Pembahasan Logaritma UM UGM 2019 Matematika Ipa Kode 624

Soal yang Akan Dibahas
Jika $ {}^{a^2} \log (3^a - 8)^{-4} . {}^3 \log \sqrt{a} = a - 2 $ , maka $ {}^a \log \left( \frac{1}{8} \right) = .... $
A). $ 0 \, $ B). $ -1 \, $ C). $ -2 \, $ D). $ -3 \, $ E). $ -4 \, $

$\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 $
$ {}^{a } \log b^n = n \, {}^a \log b $
$ {}^a \log b . {}^b \log c = {}^a \log c $
*). Sifat eksponen :
$ a^{m-n} = \frac{a^m}{a^n} $
*). Persamaan eksponen :
$ a^{f(x)} = a^{g(x)} \rightarrow f(x) = g(x) $

$\clubsuit $ Pembahasan
*). Misalkan $ 3^a = p > 0 $
*). Mengubah persamaannya :
$\begin{align} {}^{a^2} \log (3^a - 8)^{-4} . {}^3 \log \sqrt{a} & = a - 2 \\ {}^3 \log \sqrt{a} . {}^{a^2} \log (3^a - 8)^{-4} & = a - 2 \\ {}^3 \log a^\frac{1}{2} . {}^{a^2} \log (3^a - 8)^{-4} & = a - 2 \\ \frac{1}{2} \, {}^3 \log a . \frac{-4}{2} \, {}^{a} \log (3^a - 8) & = a - 2 \\ \frac{1}{2} . \frac{-4}{2} \, {}^3 \log a . {}^{a} \log (3^a - 8) & = a - 2 \\ (-1) \, {}^3 \log (3^a - 8) & = a - 2 \\ {}^3 \log (3^a - 8) & = -(a - 2) \\ {}^3 \log (3^a - 8) & = 2 - a \\ 3^a - 8 & = 3^{2 - a} \\ 3^a - 8 & = \frac{3^2}{3^a} \\ p - 8 & = \frac{9}{p} \\ p^2 - 8p & = 9 \\ p^2 - 8p - 9 & = 0 \\ (p +1)(p-9) & = 0 \\ p = -1 \vee p & = 9 \end{align} $
Yang memenuhi $ p = 9 $
*). Menentukan nilai $ a $ :
$\begin{align} p & = 9 \\ 3^a & = 3^2 \\ a & = 2 \end{align} $
*). Menentukan nilai $ {}^a \log \left( \frac{1}{8} \right) $ :
$\begin{align} {}^a \log \left( \frac{1}{8} \right) & = {}^2 \log \left( \frac{1}{2^3} \right) \\ & = {}^2 \log \left( 2^{-3} \right) \\ & = (-3) \, {}^2 \log 2 \\ & = -3 . 1 = -3 \end{align} $
Jadi, nilai $ {}^a \log \left( \frac{1}{8} \right) = -3 . \, \heartsuit $

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