Soal yang Akan Dibahas
Jika $ a $ dan $ b $ memenuhi sistem persamaan :
$\left\{ \begin{array}{c} \frac{3}{\log a} + \frac{4}{\log b} = 7 \\ -\frac{1}{\log a} + \frac{2}{\log b} = 11 \end{array} \right. $
maka $ {}^a \log \frac{1}{b} + {}^b \log \frac{1}{a} = ... $
A). $ \frac{1}{6} \, $ B). $ \frac{7}{12} \, $ C). $ 1\frac{1}{6} \, $ D). $ 2\frac{1}{12} \, $ E). $ 2\frac{1}{4} $
$\left\{ \begin{array}{c} \frac{3}{\log a} + \frac{4}{\log b} = 7 \\ -\frac{1}{\log a} + \frac{2}{\log b} = 11 \end{array} \right. $
maka $ {}^a \log \frac{1}{b} + {}^b \log \frac{1}{a} = ... $
A). $ \frac{1}{6} \, $ B). $ \frac{7}{12} \, $ C). $ 1\frac{1}{6} \, $ D). $ 2\frac{1}{12} \, $ E). $ 2\frac{1}{4} $
$\spadesuit $ Konsep Dasar :
*). Untuk menyelesaikan sistem persamaan bisa menggunakan metode eliminasi dan substitusi
*). Definisi logaritma :
$ {}^a \log b = c \rightarrow b = a^c $
*). Sifat-sifat logaritma :
$ {}^{a^m} \log b^n = \frac{n}{m} . {}^a \log b $
$ {}^a \log a = 1 $
*). Sifat eksponen : $ (a^n)^m = a^{n.m} $
*). Untuk menyelesaikan sistem persamaan bisa menggunakan metode eliminasi dan substitusi
*). Definisi logaritma :
$ {}^a \log b = c \rightarrow b = a^c $
*). Sifat-sifat logaritma :
$ {}^{a^m} \log b^n = \frac{n}{m} . {}^a \log b $
$ {}^a \log a = 1 $
*). Sifat eksponen : $ (a^n)^m = a^{n.m} $
$\clubsuit $ Pembahasan
*). Eliminasi sistem persamaannya :
$\begin{array}{c|c|cc} \frac{3}{\log a} + \frac{4}{\log b} = 7 & \times 1 & \frac{3}{\log a} + \frac{4}{\log b} = 7 & \\ -\frac{1}{\log a} + \frac{2}{\log b} = 11 & \times 3 & -\frac{3}{\log a} + \frac{6}{\log b} = 33 & + \\ \hline & & \frac{10}{\log b} = 40 & \\ & & \log b = \frac{1}{4} & \\ & & b = 10^\frac{1}{4} & \end{array} $
*). Substitusi $ \log b = \frac{1}{4} $ ke pers(i) :
$\begin{align} \frac{3}{\log a} + \frac{4}{\log b} & = 7 \\ \frac{3}{\log a} + \frac{4}{\frac{1}{4}} & = 7 \\ \frac{3}{\log a} + 16 & = 7 \\ \frac{3}{\log a} & = -9 \\ \log a & = -\frac{1}{3} \\ a & = 10^{-\frac{1}{3}} \end{align} $
*). Menentukan nilai $ {}^a \log \frac{1}{b} + {}^b \log \frac{1}{a}$ :
$\begin{align} & {}^a \log \frac{1}{b} + {}^b \log \frac{1}{a} \\ & = {}^a \log b^{-1} + {}^b \log a^{-1} \\ & = {}^{10^{-\frac{1}{3}}} \log (10^\frac{1}{4})^{-1} + {}^{10^\frac{1}{4}} \log (10^{-\frac{1}{3}})^{-1} \\ & = {}^{10^{-\frac{1}{3}}} \log 10^{-\frac{1}{4}} + {}^{10^\frac{1}{4}} \log 10^{\frac{1}{3}} \\ & = \frac{-\frac{1}{4}}{ -\frac{1}{3} } . {}^{10} \log 10 + \frac{\frac{1}{3}}{\frac{1}{4}} . {}^{10} \log 10 \\ & = \frac{3}{4} . 1 + \frac{4}{3} . 1 \\ & = \frac{3}{4} + \frac{4}{3} \\ & = \frac{9}{12} + \frac{16}{12} = \frac{25}{12} = 2 \frac{1}{12} \end{align} $
Jadi, nilai $ {}^a \log \frac{1}{b} + {}^b \log \frac{1}{a} = 2\frac{1}{12} . \, \heartsuit $
*). Eliminasi sistem persamaannya :
$\begin{array}{c|c|cc} \frac{3}{\log a} + \frac{4}{\log b} = 7 & \times 1 & \frac{3}{\log a} + \frac{4}{\log b} = 7 & \\ -\frac{1}{\log a} + \frac{2}{\log b} = 11 & \times 3 & -\frac{3}{\log a} + \frac{6}{\log b} = 33 & + \\ \hline & & \frac{10}{\log b} = 40 & \\ & & \log b = \frac{1}{4} & \\ & & b = 10^\frac{1}{4} & \end{array} $
*). Substitusi $ \log b = \frac{1}{4} $ ke pers(i) :
$\begin{align} \frac{3}{\log a} + \frac{4}{\log b} & = 7 \\ \frac{3}{\log a} + \frac{4}{\frac{1}{4}} & = 7 \\ \frac{3}{\log a} + 16 & = 7 \\ \frac{3}{\log a} & = -9 \\ \log a & = -\frac{1}{3} \\ a & = 10^{-\frac{1}{3}} \end{align} $
*). Menentukan nilai $ {}^a \log \frac{1}{b} + {}^b \log \frac{1}{a}$ :
$\begin{align} & {}^a \log \frac{1}{b} + {}^b \log \frac{1}{a} \\ & = {}^a \log b^{-1} + {}^b \log a^{-1} \\ & = {}^{10^{-\frac{1}{3}}} \log (10^\frac{1}{4})^{-1} + {}^{10^\frac{1}{4}} \log (10^{-\frac{1}{3}})^{-1} \\ & = {}^{10^{-\frac{1}{3}}} \log 10^{-\frac{1}{4}} + {}^{10^\frac{1}{4}} \log 10^{\frac{1}{3}} \\ & = \frac{-\frac{1}{4}}{ -\frac{1}{3} } . {}^{10} \log 10 + \frac{\frac{1}{3}}{\frac{1}{4}} . {}^{10} \log 10 \\ & = \frac{3}{4} . 1 + \frac{4}{3} . 1 \\ & = \frac{3}{4} + \frac{4}{3} \\ & = \frac{9}{12} + \frac{16}{12} = \frac{25}{12} = 2 \frac{1}{12} \end{align} $
Jadi, nilai $ {}^a \log \frac{1}{b} + {}^b \log \frac{1}{a} = 2\frac{1}{12} . \, \heartsuit $
Tidak ada komentar:
Posting Komentar
Catatan: Hanya anggota dari blog ini yang dapat mengirim komentar.