} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&x^{\displaystyle x}=3^{\displaystyle x+3^{\displaystyle 2}}\Leftrightarrow x^{\displaystyle x}=3^{\displaystyle x+9}\\ &\Leftrightarrow x^{\displaystyle x}=3^{\displaystyle x}.3^{\displaystyle 9}\\ &\Leftrightarrow \displaystyle \frac{x^{\displaystyle x}}{3^{\displaystyle x}}=3^{\displaystyle 9}\Leftrightarrow \left( \displaystyle \frac{x}{3} \right)^{\displaystyle x}=3^{\displaystyle 9}\\ &\textrm{masing-masing ruas dipangkatkan}\quad \displaystyle \frac{1}{3},\\ &\textrm{selanjutnya}\\ &\Leftrightarrow \left( \left( \displaystyle \frac{x}{3} \right)^{\displaystyle x} \right)^{\displaystyle \frac{1}{3}}=\left( 3^{\displaystyle 9} \right)^{\displaystyle \frac{1}{3}}\\ &\Leftrightarrow \left( \displaystyle \frac{x}{3} \right)^{\left( \displaystyle \frac{x}{3} \right)}=3^{\displaystyle 3}\\ &\Leftrightarrow \displaystyle \frac{x}{3}=3\Leftrightarrow x=9\\ &\textrm{Jadi, nilai}\quad x=9 \end{aligned} \end{array}$.

EKSPONEN-ANEKA CONTOH SOAL LANJUTAN (BAGIAN 10)

 $\begin{array}{ll}\\ 46.&\textrm{Jika diketahui}\: \quad \displaystyle x^{\displaystyle x^{\displaystyle x^{\displaystyle 2}}}=2\\ &\textrm{maka nilai}\quad x^{\displaystyle x^{\displaystyle 2}}+x^{\displaystyle 2}\:\: \textrm{adalah}\:....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{2}\sqrt{2}\\ \textrm{b}.&\displaystyle \sqrt{2}\\ \textrm{c}.&\displaystyle 2\\ \textrm{d}.&\displaystyle 2\sqrt{2}\\ \textrm{e}.&\displaystyle 4 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\textrm{Diketahui bahwa}\quad x^{\displaystyle x^{\displaystyle x^{\displaystyle 2}}}=2,\quad \textrm{maka nilai}\\ &x^{\displaystyle x^{\displaystyle 2}}+x^{\displaystyle 2}=2+2=4\\ &\textrm{perhatikan cara penyelesaiannya pada uraian}\\ &\textrm{jawaban pada nomor soal sebelumnya} \end{array}$.

$\begin{array}{ll}\\ 47.&\textrm{Nilai}\: \quad x\quad \textrm{pada}\quad\displaystyle 9^{\displaystyle 2^{\displaystyle x}}=3^{\displaystyle 8^{\displaystyle x}}\:\: \textrm{adalah}\:....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 3\\ \textrm{b}.&\displaystyle 1\\ \textrm{c}.&\displaystyle 6\\ \textrm{d}.&\displaystyle \frac{1}{2}\\ \textrm{e}.&\displaystyle \frac{1}{3} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&9^{\displaystyle 2^{\displaystyle x}}=3^{\displaystyle 8^{\displaystyle x}}\Leftrightarrow \left( 3^{\displaystyle 2} \right)^{\displaystyle 2^{\displaystyle x}}=3^{\displaystyle \left( 2^{\displaystyle 3} \right)^{\displaystyle x}}\\ &\Leftrightarrow 3^{\displaystyle 2^{\displaystyle 1}.2^{\displaystyle x}}=3^{\displaystyle 2^{\displaystyle 3x}}\Leftrightarrow 3^{\displaystyle 2^{\displaystyle (1+x)}}=3^{\displaystyle 2^{\displaystyle 3x}}\\ &\textrm{Selanjutnya pangkatnya tinggal disamakan}\\ &\textrm{yaitu}:\\ & 1+x=3x\Leftrightarrow 2x=1\Leftrightarrow x=\displaystyle \frac{1}{2} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 48.&\textrm{Nilai}\: \quad x\quad \textrm{pada}\quad\displaystyle (x+2)x^{\displaystyle x^{\displaystyle 2}}=4x^{\displaystyle 4(3-x)}\\ & \textrm{adalah}\:....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 1\\ \textrm{b}.&\displaystyle 2\\ \textrm{c}.&\displaystyle 3\\ \textrm{d}.&\displaystyle 4\\ \textrm{e}.&\displaystyle 6 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&(x+2)x^{\displaystyle x^{\displaystyle 2}}=4x^{\displaystyle 4(3-x)}\\ &\textrm{kalikan masing-masing ruas dengan}\quad x^{\displaystyle 4x}\\ &\Leftrightarrow (x+2)x^{\displaystyle x^{\displaystyle 2}}.\left( x^{\displaystyle 4x} \right)=4x^{\displaystyle 4(3-x)}.\left( x^{\displaystyle 4x} \right)\\ &\Leftrightarrow (x+2)x^{\displaystyle x^{\displaystyle 2}+4x}=4x^{\displaystyle (12-4x)+4x}\\ &\Leftrightarrow (x+2)x^{\displaystyle x(x+4)}=4x^{\displaystyle 12}\\ &\Leftrightarrow (x+2)x^{\displaystyle x(x+4)}=(2+2)x^{\displaystyle 2(2+4)}\\ &\: \textrm{Jadi},\quad x=2 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 49.&\textrm{Jika}\: \quad 3^{\displaystyle m}=a,\:\: 3^{\displaystyle n}=b\quad \textrm{maka nilai}\quad x\\ &\textrm{pada persamaan}\quad \displaystyle 9^{\displaystyle m+n}=a^{\displaystyle x}.b^{\displaystyle x}\\ & \textrm{adalah}\:....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 1\\ \textrm{b}.&\displaystyle 2\\ \textrm{c}.&\displaystyle 3\\ \textrm{d}.&\displaystyle 4\\ \textrm{e}.&\displaystyle 6 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&9^{\displaystyle m+n}=a^{\displaystyle x}.b^{\displaystyle x}\Leftrightarrow \left( 3^{\displaystyle 2} \right)^{\displaystyle m+n}=a^{\displaystyle x}.b^{\displaystyle x}\\ &\Leftrightarrow \left( 3^{\displaystyle 2} \right)^{\displaystyle m+n}=\left( 3^{\displaystyle m} \right)^{\displaystyle x}.\left( 3^{\displaystyle n} \right)^{\displaystyle x}\\ &\Leftrightarrow  \left( 3^{\displaystyle 2} \right)^{\displaystyle m+n}=3^{\displaystyle mx+nx}=\left( 3^{\displaystyle x} \right)^{\left( \displaystyle m+n \right)}\\ &\: \textrm{Jadi, nilai}\quad  x=2\end{aligned} \end{array}$.

$\begin{array}{ll}\\ 50.&\textrm{Jika}\: \quad x^{\displaystyle x}=\sqrt[\displaystyle 3]{20+14\sqrt{2}}+\sqrt[\displaystyle 3]{20-14\sqrt{2}}\\ & \textrm{maka nilai}\quad x^{\displaystyle 2}+2^{\displaystyle x}\quad\textrm{adalah}\:....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 2+2^{\displaystyle \sqrt{2}}\\ \textrm{b}.&\displaystyle 17\\ \textrm{c}.&\displaystyle 32\\ \textrm{d}.&\displaystyle 8\\ \textrm{e}.&\displaystyle 16 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&x^{\displaystyle x}=\sqrt[\displaystyle 3]{20+14\sqrt{2}}+\sqrt[\displaystyle 3]{20-14\sqrt{2}}\\ &\Leftrightarrow x^{\displaystyle x}=\sqrt[\displaystyle 3]{\left( 2+\sqrt{2} \right)^{\displaystyle 3}}+\sqrt[\displaystyle 3]{\left( 2-\sqrt{2} \right)^{\displaystyle 3}}\\ &\Leftrightarrow x^{\displaystyle x}=2+\sqrt{2}+2-\sqrt{2}=4=2^{\displaystyle 2}\\ &\textrm{Sehingga nilai}\\ &x^{\displaystyle 2}+2^{\displaystyle x}=2^{\displaystyle 2}+2^{\displaystyle 2}=4+4=8\end{aligned} \end{array}$.

EKSPONEN-ANEKA CONTOH SOAL LANJUTAN (BAGIAN 9)

$\begin{array}{ll}\\ 41.&\textrm{Nilai}\: \: x\quad \textrm{jika}\quad x^{\displaystyle x^{\displaystyle 16}}=\sqrt[\displaystyle 8]{\displaystyle 2}\quad \textrm{adalah}\:....\\ &\begin{array}{llll}\\ \textrm{a}.&\sqrt[\displaystyle 4]{2}\\ \textrm{b}.&\displaystyle \sqrt[\displaystyle 8]{2}\\ \textrm{c}.&\displaystyle \sqrt{32}\\ \textrm{d}.&\displaystyle \sqrt{8}\\ \textrm{e}.&\sqrt[\displaystyle 16]{2} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\textrm{Diketahui bahwa}\quad x^{\displaystyle x^{\displaystyle 16}}=\sqrt[\displaystyle 8]{\displaystyle 2}\:=\left( \displaystyle 2 \right)^{\displaystyle \frac{1}{8}}=2^{\displaystyle \frac{2}{16}}.\\ &\textrm{Perhatikan bahwa dengan memangkatkan 16}\\ &\textrm{di masing-masing ruas kita akan mendapatkan}\\ &\begin{aligned}&\left( x^{\displaystyle x^{\displaystyle 16}} \right)^{\displaystyle 16}=\left( \left( \displaystyle 2 \right)^{\displaystyle \frac{2}{16}} \right)^{\displaystyle 16}\\ &\Leftrightarrow x^{\displaystyle 16.x^{\displaystyle 16}}=\left( \displaystyle 2 \right)^{\left( \displaystyle 2 \right)}\\ &\Leftrightarrow \left( x^{\displaystyle 16} \right)^{\displaystyle x^{\displaystyle 16}}=\left( \displaystyle 2 \right)^{\left( \displaystyle 2 \right)}\\ &\Leftrightarrow x^{\displaystyle 16}=\displaystyle 2\Leftrightarrow x=\sqrt[\displaystyle 16]{2}  \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 42.&\textrm{Jika diketahui}\: \quad \displaystyle x^{\displaystyle x^{\displaystyle x+1}}=2^{\displaystyle -2^{\displaystyle -\frac{3}{2}}}\\ &\textrm{maka nilai}\quad x+1\quad \textrm{adalah}\:....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{3}{2}\\ \textrm{b}.&\displaystyle \frac{2}{3}\\ \textrm{c}.&\displaystyle \frac{4}{3}\\ \textrm{d}.&\displaystyle \frac{1}{3}\\ \textrm{e}.&\displaystyle \frac{1}{2} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\textrm{Diketahui bahwa}\\ &\begin{aligned}&\displaystyle x^{\displaystyle x^{\displaystyle x+1}}=2^{\displaystyle -2^{\displaystyle -\frac{3}{2}}}\\ &\Leftrightarrow \displaystyle x^{\displaystyle x^{\displaystyle x+1}}=\left( \displaystyle \frac{1}{2} \right)^{\displaystyle \left( \displaystyle \frac{1}{2} \right)^{\displaystyle \frac{3}{2}}}\\ &\Leftrightarrow \displaystyle x^{\displaystyle x^{\displaystyle x+1}}=\left( \displaystyle \frac{1}{2} \right)^{\displaystyle \left( \displaystyle \frac{1}{2} \right)^{\left( \displaystyle \frac{1}{2}+1 \right)}}\\ &\textrm{Sehingga nilai dari}\\ &x+1=\displaystyle \frac{1}{2}+1=\frac{3}{2} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 43.&\textrm{Jika diketahui}\: \quad \displaystyle x^{\displaystyle -x^{\displaystyle 1-x}}=3^{\displaystyle 18}\\ &\textrm{maka nilai}\quad x\displaystyle ^{\displaystyle x}\quad \textrm{adalah}\:....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{3}\\ \textrm{b}.&\displaystyle -\frac{1}{3}\\ \textrm{c}.&\displaystyle \frac{1}{9}\\ \textrm{d}.&\displaystyle -\frac{1}{9}\\ \textrm{e}.&\displaystyle \frac{1}{27} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\textrm{Diketahui bahwa}\\ &\begin{aligned}&\displaystyle x^{\displaystyle -x^{\displaystyle 1-x}}=3^{\displaystyle 18}\Leftrightarrow x^{\displaystyle -x^{\displaystyle 1}.x^{\displaystyle -x}}=3^{\displaystyle 2.9}\\ &\Leftrightarrow x^{\displaystyle -x.x^{\displaystyle -x}}=3^{\displaystyle 2.3^{\displaystyle 2}}\\ &\Leftrightarrow \left( x^{\displaystyle -x} \right)^{\displaystyle \left( x^{\displaystyle -x} \right)}=\left( 3^{\displaystyle 2} \right)^{\left( \displaystyle 3^{\displaystyle 2} \right)}\\ &\textrm{Sehingga kita mendapatkan persamaan}\\ &x^{\displaystyle -x}=3^{\displaystyle 2}\\ &\textrm{Selanjutnya pangkatkan -1 masing-masing ruas}\\ & \left( x^{\displaystyle -x} \right)^{\displaystyle -1}=\left( 3^{\displaystyle 2} \right)^{\displaystyle -1}\Leftrightarrow x^{\displaystyle x}=\displaystyle \frac{1}{3^{\displaystyle 2}}=\displaystyle \frac{1}{9} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 44.&\textrm{Jika diketahui}\: \quad \displaystyle x^{\displaystyle x^{\displaystyle x}}=2\\ &\textrm{maka nilai}\quad x\displaystyle ^{\displaystyle x^{\displaystyle x}+x^{\displaystyle x+x^{\displaystyle x+x^{\displaystyle x}}}}\:\: \textrm{adalah}\:....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 8\\ \textrm{b}.&\displaystyle 16\\ \textrm{c}.&\displaystyle 32\\ \textrm{d}.&\displaystyle 48\\ \textrm{e}.&\displaystyle 81 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}& x\displaystyle ^{\displaystyle x^{\displaystyle x}+x^{\displaystyle x+x^{\displaystyle x+x^{\displaystyle x}}}}\\ &=x^{\displaystyle x^{\displaystyle x}}.x^{\displaystyle x^{\displaystyle x}.x^{\displaystyle x^{\displaystyle x}.x^{\displaystyle x^{\displaystyle x}}}}\\ &=2.\left( 2^{\displaystyle 2^{\displaystyle 2}} \right)=2.\left( 2^{\displaystyle 4} \right)=2.16=32 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 45.&\textrm{Jika diketahui}\: \quad \displaystyle x^{\displaystyle x^{\displaystyle x^{\displaystyle 3}}}=3\\ &\textrm{maka nilai}\quad x^{\displaystyle 3}\:\: \textrm{adalah}\:....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{3}\sqrt{3}\\ \textrm{b}.&\displaystyle \sqrt{3}\\ \textrm{c}.&\displaystyle 3\\ \textrm{d}.&\displaystyle 3\sqrt{3}\\ \textrm{e}.&\displaystyle 27 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&x^{\displaystyle x^{\displaystyle x^{\displaystyle 3}}}=3\\ &\Leftrightarrow x^{\displaystyle x^{\displaystyle x^{\displaystyle 3}}}=\left( \sqrt[\displaystyle 3]{3} \right)^{\displaystyle 3}=\left( \sqrt[\displaystyle 3]{3} \right)^{\displaystyle \left( \sqrt[\displaystyle 3]{3} \right)^{\displaystyle 3}}\\ &\Leftrightarrow x^{\displaystyle x^{\displaystyle x^{\displaystyle 3}}}=\left( \sqrt[\displaystyle 3]{3} \right)^{\displaystyle \left( \sqrt[\displaystyle 3]{3} \right)^{\displaystyle \left( \sqrt[\displaystyle 3]{3} \right)^{\displaystyle 3}}}\\ &\Leftrightarrow x=\sqrt[\displaystyle 3]{3},\quad \textrm{maka nilai}\quad x^{\displaystyle 3}=\left( \sqrt[\displaystyle 3]{3} \right)^{\displaystyle 3}=3  \end{aligned} \end{array}$.

EKSPONEN-ANEKA CONTOH SOAL LANJUTAN (BAGIAN 8)

 $\begin{array}{ll}\\ 36.&\textrm{Nilai dari}\: \: \displaystyle \left( \frac{1}{125} \right)^{\displaystyle -9^{\displaystyle -2^{\displaystyle -1}}}=....\\ &\begin{array}{llll}\\ \textrm{a}.&1\\ \textrm{b}.&5\\ \textrm{c}.&\displaystyle \frac{1}{5}\\ \textrm{d}.&-\displaystyle \frac{1}{5}\\ \textrm{e}.&-5 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&\displaystyle \left( \frac{1}{125} \right)^{\displaystyle -9^{\displaystyle -2^{\displaystyle -1}}}=\displaystyle \left( \frac{1}{125} \right)^{\displaystyle -9^{\displaystyle -\frac{1}{2}}}\\ &=\displaystyle \left( \frac{1}{125} \right)^{\displaystyle -\left( \displaystyle \frac{1}{9} \right)^{\displaystyle \frac{1}{2}}}=\displaystyle \left( \frac{1}{125} \right)^{\displaystyle -\sqrt{\displaystyle \frac{1}{9}}}\\ &=\displaystyle \left( \frac{1}{64} \right)^{\displaystyle -\frac{1}{3}}=125^{\displaystyle \frac{1}{3}}=\left( 5^{\displaystyle 3} \right)^{\displaystyle \frac{1}{3}}=5^{\displaystyle \frac{3}{3}}=5  \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 37.&\textrm{Nilai dari}\: \quad \displaystyle 9^{\displaystyle 4^{\displaystyle -2^{\displaystyle -1}}}+\: 8^{\displaystyle 3^{\displaystyle -1^{\displaystyle 2}}}\:\: \textrm{adalah}\:....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 2\\ \textrm{b}.&\displaystyle 3\\ \textrm{c}.&\displaystyle 4\\ \textrm{d}.&\displaystyle 5\\ \textrm{e}.&\displaystyle 6 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&9^{\displaystyle 4^{\displaystyle -2^{\displaystyle -1}}}+\: 8^{\displaystyle 3^{\displaystyle -1^{\displaystyle 2}}}=9^{\displaystyle 4^{\displaystyle -\left( \displaystyle \frac{1}{2} \right)}}+\: 8^{\displaystyle 3^{\displaystyle -1}}\\ &=9^{\displaystyle \left( \displaystyle \frac{1}{4} \right)^{\displaystyle \left( \displaystyle \frac{1}{2} \right)}}+\: 8^{\displaystyle \left( \displaystyle \frac{1}{3} \right)}\\ &=9^\left( \sqrt{\displaystyle \frac{1}{4}} \right)+\: \left( 2^{\displaystyle 3} \right)^{\displaystyle \left( \displaystyle \frac{1}{3} \right)}\\ &=\left( 3^{\displaystyle 2} \right)^\left( \displaystyle \frac{1}{2} \right)+\: 2=3+2=5 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 38.&\textrm{Nilai dari}\\ &\left( \displaystyle \frac{1}{2} \right)^{\displaystyle -\left( \displaystyle \frac{1}{2} \right)^{\displaystyle -1}}+\left( \displaystyle \frac{1}{3} \right)^{\displaystyle -\left( \displaystyle \frac{1}{3} \right)^{\displaystyle -1}}+\left( \displaystyle \frac{1}{4} \right)^{\displaystyle -\left( \displaystyle \frac{1}{4} \right)^{\displaystyle -1}}+\left( \displaystyle \frac{1}{5} \right)^{\displaystyle -\left( \displaystyle \frac{1}{5} \right)^{\displaystyle -1}}\\ &\textrm{adalah}\:....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 3142\\ \textrm{b}.&\displaystyle 285\\ \textrm{c}.&\displaystyle 3412\\ \textrm{d}.&\displaystyle 4116\\ \textrm{e}.&\displaystyle 4096 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&\left( \displaystyle \frac{1}{2} \right)^{\displaystyle -\left( \displaystyle \frac{1}{2} \right)^{\displaystyle -1}}+\left( \displaystyle \frac{1}{3} \right)^{\displaystyle -\left( \displaystyle \frac{1}{3} \right)^{\displaystyle -1}}+\left( \displaystyle \frac{1}{4} \right)^{\displaystyle -\left( \displaystyle \frac{1}{4} \right)^{\displaystyle -1}}+\left( \displaystyle \frac{1}{5} \right)^{\displaystyle -\left( \displaystyle \frac{1}{5} \right)^{\displaystyle -1}}\\ &= 2^{\displaystyle 2}+3^{\displaystyle 3}+4^{\displaystyle 4}+5^{\displaystyle 5}\\ &=4+27+256+3125=3412 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 39.&\textrm{Nilai dari}\: \: x\quad \textrm{jika}\quad 8^{\displaystyle -9^{\displaystyle -32^{\displaystyle x}}}=\displaystyle \frac{1}{2}\quad \textrm{adalah}\:....\\ &\begin{array}{llll}\\ \textrm{a}.&-5\\ \textrm{b}.&-\displaystyle \frac{1}{5}\\ \textrm{c}.&\displaystyle \frac{1}{5}\\ \textrm{d}.&\displaystyle \frac{1}{3}\\ \textrm{e}.&-\displaystyle \frac{1}{4} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\textrm{Diketahui bahwa}\quad 8^{\displaystyle -9^{\displaystyle -32^{\displaystyle x}}}\\ &\textrm{Tak ada cara khusus untuk menyelesaikannya, tetapi}\\ &\textrm{kita mendapat petunjuk dari}\:\: 32^{\displaystyle x}=2^{\displaystyle 5x}.\\ &\textrm{Selanjutnya pilihannya jika tidak b ya c},\\ &\textrm{supaya 2 menjadi}\:\: \displaystyle \frac{1}{2}, \:\: \textrm{maka pilih b}.\\ &\textrm{Pengecekan}\\ &\begin{aligned}&\displaystyle 8^{\displaystyle -9^{\displaystyle -32^{\displaystyle -\frac{1}{5}}}}=\displaystyle 8^{\displaystyle -9^{\displaystyle -\left( 2^{\displaystyle 5} \right)^{\displaystyle -\frac{1}{5}}}}\\ &=\displaystyle 8^{\displaystyle -9^{\displaystyle -2^{\displaystyle -1}}}=\displaystyle 8^{\displaystyle -9^{\displaystyle -\frac{1}{2}}}=\displaystyle 8^{\displaystyle -\left( 3^{\displaystyle 2} \right)^{\displaystyle -\frac{1}{2}}}\\ &=8^{\displaystyle -3^{\displaystyle -1}}=8^{\displaystyle -\frac{1}{3}}=\left( 2^{\displaystyle 3} \right)^{\displaystyle -\frac{1}{3}}=2^{\displaystyle -1}=\displaystyle \frac{1}{2}  \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 40.&\textrm{Nilai dari}\: \: x^{\displaystyle 6}\quad \textrm{jika}\quad x^{\displaystyle x^{\displaystyle 6}}=\sqrt[\displaystyle 12]{\displaystyle \frac{1}{2}}\quad \textrm{adalah}\:....\\ &\begin{array}{llll}\\ \textrm{a}.&2\\ \textrm{b}.&\displaystyle \frac{1}{4}\\ \textrm{c}.&\displaystyle \frac{1}{2}\\ \textrm{d}.&-\displaystyle \frac{1}{2}\\ \textrm{e}.&-2 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\textrm{Diketahui bahwa}\quad x^{\displaystyle x^{\displaystyle 6}}=\sqrt[\displaystyle 12]{\displaystyle \frac{1}{2}}\:=\left( \displaystyle \frac{1}{2} \right)^{\displaystyle \frac{1}{12}}.\\ &\textrm{Perhatikan bahwa dengan memangkatkan 6}\\ &\textrm{di masing-masing ruas kita akan mendapatkan}\\ &\begin{aligned}&\left( x^{\displaystyle x^{\displaystyle 6}} \right)^{\displaystyle 6}=\left( \left( \displaystyle \frac{1}{2} \right)^{\displaystyle \frac{1}{12}} \right)^{\displaystyle 6}\\ &\Leftrightarrow x^{\displaystyle 6.x^{\displaystyle 6}}=\left( \displaystyle \frac{1}{2} \right)^{\left( \displaystyle \frac{1}{2} \right)}\\ &\Leftrightarrow \left( x^{\displaystyle 6} \right)^{\displaystyle x^{\displaystyle 6}}=\left( \displaystyle \frac{1}{2} \right)^{\left( \displaystyle \frac{1}{2} \right)}\\ &\Leftrightarrow x^{\displaystyle 6}=\displaystyle \frac{1}{2}  \end{aligned} \end{array}$.

EKSPONEN-ANEKA CONTOH SOAL LANJUTAN (BAGIAN 7)

 $\begin{array}{ll}\\ 31.&\textrm{Nilai dari}\: \: \displaystyle \frac{-1^{\displaystyle 0}+1^{\displaystyle 0}-2^{\displaystyle 0}+2^{\displaystyle 0}}{(-1)^{\displaystyle 0}+1^{\displaystyle 0}}=....\\ &\begin{array}{llll}\\ \textrm{a}.&-2\\ \textrm{b}.&-1\\ \textrm{c}.&0\\ \textrm{d}.&1\\ \textrm{e}.&2 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&\displaystyle \frac{-1^{\displaystyle 0}+1^{\displaystyle 0}-2^{\displaystyle 0}+2^{\displaystyle 0}}{(-1)^{\displaystyle 0}+1^{\displaystyle 0}}\\ &=\displaystyle \frac{-1+1-1+1}{1+1}=\frac{0}{2}=0  \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 32.&\textrm{Nilai dari}\: \: \displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 0}}}}-\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 0^{\displaystyle 2}}}}-\displaystyle 2^{\displaystyle 2^{\displaystyle 0^{\displaystyle 2^{\displaystyle 2}}}}=....\\ &\begin{array}{llll}\\ \textrm{a}.&4\\ \textrm{b}.&6\\ \textrm{c}.&8\\ \textrm{d}.&10\\ \textrm{e}.&12 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 0}}}}=\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 1}}}=\displaystyle 2^{\displaystyle 2^{\displaystyle 2}}=\displaystyle 2^{\displaystyle 4}=16\\ &\textrm{Sedangkan untuk}\\ &\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 0^{\displaystyle 2}}}}=\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 0}}}=\displaystyle 2^{\displaystyle 2^{\displaystyle 1}}=\displaystyle 2^{\displaystyle 2}=4\\ &\textrm{Selanjutnya untuk}\\ &\displaystyle 2^{\displaystyle 2^{\displaystyle 0^{\displaystyle 2^{\displaystyle 2}}}}=\displaystyle 2^{\displaystyle 2^{\displaystyle 0^{\displaystyle 4}}}=\displaystyle 2^{\displaystyle 2^{\displaystyle 0}}=2^{\displaystyle 1}=2\\ &\textrm{Sehingga}\\ &\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 0}}}}-\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 0^{\displaystyle 2}}}}-\displaystyle 2^{\displaystyle 2^{\displaystyle 0^{\displaystyle 2^{\displaystyle 2}}}}=16-4-2=10 \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 33.&\textrm{Nilai dari}\: \: \displaystyle x\displaystyle ^{\displaystyle x}=256,\quad \textrm{maka nilai}\quad x^{\displaystyle 2}-2^{\displaystyle x}=....\\ &\begin{array}{llll}\\ \textrm{a}.&-4\\ \textrm{b}.&-1\\ \textrm{c}.&0\\ \textrm{d}.&1\\ \textrm{e}.&4 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\textrm{Diketahui bahwa}\\ &\begin{aligned}&\displaystyle x\displaystyle ^{\displaystyle x}=256= 16^{\displaystyle 2}=\left( 4 ^{\displaystyle 2} \right)^{\displaystyle 2}=4^{\displaystyle 4}\Leftrightarrow x=4\\ &\textrm{Sehingga untuk}\\ &x^{\displaystyle 2}-2^{\displaystyle x}=4^{\displaystyle 2}-2^{\displaystyle 4}=16-16=0  \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 34.&\textrm{Nilai dari}\quad \displaystyle \frac{2^{\displaystyle 2^{\displaystyle n}+2^{\displaystyle n+1}}}{2^{\displaystyle 3.2^{\displaystyle n}}}\quad \textrm{adalah}\:....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{8}\\ \textrm{b}.&\displaystyle \frac{1}{4}\\ \textrm{c}.&\displaystyle \frac{1}{2}\\ \textrm{d}.&1\\ \textrm{e}.&2 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\textrm{Diketahui bahwa}\\ &\begin{aligned}&\displaystyle \frac{2^{\displaystyle 2^{\displaystyle n}+2^{\displaystyle n+1}}}{2^{\displaystyle 3.2^{\displaystyle n}}}=\displaystyle \frac{2^{\displaystyle 2^{\displaystyle n}+2^{\displaystyle n}.2^{\displaystyle 1}}}{2^{\displaystyle 3.2^{\displaystyle n}}}\\ &=\displaystyle \frac{2^{\displaystyle 2^{\displaystyle n}+2.2^{\displaystyle n}}}{2^{\displaystyle 3.2^{\displaystyle n}}}=\displaystyle \frac{2^{\displaystyle (1+2).2^{\displaystyle n}}}{2^{\displaystyle 3.2^{\displaystyle n}}}\\ &=\displaystyle \frac{2^{\displaystyle 3.2^{\displaystyle n}}}{2^{\displaystyle 3.2^{\displaystyle n}}}=1 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 35.&\textrm{Nilai dari}\: \: \displaystyle \left( \frac{1}{64} \right)^{\displaystyle -9^{\displaystyle -2^{\displaystyle -1}}}=....\\ &\begin{array}{llll}\\ \textrm{a}.&1\\ \textrm{b}.&4\\ \textrm{c}.&\displaystyle \frac{1}{4}\\ \textrm{d}.&-\displaystyle \frac{1}{4}\\ \textrm{e}.&-4 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&\displaystyle \left( \frac{1}{64} \right)^{\displaystyle -9^{\displaystyle -2^{\displaystyle -1}}}=\displaystyle \left( \frac{1}{64} \right)^{\displaystyle -9^{\displaystyle -\frac{1}{2}}}\\ &=\displaystyle \left( \frac{1}{64} \right)^{\displaystyle -\left( \displaystyle \frac{1}{9} \right)^{\displaystyle \frac{1}{2}}}=\displaystyle \left( \frac{1}{64} \right)^{\displaystyle -\sqrt{\displaystyle \frac{1}{9}}}\\ &=\displaystyle \left( \frac{1}{64} \right)^{\displaystyle -\frac{1}{3}}=64^{\displaystyle \frac{1}{3}}=\left( 4^{\displaystyle 3} \right)^{\displaystyle \frac{1}{3}}=4^{\displaystyle \frac{3}{3}}=4  \end{aligned}  \end{array}$.