Trik Menyelesaikan Soal Persamaan yang Melibatkan Bentuk Gabungan Eksponen dan Logaritma

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

$\begin{array}{l}\\ 1.& \mathbf{\text{(UMPTN '94)}}\\ & \text{Hasil kali akar-akar persamaan}\\ & {}^3\log x^{.\left({}^{2+{}^3\log x}\right)}=15\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{9}}\\ \text{b}.& {\displaystyle \frac{1}{3}}\\ \text{c}.& {\displaystyle 1}\\ \text{d}.& {\displaystyle 3}\\ \text{e}.& {\displaystyle 9}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}& {}^3\log x^{.\left({}^{2+{}^3\log x}\right)}=15\\ & \iff {\left(2+{}^3\log x\right)}^3\log x-15=0\\ & \iff 2{}^3\log x+{({}^3\log x)}^2-15=0\\ & \iff {({}^3\log x)}^2+2{}^3\log x-15=0\\ & \iff ({}^3\log x_1+5)({}^3\log x_2-3)=0\\ & \iff {}^3\log x_1+5=0\text{atau}{}^3\log x_1-3=0\\ & \iff {}^3\log x_1=-5\text{atau}{}^3\log x_2=3\\ & \iff x_1=3^{-5}\text{atau}{x}_2=3^3\\ & \text{maka}\\ & \iff x_1\times x_2=3^{-5}\times 3^3=3^{-5+3}=3^{-2}\\ & \iff ={\displaystyle \frac{1}{3^2}}\\ & \iff =\frac{1}{9}\end{array}\end{array}$.

Soal kedua juga saya pilihkan ada di blog ini, tautannya klik di sini

$\begin{array}{l}\\ 2.& \text{Persamaan}\\ & {10}^{{}^{{}^2}\log x^2}-7\left({10}^{{}^{{}^2}\log x}\right)+10=0\\ & \text{mempunyai dua akar yaitu}{x}_1\text{dan}{x}_2\\ & \text{Nilai}{x}_1\times x_2=....\\ & \begin{array}{l}\\ \text{a}.& -2\\ \text{b}.& -5\\ \text{c}.& 2\\ \text{d}.& 5\\ \text{e}.& 10\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}& {10}^{{}^{{}^2}\log x^2}-7\left({10}^{{}^{{}^2}\log x}\right)+10=0\\ & {10}^{2^{{}^2}\log x}-7\left({10}^{{}^{{}^2}\log x}\right)+10=0\\ & \text{adalah persamaan kuadrat dalam}{10}^{{}^{{}^2}\log x}\\ & \text{Misalkan}p={10}^{{}^{{}^2}\log x},\text{maka persamaan}\\ & \text{menjadi}{p}^2-7p+10=0\{\begin{array}{ll}a& =1\\ b& =-7\\ c& =10\end{array}\\ & \text{Karena nilai}{p}_1\times p_2={\displaystyle \frac{c}{a}\text{maka}}\\ & {10}^{{}^{{}^2}\log x_1}\times {10}^{{}^{{}^2}\log x_2}={\displaystyle \frac{10}{1}=10}\\ & {10}^{{}^{{}^2}\log x_1+{}^2\log x_2}=10\\ & {10}^{{}^{{}^2}\log x_1+{}^2\log x_2}={10}^1\\ & \iff {}^2\log x_1+{}^2\log x_2=1\\ & \iff {}^2\log x_1\times x_2=1\\ & \iff x_1\times x_2=2^1=2\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Hasil kali semua akar real persamaan }\\ & \sqrt{10}{(x^2-x+4)}^{{.}^{\log (x^2-x+4)}}={(x^2-x+4)}^{\frac{3}{2}}\\ & \text{adalah}....\\ & \text{UM UGM 2016 Mat IPA}\\ & \begin{array}{l}\\ \text{a}.& -18\\ \text{b}.& -6\\ \text{c}.& 1\\ \text{d}.& 6\\ \text{e}.& 18\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \sqrt{10}{(x^2-x+4)}^{{.}^{\log (x^2-x+4)}}={(x^2-x+4)}^{\frac{3}{2}}\\ & \iff \log \sqrt{10}{(x^2-x+4)}^{{.}^{\log (x^2-x+4)}}=\log {(x^2-x+4)}^{\frac{3}{2}}\\ & \iff \log \sqrt{10}+\log {(x^2-x+4)}^{{.}^{\log (x^2-x+4)}}={\displaystyle \frac{3}{2}\log (x^2-x+4)}\\ & \iff \log {10}^{{.}^{\frac{1}{2}}}+\log {(x^2-x+4)}^{{.}^{\log (x^2-x+4)}}={\displaystyle \frac{3}{2}\log (x^2-x+4)}\\ & \iff {\displaystyle \frac{1}{2}+\log (x^2-x+4)\times \log (x^2-x+4)={\displaystyle \frac{3}{2}\log (x^2-x+4)}}\\ & {\displaystyle \frac{1}{2}+{\log }^2(x^2-x+4)={\displaystyle \frac{3}{2}\log (x^2-x+4)}}\\ & \text{misalkan}\log (x^2-x+4)=p,\text{maka}\\ & {\displaystyle \frac{1}{2}+p^2={\displaystyle \frac{3}{2}p}}\\ & \iff 2p^2-3p+1=0\\ & \iff (p-1)(2p-1)=0\\ & \iff p=1\text{atau}p={\displaystyle \frac{1}{2}}\\ & \iff \log (x^2-x+4)=1\text{atau}\log (x^2-x+4)={\displaystyle \frac{1}{2}}\\ & \iff (x^2-x+4)={10}^1\text{atau}(x^2-x+4)={10}^{{.}^{\frac{1}{2}}}\\ & \iff x^2-x+4=10\text{atau}{x}^2-x+4=\sqrt{10}\\ & \iff x^2-x-6=0\text{atau}{x}^2-x+4-\sqrt{10}=0\\ & \text{Jelas bahwa yang ada akar real adalah}\\ & \text{persamaan}:x^2-x-6=0\\ & \text{dan hasil kali semua akarnya adalah}:x_1.x_2.\\ & \Rightarrow x_1.x_2={\displaystyle \frac{c}{a}=\frac{-6}{1}=-6}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Nilai}x\text{yang memenuhi persamaan}\\ & 2x^{{.}^{{}^6}\log x}+72x^{{.}^{{}^{{.}^{\frac{1}{6}}}}\log x}=24\\ & \text{adalah}....\\ & \begin{array}{ll}\text{a}.& 18\text{dan}{\displaystyle \frac{1}{36}}\\ \text{b}.& 24\text{dan}{\displaystyle 2}\\ \text{c}.& 6\text{dan}{\displaystyle \frac{1}{6}}\\ \text{d}.& 36\text{dan}{\displaystyle \frac{1}{6}}\\ \text{e}.& {\displaystyle \frac{1}{6}\text{dan}{\displaystyle \frac{1}{18}}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}& 2x^{{.}^{{}^6}\log x}+72x^{{.}^{{}^{{.}^{\frac{1}{6}}}}\log x}=24\\ & \iff 2x^{{.}^{{}^6}\log x}+72x^{-1.({}^{{}^6}\log x)}=24\\ & \iff 2x^{{.}^{{}^6}\log x}+{\displaystyle \frac{72}{x^{{.}^{{}^6}\log x}}-24=0}\\ & \text{misalkan}p=x^{{.}^{{}^6}\log x},\text{maka}\\ & 2p+{\displaystyle \frac{72}{p}-24=0\iff p^2-12p+36=0}\\ & \iff (p-6{)}^2=0\iff p=\pm 6\\ & \iff x^{{.}^{{}^6}\log x}=6\text{saja},\\ & \mathbf{\text{disebabkan}}\text{syarat numerus}{x}^{{.}^{{}^6}\log x}\ne -6.\\ & \text{Selanjutnya}\\ & \text{untuk}{x}^{{.}^{{}^6}\log x}=6,\text{nilai}x\text{yang memenuhi}\\ & \text{ada}2,\text{yaitu}:6\text{dan}{\displaystyle \frac{1}{6}}\\ \\ & \text{Karena}\{\begin{array}{ll}{x}^{{.}^{{}^6}\log x}& \Rightarrow 6^{{.}^{{}^6}\log 6}=6\\ \\ x^{{.}^{{}^6}\log x}& \Rightarrow {\left(\frac{1}{6}\right)}^{{.}^{{}^6}\log \left(\frac{1}{6}\right)}=6^{-{1.}^{{}^6}\log 6^{-1}}=6^{{.}^{{}^6}\log 6^1}=6\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Nilai}x\text{yang memenuhi persamaan}\\ & {}^{2x-3}{\log }^{-1}{2.}^x\log 2-{}^{x+6}{\log }^{-1}x+{\displaystyle \frac{1}{{}^{x+2}\log x}=1}\\ & \text{adalah}....\\ & \begin{array}{ll}\text{a}.& 8\\ \text{b}.& 9\\ \text{c}.& -1\\ \text{d}.& 6\\ \text{e}.& 5\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{rl}& {}^{2x-3}{\log }^{-1}{2.}^x\log 2-{}^{x+6}{\log }^{-1}x+{\displaystyle \frac{1}{{}^{x+2}\log x}=1}\\ & \iff {}^2\log (2x-3){.}^x\log 2-{}^x\log (x+6)+{}^x\log (x+2)=1\\ & \iff {}^x\log 2.{}^2\log (2x-3)-{}^x\log (x+6)+{}^x\log (x+2)=1\\ & \iff {}^x\log (2x-3)-{}^x\log (x+6)+{}^x\log (x+2)=1\\ & \iff {}^x\log {\displaystyle \frac{(2x-3)(x+2)}{(x+6)}=1}\\ & \begin{array}{|cc|}\hline \text{Basis}& \text{Numerus}\\ \begin{array}{rl}& (1).x>{\displaystyle \frac{3}{2}}\\ & (2).x>0\\ & (3).x>-2& \\ & \text{dipilih}\\ & x>\frac{3}{2}\end{array}& \begin{array}{rl}& x>{\displaystyle 0}\\ & \\ & \\ & \\ & \end{array}\\ \hline\end{array}\\ & \iff {\displaystyle \frac{(2x-3)(x+2)}{(x+6)}=x^1}\\ & \iff (2x-3)(x+2)=x(x+6)\\ & \iff 2x^2+x-6=x^2+6x\\ & \iff x^2-5x-6=0\\ & \iff (x-6)(x+1)=0\\ & \iff x=6\text{atau}x=-1\\ & \text{Jadi, nilai}x=6\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Himpunan penyelesaian untuk}x,y,z\\ & \text{yang memenuhi persamaan berikut}\\ & \{\begin{array}{ll}(2x+3y{)}^{\log (x-y+2z)}=1& \\ 3^{2x+y+z}\times {27}^{3z+2y+x}=81& \\ 5x+3y+8z=2& \end{array}\\ & \text{adalah}....\\ & \begin{array}{ll}\text{a}.& \left\{{\displaystyle \frac{17}{12},-\frac{1}{12},\frac{7}{6}}\right\}\\ \text{b}.& \{-{\displaystyle \frac{17}{12},\frac{1}{2},\frac{7}{6}}\}\\ \text{c}.& \{-{\displaystyle \frac{17}{12},-\frac{1}{2},-\frac{7}{6}}\}\\ \text{d}.& \left\{{\displaystyle \frac{17}{12},\frac{1}{12},\frac{7}{6}}\right\}\\ \text{e}.& \{-{\displaystyle \frac{17}{12},-\frac{1}{12},\frac{7}{6}}\}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \text{Diketahui bahwa}\\ & \{\begin{array}{ll}(2x+3y{)}^{\log (x-y+2z)}=1& \\ 3^{2x+y+z}\times {27}^{3z+2y+x}=81& \\ 5x+3y+8z=2& \end{array}\\ & \text{Perhatikan bahwa soal dapat dituliskan ulang}\\ & \text{menjadi}\\ & \{\begin{array}{ll}\log (x-y+2z)=0& \\ 3^{2x+y+z+3(3z+2y+x)}=3^4& \\ 5x+3y+8z=2& \end{array}\\ & \text{berubah lagi menjadi}\\ & \{\begin{array}{ll}x-y+2z=1& ----(1)\\ 5x+7y+10z=4& ----(2)\\ 5x+3y+8z=2& ----(3)\end{array}\\ & \text{Dengan eliminasi berikut akan didapatkan}\\ & \begin{array}{|cc|}\hline \mathbf{\text{Persamaan 1\&2}}& \mathbf{\text{Persamaan 2\&3}}\\ \begin{array}{ll}5x-5y+10z& =5\\ 5x+7y+10z& =4-\\ -12y& =1\\ y& =-{\displaystyle \frac{1}{12}}\\ & \\ & ---(4)\end{array}& \begin{array}{ll}5x+7y+10z& =4\\ 5x+3y+8z& =2-\\ 4y+2z& =2\\ & ---(5)\\ & \\ & \end{array}\\ \hline\end{array}\\ & \text{Dari persamaan}4\mathrm{\&}5\text{didapatkan}\\ & \text{nilai}z={\displaystyle \frac{7}{6},\text{selanjutnya juga}}\\ & \text{akan didapatkan nilai}x=-{\displaystyle \frac{17}{12}.}\\ & \text{Jadi, HP}=\{-{\displaystyle \frac{17}{12},-\frac{1}{12},\frac{7}{6}}\}\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Kanginan, M., Nurdiansyah, H., Akhmad, G. 2016. Matematika untuk SMA/MA Kelas XKelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: YRAMA WIDYA.
2. Susianto, B. 2011. Olimpiade Matematika dengan Proses Berpikir Aljabar dan Bilangan. Cet. II. Jakarta: GRASINDO