Soal yang Akan Dibahas
Jika $ x_1 $ dan $ x_2 $ memenuhi
$ \left( {}^{2} \log \frac{1}{3x-1} \right)^2 = 9 $ , maka nilai
$ x_1 + x_2 $ adalah ...
A). $ \frac{27}{8} \, $ B). $ \frac{25}{8} \, $ C). $ \frac{1}{8} \, $ D). $ -\frac{25}{8} \, $ E). $ -\frac{27}{8} $
A). $ \frac{27}{8} \, $ B). $ \frac{25}{8} \, $ C). $ \frac{1}{8} \, $ D). $ -\frac{25}{8} \, $ E). $ -\frac{27}{8} $
$\spadesuit $ Konsep Dasar
*). Definisi logaritma :
$ {}^a \log b = c \rightarrow b = a^c $
*). Sifat logaritma :
$ {}^{a^m} \log b^n = \frac{n}{m} {}^a \log b $
*). Sifat eksponen : $ (a.b)^n = a^n . b^n $ dan $ a^{-n} = \frac{1}{a^n} $
*). Definisi logaritma :
$ {}^a \log b = c \rightarrow b = a^c $
*). Sifat logaritma :
$ {}^{a^m} \log b^n = \frac{n}{m} {}^a \log b $
*). Sifat eksponen : $ (a.b)^n = a^n . b^n $ dan $ a^{-n} = \frac{1}{a^n} $
$\clubsuit $ Pembahasan
*). Menyelesaikan persamaannya :
$\begin{align} \left( {}^{2} \log \frac{1}{3x-1} \right)^2 & = 9 \\ \left( {}^{2} \log (3x-1)^{-1} \right)^2 & = 9 \\ \left( (-1). {}^{2} \log (3x-1) \right)^2 & = 9 \\ (-1)^2. \left( {}^{2} \log (3x-1) \right)^2 & = 9 \\ \left( {}^{2} \log (3x-1) \right)^2 & = 9 \\ {}^{2} \log (3x-1) & = \pm \sqrt{ 9 } \\ {}^{2} \log (3x-1) & = \pm 3 \\ {}^{2} \log (3x-1) = 3 \vee {}^{2} \log (3x-1) & = - 3 \\ 3x - 1 = 2^3 \vee 3x-1 & = 2^{-3} \\ 3x - 1 = 8 \vee 3x-1 & = \frac{1}{8} \\ 3x = 8 + 1 \vee 3x & = \frac{1}{8} + 1 \\ 3x = 9 \vee 3x & = \frac{9}{8} \\ x = 3 \vee x & = \frac{3}{8} \\ x_1 = 3 \vee x_2 & = \frac{3}{8} \end{align} $
*). Menentukan nilai $ x_1 + x_2 $ :
$\begin{align} x_1 + x_2 & = 3 + \frac{3}{8} = \frac{27}{8} \end{align} $
Jadi, nilai $ x_1 + x_2 = \frac{27}{8} . \, \heartsuit $
*). Menyelesaikan persamaannya :
$\begin{align} \left( {}^{2} \log \frac{1}{3x-1} \right)^2 & = 9 \\ \left( {}^{2} \log (3x-1)^{-1} \right)^2 & = 9 \\ \left( (-1). {}^{2} \log (3x-1) \right)^2 & = 9 \\ (-1)^2. \left( {}^{2} \log (3x-1) \right)^2 & = 9 \\ \left( {}^{2} \log (3x-1) \right)^2 & = 9 \\ {}^{2} \log (3x-1) & = \pm \sqrt{ 9 } \\ {}^{2} \log (3x-1) & = \pm 3 \\ {}^{2} \log (3x-1) = 3 \vee {}^{2} \log (3x-1) & = - 3 \\ 3x - 1 = 2^3 \vee 3x-1 & = 2^{-3} \\ 3x - 1 = 8 \vee 3x-1 & = \frac{1}{8} \\ 3x = 8 + 1 \vee 3x & = \frac{1}{8} + 1 \\ 3x = 9 \vee 3x & = \frac{9}{8} \\ x = 3 \vee x & = \frac{3}{8} \\ x_1 = 3 \vee x_2 & = \frac{3}{8} \end{align} $
*). Menentukan nilai $ x_1 + x_2 $ :
$\begin{align} x_1 + x_2 & = 3 + \frac{3}{8} = \frac{27}{8} \end{align} $
Jadi, nilai $ x_1 + x_2 = \frac{27}{8} . \, \heartsuit $
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