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)
$\begin{aligned}&\begin{array}{|l|c|c|}\hline \begin{aligned}&\textrm{Jenis}\\ &\textrm{Transformasi} \end{aligned}&\textrm{Rumus}&\textrm{Matriks}\\\hline \textrm{Dilatasi}&&\\\hline \textrm{Pusat}\: \left [ O,k \right ]&(x,y)\rightarrow (kx,ky)&\begin{pmatrix} k & 0\\ 0 & k \end{pmatrix}\\\hline \begin{aligned}&\textrm{Pusat}\: (a,b)\\ & \textrm{faktor skala}\: k \end{aligned}&\begin{pmatrix} x'-a\\ y'-b \end{pmatrix}&\begin{aligned}&\colorbox{yellow}{Lihat}\\ &\colorbox{yellow}{di bawah}\\ &\colorbox{yellow}{tulisan}\\ &\colorbox{yellow}{warna}\\ &\colorbox{yellow}{merah} \end{aligned}\\\hline \begin{aligned}&\textrm{Luas bangun}\\ &\textrm{ datar} \end{aligned}&\textrm{Misal bangun A}&\textrm{T}=\begin{pmatrix} a & b\\ c & d \end{pmatrix}\\\hline &\textbf{Bangun A}'&= \textrm{det T}\times \textrm{A}\\\hline \end{array}\\ &\color{red}\triangleright \triangleright \triangleright \triangleright \triangleright \triangleright \begin{pmatrix} x'-a\\ y'-b \end{pmatrix}=\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}.\begin{pmatrix} x-a\\ y-b \end{pmatrix} \end{aligned}$.
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
- Johanes, Kastolan, Sulasim, 2006. Kompetensi Matematika 3A SMA Kelas XII Program IPA Semester Pertama. Jakarta: YUDHISTIRA.
- Mastd, A. dkk. 2021. Matematika Tingkat Lanjut untuk SMA Kelas XI (Kurikulum Merdeka). Jakarta: Pusat Perbukuan Kemendikbudristek.
- Nugroho, P. A. Gunarto, D. 2013. Big Bank Soal-Bahas MAtematika SMA/MA. Jakarta: WAHYUMEDIA.
EKSPONEN-ANEKA CONTOH SOAL LANJUTAN (BAGIAN 12)
$\begin{array}{ll}\\ 53.&\textrm{Bentuk sederhana dari}\\ &\left[ \sqrt[\displaystyle n^{\displaystyle n+1}]{\left( n^{\displaystyle n^{\displaystyle n^{\displaystyle 5n}}}\qquad \right)^{\displaystyle n\:\qquad}}\qquad \right]^{\displaystyle n^{\displaystyle n-\left( \displaystyle n^{n} \right)^{\displaystyle 5}}}\\ &\textrm{adalah}\:....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 1\\ \textrm{b}.&\displaystyle n\\ \textrm{c}.&\displaystyle n^{\displaystyle n}\\ \textrm{d}.&\displaystyle \sqrt[\displaystyle n]{n}\\ \textrm{e}.&\displaystyle \sqrt[\displaystyle n^{n}]{n} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&\left[ \sqrt[\displaystyle n^{\displaystyle n+1}]{\left( n^{\displaystyle n^{\displaystyle n^{\displaystyle 5n}}}\qquad \right)^{\displaystyle n\:\qquad}}\qquad \right]^{\displaystyle n^{\displaystyle n-\left( \displaystyle n^{n} \right)^{\displaystyle 5}}}\\ &=\left( n^{\displaystyle n^{\displaystyle n^{\displaystyle 5n}}} \right)^{\left( \displaystyle \frac{n}{n^{\displaystyle n+1}} \right).\displaystyle n^{\displaystyle n-\left( n^{\displaystyle n} \right)^{\displaystyle 5}}}\\ &=\left( n^{\displaystyle n^{\displaystyle n^{\displaystyle 5n}}} \right)^{\left( \displaystyle \frac{n}{n^{\displaystyle n}.n} \right).\displaystyle n^{\displaystyle n-\left( n^{\displaystyle n} \right)^{\displaystyle 5}}}\\ &=n^{\displaystyle n^{\displaystyle n^{\displaystyle 5n}}.n^{\displaystyle -n}.n^{\displaystyle n}.n^{\displaystyle -n^{\displaystyle 5n}}}=n^{\displaystyle n^{\displaystyle 0}}=n^{\displaystyle 1}=n\\ \end{aligned} \end{array}$.
EKSPONEN-ANEKA CONTOH SOAL LANJUTAN (BAGIAN 11)
$\begin{array}{ll}\\ 51.&\textrm{Penyelesaian dari}\\ &\left(\displaystyle \frac{1}{x^{\displaystyle 3}} \right)^{\left( \displaystyle \frac{1}{x^{\displaystyle 4}} \right)}=\left(\displaystyle \frac{1}{\sqrt[\displaystyle 3x]{x}\:} \right)^{\left( \displaystyle \frac{1}{x} \right)}\\ &\textrm{untuk}\quad x>1\quad\textrm{adalah}\:....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \displaystyle \frac{4}{3}\\ \textrm{b}.&\displaystyle 2\\ \textrm{c}.&\displaystyle \frac{3}{2}\\ \textrm{d}.&\displaystyle 5\\ \textrm{e}.&\displaystyle 3 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&\left(\displaystyle \frac{1}{x^{\displaystyle 3}} \right)^{\left( \displaystyle \frac{1}{x^{\displaystyle 4}} \right)}=\left(\displaystyle \frac{1}{\sqrt[\displaystyle 3x]{x}\:} \right)^{\left( \displaystyle \frac{1}{x} \right)}\:\textrm{dengan}\quad x>1\\ &\Leftrightarrow \left(\displaystyle \frac{1}{x^{\displaystyle 3}} \right)^{\displaystyle x}=\left(\displaystyle \frac{1}{x^{\displaystyle \frac{1}{3x}}} \right)^{\displaystyle x^{\displaystyle 4}}\\ &\Leftrightarrow \left(\displaystyle x^{\displaystyle -3} \right)^{\displaystyle x}=\left(x^{\displaystyle -\frac{1}{3x}} \right)^{\displaystyle x^{\displaystyle 4}}\\ &\Leftrightarrow x^{\displaystyle -3x}=x^{\displaystyle -\frac{x^{\displaystyle 4}}{3x}}\\ &\Leftrightarrow -3x=\displaystyle -\frac{x^{\displaystyle 4}}{3x}\Leftrightarrow 9x^{\displaystyle 2}=x^{\displaystyle 4}\\ &\Leftrightarrow x^{\displaystyle 2}=9\Leftrightarrow x=\left| 3 \right|=\pm 3,\quad\textrm{karena}\quad x>1\\ &\textrm{maka nilai}\quad x=3 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 52.&\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}$.
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)
$\begin{array}{ll}\\ 31.&\textrm{Nilai dari}\: \: \displaystyle \frac{-1^{\displaystyle 0}+1^{\displaystyle 0}-2^{\displaystyle 0}+2^{\displaystyle 0}}{(-1)^{\displaystyle 0}+1^{\displaystyle 0}}=....\\ &\begin{array}{llll}\\ \textrm{a}.&-2\\ \textrm{b}.&-1\\ \textrm{c}.&0\\ \textrm{d}.&1\\ \textrm{e}.&2 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&\displaystyle \frac{-1^{\displaystyle 0}+1^{\displaystyle 0}-2^{\displaystyle 0}+2^{\displaystyle 0}}{(-1)^{\displaystyle 0}+1^{\displaystyle 0}}\\ &=\displaystyle \frac{-1+1-1+1}{1+1}=\frac{0}{2}=0 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 32.&\textrm{Nilai dari}\: \: \displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 0}}}}-\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 0^{\displaystyle 2}}}}-\displaystyle 2^{\displaystyle 2^{\displaystyle 0^{\displaystyle 2^{\displaystyle 2}}}}=....\\ &\begin{array}{llll}\\ \textrm{a}.&4\\ \textrm{b}.&6\\ \textrm{c}.&8\\ \textrm{d}.&10\\ \textrm{e}.&12 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 0}}}}=\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 1}}}=\displaystyle 2^{\displaystyle 2^{\displaystyle 2}}=\displaystyle 2^{\displaystyle 4}=16\\ &\textrm{Sedangkan untuk}\\ &\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 0^{\displaystyle 2}}}}=\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 0}}}=\displaystyle 2^{\displaystyle 2^{\displaystyle 1}}=\displaystyle 2^{\displaystyle 2}=4\\ &\textrm{Selanjutnya untuk}\\ &\displaystyle 2^{\displaystyle 2^{\displaystyle 0^{\displaystyle 2^{\displaystyle 2}}}}=\displaystyle 2^{\displaystyle 2^{\displaystyle 0^{\displaystyle 4}}}=\displaystyle 2^{\displaystyle 2^{\displaystyle 0}}=2^{\displaystyle 1}=2\\ &\textrm{Sehingga}\\ &\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 0}}}}-\displaystyle 2^{\displaystyle 2^{\displaystyle 2^{\displaystyle 0^{\displaystyle 2}}}}-\displaystyle 2^{\displaystyle 2^{\displaystyle 0^{\displaystyle 2^{\displaystyle 2}}}}=16-4-2=10 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 33.&\textrm{Nilai dari}\: \: \displaystyle x\displaystyle ^{\displaystyle x}=256,\quad \textrm{maka nilai}\quad x^{\displaystyle 2}-2^{\displaystyle x}=....\\ &\begin{array}{llll}\\ \textrm{a}.&-4\\ \textrm{b}.&-1\\ \textrm{c}.&0\\ \textrm{d}.&1\\ \textrm{e}.&4 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\textrm{Diketahui bahwa}\\ &\begin{aligned}&\displaystyle x\displaystyle ^{\displaystyle x}=256= 16^{\displaystyle 2}=\left( 4 ^{\displaystyle 2} \right)^{\displaystyle 2}=4^{\displaystyle 4}\Leftrightarrow x=4\\ &\textrm{Sehingga untuk}\\ &x^{\displaystyle 2}-2^{\displaystyle x}=4^{\displaystyle 2}-2^{\displaystyle 4}=16-16=0 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 34.&\textrm{Nilai dari}\quad \displaystyle \frac{2^{\displaystyle 2^{\displaystyle n}+2^{\displaystyle n+1}}}{2^{\displaystyle 3.2^{\displaystyle n}}}\quad \textrm{adalah}\:....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{8}\\ \textrm{b}.&\displaystyle \frac{1}{4}\\ \textrm{c}.&\displaystyle \frac{1}{2}\\ \textrm{d}.&1\\ \textrm{e}.&2 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\textrm{Diketahui bahwa}\\ &\begin{aligned}&\displaystyle \frac{2^{\displaystyle 2^{\displaystyle n}+2^{\displaystyle n+1}}}{2^{\displaystyle 3.2^{\displaystyle n}}}=\displaystyle \frac{2^{\displaystyle 2^{\displaystyle n}+2^{\displaystyle n}.2^{\displaystyle 1}}}{2^{\displaystyle 3.2^{\displaystyle n}}}\\ &=\displaystyle \frac{2^{\displaystyle 2^{\displaystyle n}+2.2^{\displaystyle n}}}{2^{\displaystyle 3.2^{\displaystyle n}}}=\displaystyle \frac{2^{\displaystyle (1+2).2^{\displaystyle n}}}{2^{\displaystyle 3.2^{\displaystyle n}}}\\ &=\displaystyle \frac{2^{\displaystyle 3.2^{\displaystyle n}}}{2^{\displaystyle 3.2^{\displaystyle n}}}=1 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 35.&\textrm{Nilai dari}\: \: \displaystyle \left( \frac{1}{64} \right)^{\displaystyle -9^{\displaystyle -2^{\displaystyle -1}}}=....\\ &\begin{array}{llll}\\ \textrm{a}.&1\\ \textrm{b}.&4\\ \textrm{c}.&\displaystyle \frac{1}{4}\\ \textrm{d}.&-\displaystyle \frac{1}{4}\\ \textrm{e}.&-4 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&\displaystyle \left( \frac{1}{64} \right)^{\displaystyle -9^{\displaystyle -2^{\displaystyle -1}}}=\displaystyle \left( \frac{1}{64} \right)^{\displaystyle -9^{\displaystyle -\frac{1}{2}}}\\ &=\displaystyle \left( \frac{1}{64} \right)^{\displaystyle -\left( \displaystyle \frac{1}{9} \right)^{\displaystyle \frac{1}{2}}}=\displaystyle \left( \frac{1}{64} \right)^{\displaystyle -\sqrt{\displaystyle \frac{1}{9}}}\\ &=\displaystyle \left( \frac{1}{64} \right)^{\displaystyle -\frac{1}{3}}=64^{\displaystyle \frac{1}{3}}=\left( 4^{\displaystyle 3} \right)^{\displaystyle \frac{1}{3}}=4^{\displaystyle \frac{3}{3}}=4 \end{aligned} \end{array}$.
EKSPONEN-ANEKA CONTOH SOAL LANJUTAN (BAGIAN 6)
$\begin{array}{ll}\\ 26.&\textrm{Nilai dari}\: \: \displaystyle \sqrt[4]{4}-\sqrt{2}=....\\ &\begin{array}{llll}\\ \textrm{a}.&-2\\ \textrm{b}.&-1\\ \textrm{c}.&0\\ \textrm{d}.&1\\ \textrm{e}.&2 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&\displaystyle \sqrt[4]{4}-\sqrt{2}=\displaystyle \sqrt[2]{\sqrt[2]{2^{2}}}-\sqrt{2}=\sqrt{2}-\sqrt{2}=0 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 27.&\textrm{Nilai dari}\: \: \displaystyle \frac{\sqrt[\displaystyle 3]{64}}{\sqrt[\displaystyle 4]{64}}=....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{2}\sqrt{2}\\ \textrm{b}.&\sqrt{2}\\ \textrm{c}.&2\sqrt{2}\\ \textrm{d}.&3\\ \textrm{e}.&4 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&\displaystyle \frac{\sqrt[\displaystyle 3]{64}}{\sqrt[\displaystyle 4]{64}}=\displaystyle \frac{\sqrt[\displaystyle 3]{4^{\displaystyle 3}}}{\sqrt[\displaystyle 4]{8^{\displaystyle 2}}}=\displaystyle \frac{\sqrt[\displaystyle 3]{4^{\displaystyle 3}}}{\sqrt[\displaystyle 4]{\sqrt{8}^{\displaystyle 4}}}=\frac{4}{\sqrt{8}}\\ &=\frac{4}{\sqrt{8}}\times \frac{\sqrt{8}}{\sqrt{8}}=\frac{4}{8}\left( \sqrt{8} \right)\\ &=\frac{1}{2}\left( 2\sqrt{2} \right)=\sqrt{2} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 28.&\textrm{Nilai dari}\: \: \displaystyle \frac{\sqrt{\sqrt{25}}}{\sqrt{5}}=....\\ &\begin{array}{llll}\\ \textrm{a}.&5\\ \textrm{b}.&\sqrt{5}\\ \textrm{c}.&\displaystyle \frac{1}{2}\sqrt{5}\\ \textrm{d}.&1\\ \textrm{e}.&\displaystyle \frac{1}{2} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&\displaystyle \frac{\sqrt{\sqrt{25}}}{\sqrt{5}}=\frac{\sqrt{\sqrt[\displaystyle 2]{5^{\displaystyle 2}}}}{\sqrt{5}}=\frac{\sqrt{5}}{\sqrt{5}}=1 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 29.&\textrm{Nilai dari}\: \: \displaystyle \frac{\sqrt{49\sqrt{7}}}{\sqrt{7\sqrt{49}}}=....\\ &\begin{array}{llll}\\ \textrm{a}.&7\\ \textrm{b}.&7\sqrt{7}\\ \textrm{c}.&\displaystyle \frac{1}{2}\sqrt{7}\\ \textrm{d}.&\sqrt{7}\\ \textrm{e}.&\displaystyle \sqrt[\displaystyle 4]{7} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&\displaystyle \frac{\sqrt{49\sqrt{7}}}{\sqrt{7\sqrt{49}}}=\displaystyle \frac{7.\sqrt[\displaystyle 4]{7}}{\sqrt{7}.\sqrt[\displaystyle 4]{49}}=\displaystyle \frac{7.\sqrt[\displaystyle 4]{7}}{\sqrt{7}.\sqrt[\displaystyle 4]{7^{\displaystyle 2}}}\\ &=\displaystyle \frac{7.\sqrt[\displaystyle 4]{7}}{\sqrt{7}.\sqrt{7}}=\displaystyle \frac{7.\sqrt[\displaystyle 4]{7}}{7}=\sqrt[\displaystyle 4]{7} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 30.&\textrm{Nilai dari}\: \: \displaystyle \left( \frac{1}{2} \right)^{\displaystyle \frac{1}{2}}-\left( \frac{1}{4} \right)^{\displaystyle \frac{1}{4}}=....\\ &\begin{array}{llll}\\ \textrm{a}.&-2\\ \textrm{b}.&-1\\ \textrm{c}.&0\\ \textrm{d}.&1\\ \textrm{e}.&2 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&\displaystyle \left( \frac{1}{2} \right)^{4}=\left( \frac{1}{4} \right)^{2}\Leftrightarrow \displaystyle \left( \frac{1}{2} \right)^{\displaystyle \frac{4}{1}}=\left( \frac{1}{4} \right)^{\displaystyle \frac{2}{1}}\\ &\Leftrightarrow \left( \frac{1}{2} \right)^{\displaystyle \frac{1}{2}}=\left( \frac{1}{4} \right)^{\displaystyle \frac{1}{4}}\\ &\Leftrightarrow \left( \frac{1}{2} \right)^{\displaystyle \frac{1}{2}}-\left( \frac{1}{4} \right)^{\displaystyle \frac{1}{4}}=0 \end{aligned} \end{array}$.
EKSPONEN-ANEKA CONTOH SOAL LANJUTAN (BAGIAN 5)
$\begin{array}{ll}\\ 21.&(\textbf{UM UGM 05})\textrm{Hasil dari}\\ &\sqrt{0,3+\sqrt{0,08}}=\sqrt{a}+\sqrt{b}\: ,\: \textrm{maka}\: \: \displaystyle \frac{1}{a}+\frac{1}{b}=....\\ &\begin{array}{llll}\\ \textrm{a}.&25\\ \textrm{b}.&20\\ \textrm{c}.&15\\ \textrm{d}.&10\\ \textrm{e}.&5 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\sqrt{0,3+\sqrt{0,08}}&=\sqrt{0,2+0,1+\sqrt{4\times 0,2\times 0,1}}\\ &=\sqrt{0,2+0,1+2\sqrt{\times 0,2\times 0,1}}\\ &=\sqrt{0,2}+\sqrt{0,1}\\ \textrm{maka},\: \: a=0,2&,\: \: b=0,1\\ \textrm{sehingga}\: \displaystyle \frac{1}{a}+\frac{1}{b}&=\displaystyle \frac{1}{0,2}+\frac{1}{0,1}=5+10=15\\ \end{aligned} \end{array}$
$\begin{array}{ll}\\ 22.&(\textbf{SPMB 06})\textrm{Jika bilangan bulat}\: \: a\: \: \: \textrm{dan}\: \: b\: \: \textrm{memenuhi}\\ &\displaystyle \frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}+\sqrt{6}}=a+b\sqrt{30}\: ,\: \textrm{maka}\: \: ab=....\\ &\begin{array}{llll}\\ \textrm{a}.&-22\\ \textrm{b}.&-11\\ \textrm{c}.&-9\\ \textrm{d}.&2\\ \textrm{e}.&13 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\displaystyle \frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}+\sqrt{6}}&=\displaystyle \frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}+\sqrt{6}}\times \displaystyle \frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}-\sqrt{6}}\\ &=\displaystyle \frac{5-2\sqrt{30}+6}{5-6}\\ &=\displaystyle \frac{11-2\sqrt{30}}{-1}\\ &=-11+2\sqrt{30}\\ \textrm{sehingga}&\: \: \: a=-11,\: \: b=2,\: \: \textrm{maka}\\ ab&=(-11)\times 2\\ &=-22 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 23.&(\textbf{OSK 2013})\textrm{Misal}\: \: a\: \: \textrm{dan}\: \: b\: \: \textrm{bilangan asli}\\ &\textrm{dengan}\: \: a>b.\: \: \textrm{Jika} \: \: \sqrt{94+2\sqrt{2013}}=\sqrt{a}+\sqrt{b}\\ &\textrm{maka nilai} \: \: a-b\: \: \textrm{adalah... .}\\\\ &\textrm{Jawab}:\\ &\begin{aligned} \sqrt{94+2\sqrt{2013}}&=\sqrt{61+33+2\sqrt{61\times 33}}\\ &=\sqrt{61}+\sqrt{33}\\ &=\sqrt{a}+\sqrt{b}\\ \textrm{Sehingga}\: \: a&=61,\: \: b=33,\: \: \textrm{maka}\\ a-b&=61-33\\ &=28 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 24.&\textrm{Daerah hasil dari fungsi eksponen}\: \: y\: =x^{- \frac{2}{3}}\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllllll}\\ \textrm{a}.&y< 0\\ \textrm{b}.&y> 0\\ \textrm{c}.&y\geq 0\\ \textrm{d}.&y\leq 0\\ \textrm{e}.&\textrm{Semua bilangan real} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\textrm{Perhatikanlah gambar berikut} \end{array}$
EKSPONEN-ANEKA CONTOH SOAL LANJUTAN (BAGIAN 4)
$\begin{array}{ll}\\ 16.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &x^{\displaystyle x^{\displaystyle x^{\displaystyle x^{\displaystyle x^{\cdots }}}}}=2026 \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\sqrt{2026}\\ \textrm{b}.&\sqrt[\displaystyle 2026]{2026}\\ \textrm{c}.&2026^{\sqrt{2026}}\\ \textrm{d}.&\sqrt{2026}^{\sqrt{2026}}\\ \textrm{e}.&\sqrt{2026\sqrt{2026}} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}x^{\displaystyle x^{\displaystyle x^{\displaystyle x^{\displaystyle x^{\cdots }}}}}&=2026\\ x^{2026}&=2026\\ x&=\sqrt[\displaystyle 2026]{2026} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 17.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ & \sqrt{x+\sqrt{x+\sqrt{x+\cdots }}}=3 \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&3\\ \textrm{b}.&6\\ \textrm{c}.&7\\ \textrm{d}.&8\\ \textrm{e}.&9 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{Misalkan}\quad A&=\sqrt{+\sqrt{x+\sqrt{+\cdots }}}\\ \sqrt{x+\sqrt{x+\sqrt{x+\cdots }}}&=3\\ \textrm{dikuadratkan}&\\ x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots }}}&=9\\ x+3&=9\\ x&=9-3\\ x&=6 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 18.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &x=\sqrt[3]{49\sqrt[3]{49\sqrt[3]{49\cdots }}} \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&7\sqrt[7]{7}\\ \textrm{b}.&7\\ \textrm{c}.&14\\ \textrm{d}.&49\\ \textrm{e}.&\sqrt[3]{81} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}x&=\sqrt[3]{49\sqrt[3]{49\sqrt[3]{49\cdots }}}\\ x^{3}&=49\sqrt[3]{49\sqrt[3]{49\sqrt[3]{49\cdots }}}\\ x^{3}&=49x\\ x^{2}&=49\\ x&=\sqrt{49}\\ &=7 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 19.&\textrm{Nilai dari}\\ &\displaystyle \frac{1+\sqrt[3]{2}}{1+\sqrt[3]{2}+\sqrt[3]{4}}\\ &\textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&1\\ \textrm{b}.&\sqrt[3]{2}+1\\ \textrm{c}.&\sqrt[3]{2}-1\\ \textrm{d}.&\sqrt[3]{4}+1\\ \textrm{e}.&\sqrt[3]{4}-1 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}&\displaystyle \frac{1+\sqrt[3]{2}}{1+\sqrt[3]{2}+\sqrt[3]{4}}\times \frac{\sqrt[3]{2}-1}{\sqrt[3]{2}-1}\\ &=\displaystyle \frac{\left ( \sqrt[3]{2} \right )^{2}-1}{\sqrt[3]{2}+\sqrt[3]{4}+\sqrt[3]{8}-1-\sqrt[3]{2}-\sqrt[3]{4}}\\ &=\displaystyle \frac{\sqrt[3]{4}-1}{\sqrt[3]{8}-1}\\ &=\displaystyle \frac{\sqrt[3]{4}-1}{2-1}\\ &=\sqrt[3]{4}-1 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 20.&\textrm{Nilai dari}\\ &\displaystyle \frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}\\ &\textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&1\\ \textrm{b}.&2\sqrt{2}-1\\ \textrm{c}.&\displaystyle \frac{1}{2}\sqrt{2}\\ \textrm{d}.&\sqrt{\displaystyle \frac{5}{3}}\\ \textrm{e}.&\sqrt{\displaystyle \frac{2}{5}} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}&\displaystyle \frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}\\ &=\displaystyle \frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\times \frac{\sqrt{\sqrt{5}+1}}{\sqrt{\sqrt{5}+1}} -\sqrt{3-2\sqrt{2}}\\ &=\displaystyle \frac{\sqrt{7+3\sqrt{5}}+\sqrt{3-\sqrt{5}}}{\sqrt{5}+1}-\left ( \sqrt{2}-1 \right )\\ &=\displaystyle \frac{\left ( \displaystyle \frac{3+\sqrt{5}}{\sqrt{2}} \right )+\left ( \displaystyle \frac{\sqrt{5}-1}{\sqrt{2}} \right )}{\sqrt{5}+1}+1-\sqrt{2}\\ &=\displaystyle \frac{\displaystyle \frac{2+2\sqrt{5}}{\sqrt{2}}}{1+\sqrt{5}}+1-\sqrt{2}\\ &=\displaystyle \frac{2}{\sqrt{2}}+1-\sqrt{2}\\ &=\sqrt{2}+1-\sqrt{2}\\ &=1 \end{aligned} \end{array}$
EKSPONEN-ANEKA CONTOH SOAL LANJUTAN (BAGIAN 3)
$\begin{array}{l}\\ 11.&\textrm{Nilai dari}\\ & \displaystyle \frac{2^{2026}+2^{2027}}{3}-2^{2026}\: \: \textrm{adalah}\: ....\\\\ &\begin{array}{lllllllll}\\ \textrm{a}.&-2&&&\\ \textrm{b}.&-1\\ \textrm{c}.&0\\ \textrm{d}.&1\\ \textrm{e}.&2 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}&\displaystyle \frac{2^{2026}+2^{2027}}{3}-2^{2026}\\ &=\displaystyle \frac{2^{2026}+2^{2027}-3.2^{2026}}{3}\\ &=\displaystyle \frac{1.2^{2026}+2^{1}.2^{2026}-3.2^{2026}}{3}\\ &=\displaystyle \frac{(3-3).2^{2026}}{3}\\ &=0 \end{aligned} \end{array}$.
$\begin{array}{l}\\ 12.&\textrm{Nilai dari}\\ &\displaystyle \frac{2^{2026}+2^{2027}+2^{2028}}{2^{2029}+2^{2030}+2^{2031}}\: \: \textrm{adalah}\: ....\\\\ &\begin{array}{lllllllll}\\ \textrm{a}.&0,125&&&\\ \textrm{b}.&0,\overline{333}\\ \textrm{c}.&0,45\\ \textrm{d}.&0,5\\ \textrm{e}.&0,\overline{666} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}&\displaystyle \frac{2^{2026}+2^{2027}+2^{2028}}{2^{2029}+2^{2030}+2^{2031}}\\ &=\displaystyle \frac{1.2^{2026}+2^{1}.2^{2026}+2^{2}.2^{2026}}{2^{3}.2^{2026}+2^{4}.2^{2026}+2^{5}.2^{2026}}\\ &=\displaystyle \frac{(1+2+4).2^{2026}}{(8+16+32).2^{2026}}\\ &=\displaystyle \frac{7}{56}=\frac{1}{8}=0,125 \end{aligned} \end{array}$.
$\begin{array}{l}\\ 13.&\textrm{Jika nilai dari}\\ &a^{\displaystyle a}=3\: ,\: \textrm{maka nilai}\quad a^{\displaystyle a^{\displaystyle a+1}}\quad \textrm{adalah}\: ....\\\\ &\begin{array}{lllllllll}\\ \textrm{a}.&9&&&\\ \textrm{b}.&18\\ \textrm{c}.&27\\ \textrm{d}.&81\\ \textrm{e}.&243 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}&\displaystyle a^{\displaystyle a^{\displaystyle a+1}}=a^{\displaystyle a^{\displaystyle a}.a}=a^{\displaystyle 3.a}=\left( a^{\displaystyle a} \right)^{3}=3^{3}=27 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 14.&\textrm{Hitunglah}\:\\\\ &\quad\quad\qquad \displaystyle \sqrt[8]{2207-\displaystyle \frac{1}{2207-\displaystyle \frac{1}{2207-\displaystyle \frac{1}{2207-\displaystyle \frac{1}{...}}}}}\\\\ &\textrm{nyatakan jawabannya dalam bentuk }\: \displaystyle \frac{a\pm b\sqrt{c}}{d}\\ &\textrm{dengan a, b, c, dan d bilangan-bilangan bulat}\\ \end{array}$
Pembahasan:
$\begin{aligned}x^{8}&=2207-\displaystyle \underset{x^{8}}{\underbrace{\displaystyle \frac{1}{2207-\frac{1}{2207-\frac{1}{2207-...}}}}}\\ x^{8}&=2207-\displaystyle \frac{1}{x^{8}}\\ x^{8}+\displaystyle \frac{1}{x^{8}}&=2207\\ \left ( x^{4}+\displaystyle \frac{1}{x^{4}} \right )^{2}&=2207+2\\ \left ( x^{4}+\displaystyle \frac{1}{x^{4}} \right )&=\sqrt{2209}=47 \end{aligned}$
$\begin{aligned}x^{4}+\displaystyle \frac{1}{x^{4}}&=47\\ \left ( x^{2}+\displaystyle \frac{1}{x^{2}} \right )^{2}&=47+2\\ x^{2}+\displaystyle \frac{1}{x^{2}}&=\sqrt{49}=7\\ \left ( x+\displaystyle \frac{1}{x} \right )^{2}&=7+2\\ x+\displaystyle \frac{1}{x}&=\sqrt{9}=3\\ x^{2}-3x+1&=0,\\ &\textrm{persamaan kuadrat dalam x,}\\ & \textbf{gunakan rumus abc}\\ x_{1,2}=&\displaystyle \frac{3\pm \sqrt{5}}{2}=\displaystyle \frac{3\pm 1\sqrt{5}}{2}=\displaystyle \frac{a\pm b\sqrt{c}}{d}\\ &\textbf{Sehingga},\quad \begin{cases} & a=3 \\ & b=1 \\ & c=5 \\ & d=2 \end{cases} \end{aligned}$.
$\begin{array}{ll}\\ 15.&\textrm{Diketahui}\\ &x=\displaystyle \frac{1+p+p^{2}+p^{3}+\cdots +p^{n-1}}{1+p+p^{2}+p^{3}+\cdots +p^{n-2}+p^{n-1}+p^{n}} \\ &y=\displaystyle \frac{1+q+q^{2}+q^{3}+\cdots +q^{n-1}}{1+q+q^{2}+q^{3}+\cdots +q^{n-2}+q^{n-1}+q^{n}}\\\\ &\textrm{dan}\: \: p>q>0\\\\ &\textrm{Tunjukkan bahwa}\: \: x<y \\\\\\ &\textbf{Bukti}:\\ &\begin{aligned}&\textrm{Perhatikan bahwa}:\: \: p>q>0\\ &\textrm{sehingga}\\ &\displaystyle \frac{1}{p}< \frac{1}{q},\: \: \displaystyle \frac{1}{p^{2}}< \frac{1}{q^{2}},\cdots , \displaystyle \frac{1}{p^{n}}< \frac{1}{q^{n}}\\ &\textrm{Jika bentuk di atas dijumlahkan, maka}\\ &\displaystyle \frac{1}{p}+\frac{1}{p^{2}}+\cdots +\frac{1}{p^{n}}< \frac{1}{q}+\frac{1}{q^{2}}+\cdots +\frac{1}{q^{n}}\\ &\Leftrightarrow \displaystyle \frac{p^{n-1}+\cdots +p^{2}+p+1}{p^{n}}< \displaystyle \frac{q^{n-1}+\cdots +q^{2}+q+1}{q^{n}}\\ &\Leftrightarrow \displaystyle \frac{p^{n}}{1+p+p^{2}+\cdots +p^{n-1}}>\displaystyle \frac{q^{n}}{1+q+q^{2}+\cdots +q^{n-1}}\\ &\Leftrightarrow \displaystyle \frac{p^{n}}{1+p+p^{2}+\cdots +p^{n-1}}+1>\displaystyle \frac{q^{n}}{1+q+q^{2}+\cdots +q^{n-1}}+1\\ &\Leftrightarrow \displaystyle \frac{1+p+p^{2}+\cdots +p^{n-1}+p^{n}}{1+p+p^{2}+\cdots +p^{n-1}}>\displaystyle \frac{1+q+q^{2}+\cdots +q^{n-1}+q^{n}}{1+q+q^{2}+\cdots +q^{n-1}}\\ &\Leftrightarrow \displaystyle \frac{1+p+p^{2}+\cdots +p^{n-1}}{1+p+p^{2}+\cdots +p^{n-1}+p^{n}}<\displaystyle \frac{1+q+q^{2}+\cdots +q^{n-1}}{1+q+q^{2}+\cdots +q^{n-1}+q^{n}}\\ &\Leftrightarrow x<y\qquad \blacksquare \end{aligned} \end{array}$.
EKSPONEN-ANEKA CONTOH SOAL LANJUTAN (BAGIAN 2)
$\begin{array}{l}\\ 06.&\textrm{Jumlah akar-akar persamaan}\\ & 5^{x+1}+5^{2-x}-30=0\: \: \textrm{adalah}\: ....\\\\ &\begin{array}{lllllllll}\\ \textrm{a}.&-2&&&\\ \textrm{b}.&-1\\ \textrm{c}.&0\\ \textrm{d}.&1\\ \textrm{e}.&2 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}5^{x+1}+5^{2-x}-30&=0\\ \left (5^{x} \right ).5^{1}+\displaystyle \frac{5^{2}}{5^{x}}-30&=0\\ 5\left ( 5^{x} \right )^{2}+25-30\left ( 5^{x} \right )&=0\\ \textrm{Persamaan kuadrat}&\: \textrm{dalam}\: \: 5^{x},\: \textrm{maka}\\ 5(5^{x})^{2}-30(5^{x})+25&=0\begin{cases} a & =5 \\ b & =-30 \\ c & =25 \end{cases}\\ (5^{x_{1}}).\left ( 5^{x_{2}} \right )&=\displaystyle \frac{c}{a}\\ 5^{x_{1}+x_{2}}&=\displaystyle \frac{25}{5}=5\\ 5^{x_{1}+x_{2}}&=5^{1}\\ x_{1}+x_{2}&=1 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 07.&\textrm{Jumlah akar-akar persamaan}\\ &2020^{x^{2}-7x+7}=2021^{x^{2}-7x+7}\: \: \textrm{adalah}\: ....\\\\ &\begin{array}{lllllllll}\\ \textrm{a}.&-7\\ \textrm{b}.&-5\\ \textrm{c}.&-3\\ \textrm{d}.&5\\ \textrm{e}.&7 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}2020^{x^{2}-7x+7}&=2021^{x^{2}-7x+7}\\ \textrm{Karena basis}&\: \textrm{tidak sama},\\ \textrm{maka harusl}&\textrm{ah pangkatnya}=0,\\ x^{2}-7x+7&=0\\ \textrm{dan jumlah}\: &\textrm{akar-akarnya adalah}:\\ x_{1}+x_{2}&=-\displaystyle \frac{b}{a}, \: \: \textrm{dari persamaan}\\ x^{2}-7x+7&=0\begin{cases} a &=1 \\ b &=-7 \\ c &=7 \end{cases}\\ \textrm{maka}\: \: x_{1}+x_{2}&=-\displaystyle \frac{b}{a}=-\frac{-7}{1}=7 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 08.&\textrm{Nilai dari}\: \: \displaystyle \frac{2^{2020}+2^{2018}}{2^{2018}+2^{2016}} \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&2\\ \textrm{b}.&5\\ \textrm{c}.&10\\ \textrm{d}.&20\\ \textrm{e}.&40 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\displaystyle \frac{2^{2020}+2^{2018}}{2^{2018}+2^{2016}}&=\displaystyle \frac{2^{4}.2^{2016}+2^{2}.2^{2016}}{2^{2}.2^{2018}+2^{2016}}\\ &=\displaystyle \frac{2^{2016}\left ( 2^{4}+2^{2} \right )}{2^{2016}\left ( 2^{2}+1 \right )}\\ &=\displaystyle \frac{16+4}{4+1}\\ &=\displaystyle \frac{20}{5}\\ &=4 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 09.&(\textbf{UM IPB})\textrm{Jika}\: \: ab=a^{b} \: \: \textrm{dan}\: \: \displaystyle \frac{a}{b}=a^{3b}\\ &\textrm{maka nilai}\: \: a\: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&0\\ \textrm{b}.&0,5\\ \textrm{c}.&1\\ \textrm{d}.&0,25\\ \textrm{e}.&0,75 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\textrm{Diketahui}&\\ ab&=a^{b}\\ b&=\displaystyle \frac{a^{b}}{a}=a^{b-1}.....\textbf{1}\\ \textrm{maka}&\\ \displaystyle \frac{a}{b}&=a^{3b}...............\textbf{2}\\ \textbf{1}&\: \: ke\: \: \textbf{2}\\ \displaystyle \frac{a}{a^{b-1}}&=a^{3b}\\ a^{2-b}&=a^{3b}\\ 2-b&=3b\\ -4b&=-2\\ b&=\displaystyle \frac{1}{2}................\textbf{3}\\ \textbf{3}&\: \: ke\: \: \textbf{1}\\ a\left ( \displaystyle \frac{1}{2} \right )&=a^{\frac{1}{2}}\\ \displaystyle \frac{1}{4}a^{2}&=a\\ a^{2}-4a&=0\\ a(a-4)&=0\\ a=0\: \: &\textrm{atau}\: \: a=4 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 10.&\textrm{Jika}\: \: \displaystyle 3.x^{^{^{^{ \frac{3}{2}}}}}=4\: ,\: \textrm{maka}\: \: x=\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&\displaystyle 1,1\\ \textrm{b}.&\displaystyle 1,2\\ \textrm{c}.&1,3\\ \textrm{d}.&\displaystyle 1,4\\ \textrm{e}.&1,5\\\\ &&&(\textbf{SAT Test Math Level 2})\\ \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\displaystyle 3.x^{^{^{^{ \frac{3}{2}}}}}&=4\\ \left (3.x^{^{^{^{ \frac{3}{2}}}}} \right )^{2}&=4^{2}\\ 3^{2}.x^{3}&=4^{2}\\ x^{3}&=\displaystyle \frac{4^{2}}{3^{2}}\\ x^{3}&=\displaystyle \frac{4^{2}}{3^{2}}\times \frac{3}{3}\\ x^{3}&\leq \displaystyle \frac{4^{2}}{3^{2}}\times \frac{4}{3}\\ x^{3}&\leq \left ( \displaystyle \frac{4^{3}}{3^{3}} \right )\\ x^{3}&\leq \left ( \displaystyle \frac{4}{3} \right )^{3}\\ x&\leq \displaystyle \frac{4}{3}\\ x&\leq 1,\overline{333}\\ x&\approx 1,3 \end{aligned} \end{array}$