Terkadang beberapa soal pada akhir semester gasal dimunculkan soal yang melibatkan bentuk ekponen dan logaritma sekaligus dalam sebuah persamaan. Bentuk soal yang dihadapi para siswa pada suatu waktu tidak hanya fokus pada satu pokok bahasan saja, terkadang tersaji soal yang menuntut siswa untuk mengkombinasikan konsep-konsep yang telah disampaikan dan diajarkan oleh para guru dan pembimbing. Berawal dari sana, di bagian ini dipaparkan beberapa soal yang yang dimaksudkan dengan harapan siswa lebih terbiasa dalam menghadapi tipe soal yang tersaji demikian.
Soal pertama saya pilihkan ada di blog ini, berikut tautannya klik di sini
$\begin{array}{ll}\\ 1.&\textbf{(UMPTN '94)}\\ &\textrm{Hasil kali akar-akar persamaan}\\ &^{3}\log x^{.\left (^{2+\: ^{3}\log x} \right )}=15\\ & \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle \frac{1}{9}\\ \textrm{b}.&\displaystyle \frac{1}{3}\\ \textrm{c}.&\displaystyle 1\\ \textrm{d}.&\displaystyle 3\\ \textrm{e}.&\displaystyle 9 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&^{3}\log x^{.\left (^{2+\: ^{3}\log x} \right )}=15\\ &\Leftrightarrow \: \left ({2+\: ^{3}\log x} \right )^{3}\log x-15=0\\ &\Leftrightarrow 2\: ^{3}\log x+\: \left ( ^{3}\log x \right )^{2}-15=0\\ &\Leftrightarrow \: \left ( ^{3}\log x \right )^{2}+2\: ^{3}\log x-15=0\\ &\Leftrightarrow \: \left (^{3}\log x_{1}+5 \right )\left ( ^{3}\log x_{2}-3 \right )=0\\ &\Leftrightarrow \: ^{3}\log x_{1}+5=0\: \: \color{purple}\textrm{atau}\: \: \color{blue}^{3}\log x_{1}-3=0\\ &\Leftrightarrow \: ^{3}\log x_{1}=-5\: \: \color{purple}\textrm{atau}\: \: \color{blue}^{3}\log x_{2}=3\\ &\Leftrightarrow \: x_{1}=3^{-5}\: \: \color{purple}\textrm{atau}\: \: \color{blue}x_{2}=3^{3}\\ &\qquad \color{black}\textrm{maka}\\ &\Leftrightarrow \: x_{1}\times x_{2}=3^{-5}\times 3^{3}=3^{-5+3}=3^{-2}\\ &\Leftrightarrow \qquad =\displaystyle \frac{1}{3^{2}}\\ &\Leftrightarrow \qquad =\color{red}\frac{1}{9} \end{aligned} \end{array}$.
Soal kedua juga saya pilihkan ada di blog ini, tautannya klik di sini
$\begin{array}{ll}\\ 2.&\textrm{Persamaan}\\ & 10^{\,^{\, ^{2}}\log x^{2} }-7\left ( 10^{\,^{\, ^{2}}\log x } \right )+10=0\\ & \textrm{mempunyai dua akar yaitu}\: \: x_{1}\: \: \textrm{dan}\: \: x_{2}\\ &\textrm{Nilai}\: \: x_{1}\times x_{2}=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-2\\ \textrm{b}.&-5\\ \color{red}\textrm{c}.&2\\ \textrm{d}.&5\\ \textrm{e}.&10 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&10^{\,^{\, ^{2}}\log x^{2} }-7\left ( 10^{\,^{\, ^{2}}\log x } \right )+10=0\\ &10^{\, 2^{\, ^{2}}\log x }-7\left ( 10^{\,^{\, ^{2}}\log x } \right )+10=0\\ &\color{black}\textrm{adalah persamaan kuadrat dalam}\: \: \color{red}10^{\,^{\, ^{2}}\log x }\\ &\color{black}\textrm{Misalkan}\: \: \color{red}p=10^{\,^{\, ^{2}}\log x },\: \: \color{black}\textrm{maka persamaan}\\ &\color{black}\textrm{menjadi}\: \: \color{purple}p^{2}-7p+10=0\: \begin{cases} a & =1 \\ b & =-7 \\ c & =10 \end{cases}\\ & \color{red}\textrm{Karena nilai}\: \: \color{black}p_{1}\times p_{2}=\displaystyle \frac{c}{a}\: \: \textrm{maka}\\ &10^{\,^{\, ^{2}}\log x_{1} }\times 10^{\,^{\, ^{2}}\log x_{2} }=\displaystyle \frac{10}{1}=10\\ &10^{\,^{\, ^{2}}\log x_{1}\: +\: ^{2}\log x_{2} }=10\\ &10^{\,^{\, ^{2}}\log x_{1}\: +\: ^{2}\log x_{2} }=10^{1}\\ &\Leftrightarrow \: ^{2}\log x_{1}\: +\: ^{2}\log x_{2}=1\\ &\Leftrightarrow \: ^{2}\log x_{1}\times x_{2}=1\\ &\Leftrightarrow \: \color{red}x_{1}\times x_{2}=2^{1}=2\\ \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 3.&\textrm{Hasil kali semua akar real persamaan }\\ &\sqrt{10}\left ( x^{2}-x+4 \right )^{.^{\log \left ( x^{2}-x+4 \right )}}=\left ( x^{2}-x+4 \right )^{\frac{3}{2}}\\ & \textrm{adalah}\: ....\\ &\qquad\qquad\qquad\textrm{UM UGM 2016 Mat IPA}\\ &\begin{array}{llll}\\ \textrm{a}.&-18\\ \color{red}\textrm{b}.&-6\\ \textrm{c}.&1\\ \textrm{d}.&6\\ \textrm{e}.&18 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&\sqrt{10}\left ( x^{2}-x+4 \right )^{.^{\color{black}\log \color{blue}\left ( x^{2}-x+4 \right )}}=\left ( x^{2}-x+4 \right )^{\frac{3}{2}}\\ &\Leftrightarrow \color{black}\log \color{red}\sqrt{10}\left ( x^{2}-x+4 \right )^{.^{\color{black}\log \color{blue}\left ( x^{2}-x+4 \right )}}=\color{black}\log \color{red}\left ( x^{2}-x+4 \right )^{\frac{3}{2}}\\ &\Leftrightarrow \log \sqrt{10}+\log \left ( x^{2}-x+4 \right )^{.^{\color{black}\log \color{blue}\left ( x^{2}-x+4 \right )}}=\displaystyle \frac{3}{2}\log \left ( x^{2}-x+4 \right )\\ &\Leftrightarrow \log 10^{.^{\frac{1}{2}}}+\log \left ( x^{2}-x+4 \right )^{.^{\color{black}\log \color{blue}\left ( x^{2}-x+4 \right )}}=\displaystyle \frac{3}{2}\log \left ( x^{2}-x+4 \right )\\ &\Leftrightarrow \displaystyle \frac{1}{2}+ \log \left ( x^{2}-x+4 \right )\times \log \left ( x^{2}-x+4 \right )=\displaystyle \frac{3}{2}\log \left ( x^{2}-x+4 \right )\\ &\displaystyle \frac{1}{2}+ \log^{2} \left ( x^{2}-x+4 \right )=\displaystyle \frac{3}{2}\log \left ( x^{2}-x+4 \right )\\ &\color{red}\textrm{misalkan}\: \: \color{black}\log \left ( x^{2}-x+4 \right )=p,\: \: \textrm{maka}\\ &\displaystyle \frac{1}{2}+p^{2}=\displaystyle \frac{3}{2}p\\ &\Leftrightarrow 2p^{2}-3p+1=0\\ &\Leftrightarrow \left ( p-1 \right )\left ( 2p-1 \right )=0\\ &\Leftrightarrow p=1\: \: \textrm{atau}\: \: p=\displaystyle \frac{1}{2}\\ &\Leftrightarrow \log \left ( x^{2}-x+4 \right )=1\: \: \textrm{atau}\: \: \log \left ( x^{2}-x+4 \right )=\displaystyle \frac{1}{2}\\ &\Leftrightarrow \left ( x^{2}-x+4 \right )=10^{1}\: \: \textrm{atau}\: \: \left ( x^{2}-x+4 \right )=10^{.^{\frac{1}{2}}}\\ &\Leftrightarrow x^{2}-x+4=10\: \: \textrm{atau}\: \: x^{2}-x+4=\sqrt{10}\\ &\Leftrightarrow x^{2}-x-6=0\: \: \textrm{atau}\: \: x^{2}-x+4-\sqrt{10}=0\\ &\textrm{Jelas bahwa yang ada akar real adalah}\\ &\textrm{persamaan}\: :\: \color{red}x^{2}-x-6=0\\ &\textrm{dan hasil kali semua akarnya adalah}:\: x_{1}.x_{2}.\\ &\Rightarrow x_{1}.x_{2}=\displaystyle \frac{c}{a}=\frac{-6}{1}=\color{red}-6 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 4.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi persamaan}\\ &2x^{.^{^{6}}\log x}+72x^{.^{^{.^{\frac{1}{6}}}}\log x}=24\\ & \textrm{adalah}\: ....\\ &\begin{array}{llll} \textrm{a}.&18\: \: \textrm{dan}\: \: \displaystyle \frac{1}{36}\\ \textrm{b}.&24\: \: \textrm{dan}\: \: \displaystyle 2\\ \color{red}\textrm{c}.&6\: \: \textrm{dan}\: \: \displaystyle \frac{1}{6}\\ \textrm{d}.&36\: \: \textrm{dan}\: \: \displaystyle \frac{1}{6}\\ \textrm{e}.&\displaystyle \frac{1}{6}\: \: \textrm{dan}\: \: \displaystyle \frac{1}{18} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&2x^{.^{^{6}}\log x}+72x^{.^{^{.^{\frac{1}{6}}}}\log x}=24\\ &\Leftrightarrow 2x^{.^{^{6}}\log x}+72x^{-1.\left (^{^{6}}\log x \right )}=24\\ &\Leftrightarrow 2x^{.^{^{6}}\log x}+\displaystyle \frac{72}{x^{.^{^{6}}\log x}}-24=0\\ &\color{red}\textrm{misalkan}\: \: \color{black}p=x^{.^{^{6}}\log x},\: \: \textrm{maka}\\ &2p+\displaystyle \frac{72}{p}-24=0\Leftrightarrow p^{2}-12p+36=0\\ &\Leftrightarrow (p-6)^{2}=0\Leftrightarrow p=\pm 6\\ &\Leftrightarrow x^{.^{^{6}}\log x}=6\: \: \textrm{saja},\\ &\textbf{disebabkan} \: \textrm{syarat numerus}\: \: \color{blue}x^{.^{^{6}}\log x}\color{black}\neq -6.\\ & \textrm{Selanjutnya}\\ &\textrm{untuk}\: \: x^{.^{^{6}}\log x}=6,\: \: \textrm{nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &\textrm{ada} \: \: 2,\: \: \textrm{yaitu} :\: \: 6\: \: \textrm{dan}\: \: \displaystyle \frac{1}{6}\\\\ &\textrm{Karena}\begin{cases} x^{.^{^{6}}\log x} & \Rightarrow 6^{.^{^{6}}\log 6}=6 \\\\ x^{.^{^{6}}\log x} & \Rightarrow \left (\frac{1}{6} \right )^{.^{^{6}}\log \left ( \frac{1}{6} \right )}=6^{-1.^{^{6}}\log 6^{-1}}=6^{.^{^{6}}\log 6^{1}}=6 \end{cases} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 5.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi persamaan}\\ &^{2x-3}\log ^{-1}2.^{x}\log 2-\, ^{x+6}\log ^{-1}x+\displaystyle \frac{1}{^{x+2}\log x}=1\\ & \textrm{adalah}\: ....\\ &\begin{array}{llll} \textrm{a}.&8\\ \textrm{b}.&9\\ \textrm{c}.&-1\\ \color{red}\textrm{d}.&6\\ \textrm{e}.&5 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&^{2x-3}\log ^{-1}2.^{x}\log 2-\, ^{x+6}\log ^{-1}x+\displaystyle \frac{1}{^{x+2}\log x}=1\\ &\Leftrightarrow \: ^{2}\log (2x-3).^{x}\log 2-\, ^{x}\log (x+6)+\, ^{x}\log (x+2)=1\\ &\Leftrightarrow \: ^{x}\log 2.\, ^{2}\log (2x-3)-\, ^{x}\log (x+6)+\, ^{x}\log (x+2)=1\\ &\Leftrightarrow \: ^{x}\log (2x-3)-\, ^{x}\log (x+6)+\, ^{x}\log (x+2)=1\\ &\Leftrightarrow \: ^{x}\log \displaystyle \frac{(2x-3)(x+2)}{(x+6)}=1\\ &\begin{array}{|c|c|}\hline \textrm{Basis}&\textrm{Numerus}\\\hline \begin{aligned}&(1).\quad x>\displaystyle \frac{3}{2}\\ &(2).\quad x>0\\ &(3).\quad x>-2 &\\ &\color{red}\textrm{dipilih}\\ &\: \: \qquad x>\frac{3}{2} \end{aligned}&\begin{aligned}&\quad x>\displaystyle 0\\ &\\ &\\ &\\ & \end{aligned}\\\hline \end{array}\\ &\Leftrightarrow \displaystyle \frac{(2x-3)(x+2)}{(x+6)}=x^{1}\\ &\Leftrightarrow (2x-3)(x+2)=x(x+6)\\ &\Leftrightarrow 2x^{2}+x-6=x^{2}+6x\\ &\Leftrightarrow x^{2}-5x-6=0\\ &\Leftrightarrow (x-6)(x+1)=0\\ &\Leftrightarrow x=6\: \: \textrm{atau}\: \: x=-1\\ &\textrm{Jadi, nilai}\: \: x=\color{red}6 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 6.&\textrm{Himpunan penyelesaian untuk}\: \: x,y,z\\ &\textrm{yang memenuhi persamaan berikut}\\ &\begin{cases} (2x+3y)^{\log (x-y+2z)}=1 & \\ 3^{2x+y+z}\times 27^{3z+2y+x}=81 & \\ 5x+3y+8z=2 & \end{cases}\\ & \textrm{adalah}\: ....\\ &\begin{array}{llll} \textrm{a}.&\left \{ \displaystyle \frac{17}{12},-\frac{1}{12},\frac{7}{6} \right \}\\ \textrm{b}.&\left \{ -\displaystyle \frac{17}{12},\frac{1}{2},\frac{7}{6} \right \}\\ \textrm{c}.&\left \{ -\displaystyle \frac{17}{12},-\frac{1}{2},-\frac{7}{6} \right \}\\ \textrm{d}.&\left \{ \displaystyle \frac{17}{12},\frac{1}{12},\frac{7}{6} \right \}\\ \color{red}\textrm{e}.&\left \{ -\displaystyle \frac{17}{12},-\frac{1}{12},\frac{7}{6} \right \} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\textrm{Diketahui bahwa}\\ &\begin{cases} (2x+3y)^{\log (x-y+2z)}=1 & \\ 3^{2x+y+z}\times 27^{3z+2y+x}=81 & \\ 5x+3y+8z=2 & \end{cases}\\ &\textrm{Perhatikan bahwa soal dapat dituliskan ulang}\\ &\textrm{menjadi}\\ &\begin{cases} \log (x-y+2z)=0 & \\ 3^{2x+y+z+3(3z+2y+x)}=3^{4} & \\ 5x+3y+8z=2 & \end{cases}\\ &\textrm{berubah lagi menjadi}\\ &\begin{cases} x-y+2z=1 &----(1) \\ 5x+7y+10z=4 &----(2) \\ 5x+3y+8z=2 &----(3) \end{cases}\\ &\textrm{Dengan eliminasi berikut akan didapatkan}\\ &\begin{array}{|c|c|}\hline \textbf{Persamaan 1&2}&\textbf{Persamaan 2&3}\\\hline \begin{array}{ll} 5x-5y+10z&=5\\ 5x+7y+10z&=4\quad -\\\hline \: \: \quad-12y&=1\\ \qquad\qquad\qquad y&=-\displaystyle \frac{1}{12}\\ &\\ &---(4) \end{array}&\begin{array}{ll} 5x+7y+10z&=4\\ 5x+3y+8z&=2\quad -\\\hline \: \: \qquad 4y+2z&=2\\ &---(5)\\ &\\ & \end{array} \\\hline \end{array}\\ &\textrm{Dari persamaan}\: 4\&5\: \: \textrm{didapatkan}\\ &\textrm{nilai}\: \: z=\displaystyle \frac{7}{6},\: \textrm{selanjutnya juga}\\ &\textrm{akan didapatkan nilai}\: \: x=-\displaystyle \frac{17}{12}.\\ &\textrm{Jadi, HP}=\color{red}\left \{ -\displaystyle \frac{17}{12},-\frac{1}{12},\frac{7}{6} \right \} \end{aligned} \end{array}$.
DAFTAR PUSTAKA
- Kanginan, M., Nurdiansyah, H., Akhmad, G. 2016. Matematika untuk SMA/MA Kelas XKelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: YRAMA WIDYA.
- Susianto, B. 2011. Olimpiade Matematika dengan Proses Berpikir Aljabar dan Bilangan. Cet. II. Jakarta: GRASINDO
Tidak ada komentar:
Posting Komentar
Informasi