Soal yang Akan Dibahas
Jika $ x_1 $ dan $ x_2 $ memenuhi
$ \left( {}^{27} \log \frac{1}{x+1} \right)^2 = \frac{1}{9} $ , maka nilai
$ x_1 x_2 $ adalah ...
A). $ \frac{5}{3} \, $ B). $ \frac{4}{3} \, $ C). $ \frac{1}{3} \, $ D). $ -\frac{2}{3} \, $ E). $ -\frac{4}{3} $
A). $ \frac{5}{3} \, $ B). $ \frac{4}{3} \, $ C). $ \frac{1}{3} \, $ D). $ -\frac{2}{3} \, $ E). $ -\frac{4}{3} $
$\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( {}^{27} \log \frac{1}{x+1} \right)^2 & = \frac{1}{9} \\ \left( {}^{3^3} \log (x+1)^{-1} \right)^2 & = \frac{1}{9} \\ \left( \frac{-1}{3} \, {}^{3} \log (x+1) \right)^2 & = \frac{1}{9} \\ \left( \frac{-1}{3} \right)^2 . \left( {}^{3} \log (x+1) \right)^2 & = \frac{1}{9} \\ \frac{1}{9} . \left( {}^{3} \log (x+1) \right)^2 & = \frac{1}{9} \, \, \, \, \, \text{(kali 9)} \\ \left( {}^{3} \log (x+1) \right)^2 & = 1 \\ {}^{3} \log (x+1) & = \pm \sqrt{ 1} \\ {}^{3} \log (x+1) & = \pm 1 \\ {}^{3} \log (x+1) = 1 & \vee {}^{3} \log (x+1) = - 1 \\ (x+1) = 3^1 & \vee (x+1) = 3^{-1} \\ (x+1) = 3 & \vee (x+1) = \frac{1}{3} \\ x = 2 & \vee x = -\frac{2}{3} \\ x_1 = 2 & \vee x_2 = -\frac{2}{3} \end{align} $
*). Menentukan nilai $ x_1x_2 $ :
$\begin{align} x_1x_2 & = 2 . \left( -\frac{2}{3} \right) = -\frac{4}{3} \end{align} $
Jadi, nilai $ x_1x_2 = -\frac{4}{3} . \, \heartsuit $
*). Menyelesaikan persamaannya :
$\begin{align} \left( {}^{27} \log \frac{1}{x+1} \right)^2 & = \frac{1}{9} \\ \left( {}^{3^3} \log (x+1)^{-1} \right)^2 & = \frac{1}{9} \\ \left( \frac{-1}{3} \, {}^{3} \log (x+1) \right)^2 & = \frac{1}{9} \\ \left( \frac{-1}{3} \right)^2 . \left( {}^{3} \log (x+1) \right)^2 & = \frac{1}{9} \\ \frac{1}{9} . \left( {}^{3} \log (x+1) \right)^2 & = \frac{1}{9} \, \, \, \, \, \text{(kali 9)} \\ \left( {}^{3} \log (x+1) \right)^2 & = 1 \\ {}^{3} \log (x+1) & = \pm \sqrt{ 1} \\ {}^{3} \log (x+1) & = \pm 1 \\ {}^{3} \log (x+1) = 1 & \vee {}^{3} \log (x+1) = - 1 \\ (x+1) = 3^1 & \vee (x+1) = 3^{-1} \\ (x+1) = 3 & \vee (x+1) = \frac{1}{3} \\ x = 2 & \vee x = -\frac{2}{3} \\ x_1 = 2 & \vee x_2 = -\frac{2}{3} \end{align} $
*). Menentukan nilai $ x_1x_2 $ :
$\begin{align} x_1x_2 & = 2 . \left( -\frac{2}{3} \right) = -\frac{4}{3} \end{align} $
Jadi, nilai $ x_1x_2 = -\frac{4}{3} . \, \heartsuit $
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