Pembahasan Logaritma Simak UI 2019 Matematika Dasar kode 521

Soal yang Akan Dibahas
Jika $ x_1 $ dan $ x_2 $ memenuhi $ {}^4 \log x - {}^x \log 16 = \frac{7}{6} - {}^x \log 8 $ , nilai $ x_1 . x_2 $ adalah ....
A). $ \sqrt[3]{2} \, $ B). $ \sqrt{3} \, $ C). $ 2\sqrt[3]{2} \, $ D). $ 2\sqrt{3} \, $ E). $ 4\sqrt[3]{2} \, $

$\spadesuit $ Konsep Dasar

*). Definisi Logaritma :
$ {}^a \log b = c \rightarrow b = a^c $

*). Sifat Logaritma :
1). $ {}^a \log b - {}^a \log c = {}^a \log \frac{b}{c} $
2). $ {}^{a^m} \log b = \frac{1}{m} \log b $
3). $ {}^a \log b = \frac{1}{{}^b \log a} $

*). Sifat Eksponen :
1). $ a^m . a^n = a^{m+n} $
2). $ a^\frac{1}{n} = \sqrt[n]{a} $

$\clubsuit $ Pembahasan
*).Mengubah persamaan logaritma yang diketahui :
$\begin{align} {}^4 \log x - {}^x \log 16 & = \frac{7}{6} - {}^x \log 8 \\ {}^{2^2} \log x - {}^x \log 16 + {}^x \log 8 & = \frac{7}{6} \\ \frac{1}{2} {}^{2} \log x - ( {}^x \log 16 - {}^x \log 8 ) & = \frac{7}{6} \\ \frac{1}{2} {}^{2} \log x - {}^x \log \frac{16}{8} & = \frac{7}{6} \\ \frac{1}{2} {}^{2} \log x - {}^x \log 2 & = \frac{7}{6} \\ \frac{1}{2} {}^{2} \log x - \frac{1}{{}^2 \log x } & = \frac{7}{6} \end{align} $
*).Misalkan $ {}^2 \log x = p $ , persamaannya menjadi :
$\begin{align} \frac{1}{2} {}^{2} \log x - \frac{1}{{}^2 \log x } & = \frac{7}{6} \\ \frac{1}{2} p - \frac{1}{p} & = \frac{7}{6} \, \, \, \, \, (\times 6p) \\ 3 p^2 - 6 & = 7p \\ 3 p^2 - 7p - 6 & = 0 \\ (3p+2)(p-3) & = 0 \\ p = -\frac{2}{3} \vee p & = 3 \\ p = -\frac{2}{3} \vee p & = 3 \\ \end{align} $
*).Menentukan nilai $ x $ dengan definisi logaritma :
$\begin{align} p = -\frac{2}{3} & \vee p = 3 \\ {}^{2} \log x = -\frac{2}{3} & \vee {}^{2} \log x = 3 \\ x = 2^{-\frac{2}{3}} & \vee x = 2^3 \\ x_1 = 2^{-\frac{2}{3}} & \vee x_2 = 2^3 \end{align} $
*).Menentukan nilai $ x_1. x_2 $ :
$\begin{align} x_1. x_2 & = 2^{-\frac{2}{3}} . 2^3 \\ & = 2^{-\frac{2}{3} + 3} = 2^{-\frac{2}{3} + \frac{9}{3}} = 2^{\frac{7}{3}} \\ & = 2^{2\frac{1}{3}} = 2^{2 + \frac{1}{3}} = 2^2.2^\frac{1}{3} = 4 \sqrt[3]{2} \end{align} $
Jadi, nilai $ x_1. x_2 = 4 \sqrt[3]{2} . \, \heartsuit $