Pembahasan Logaritma UM UGM 2019 Matematika Dasar Kode 633

Soal yang Akan Dibahas
Jika $ {}^{(p^2 + 4)} \log (p+1) = \frac{{}^2\log 5}{{}^3\log \sqrt{5} . {}^2\log 81} $, maka $ 4p^2 = .... $
A). $ \frac{3}{2} \, $ B). $ 3 \, $ C). $ 6 \, $ D). $ 9 \, $ E). $ 12 $

$\spadesuit $ Konsep Dasar :
*). Definisi Logaritma :
$ {}^a \log b = c \rightarrow b = a^c $
*). Sifat-sifat Logaritma :
(i). $ \frac{1}{{}^a \log b} = {}^b \log a $
(ii). $ {}^{a^m} \log b^n = \frac{n}{m} .{}^a \log b $
(iii). $ {}^a \log b . {}^b \log c = {}^a \log c $

$\clubsuit $ Pembahasan
*). Menentukan nilai $ p $ :
$\begin{align} {}^{(p^2 + 4)} \log (p+1) & = \frac{{}^2\log 5}{{}^3\log \sqrt{5} . {}^2\log 81} \\ {}^{(p^2 + 4)} \log (p+1) & = {}^2\log 5. {}^\sqrt{5} \log 3 . {}^{81} \log 2 \\ {}^{(p^2 + 4)} \log (p+1) & = {}^2\log 5. {}^{5^\frac{1}{2}} \log 3 . {}^{3^4} \log 2 \\ {}^{(p^2 + 4)} \log (p+1) & = ({}^2\log 5). \frac{1}{\frac{1}{2}} .( {}^{5} \log 3) . \frac{1}{4} . ({}^{3 } \log 2 ) \\ {}^{(p^2 + 4)} \log (p+1) & = ({}^2\log 5). 2 . ({}^{5} \log 3 ). \frac{1}{4} . ({}^{3 } \log 2 ) \\ {}^{(p^2 + 4)} \log (p+1) & = \frac{2}{4} . ({}^2\log 5). ({}^{5} \log 3 ) . ({}^{3 } \log 2 ) \\ {}^{(p^2 + 4)} \log (p+1) & = \frac{1}{2} . ({}^2\log 2) \\ {}^{(p^2 + 4)} \log (p+1) & = \frac{1}{2} . 1 \\ {}^{(p^2 + 4)} \log (p+1) & = \frac{1}{2} \\ (p+1) & = (p^2 + 4) ^ \frac{1}{2} \, \, \, \, \, \text{(kuadratkan)} \\ (p+1)^2 & = (p^2 + 4) \\ p^2 + 2p + 1 & = p^2 + 4 \\ 2p & = 3 \\ p & = \frac{3}{2} \end{align} $
*). Menentukan nilai $ 4p^2 $ :
$\begin{align} 4p^2 & = 4. (\frac{3}{2})^2 = 4 \times \frac{9}{4} = 9 \end{align} $
Jadi, nilai $ 4p^2 = 9 . \, \heartsuit $

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