Question

evaluate:
$log_{8} \frac{1}{4}$
answer

Explanation:

Step1: Recall log properties

Use the property $\log_a \frac{1}{x}=-\log_a x$ and change - of - base formula $\log_a b=\frac{\ln b}{\ln a}$ (or $\frac{\log_c b}{\log_c a}$ for any $c>0,c
eq1$). First, rewrite $\log_{8}\frac{1}{4}$ as $-\log_{8}4$.

Step2: Express in terms of common base

Note that $8 = 2^3$ and $4=2^2$. So $\log_{2^3}2^2$. Use the property $\log_{a^m}b^n=\frac{n}{m}\log_a b$. Here $a = 2$, $m = 3$, $n = 2$, and $\log_2 2 = 1$. So $\log_{2^3}2^2=\frac{2}{3}\log_2 2=\frac{2}{3}\times1=\frac{2}{3}$? Wait, no, wait. Wait, we had $-\log_{8}4$. Let's do it again. $\log_{8}4=\frac{\ln 4}{\ln 8}=\frac{\ln 2^2}{\ln 2^3}=\frac{2\ln 2}{3\ln 2}=\frac{2}{3}$. So $-\log_{8}4=-\frac{2}{3}$.
Wait, let's check with the change - of - base formula properly. $\log_{8}\frac{1}{4}=\frac{\ln\frac{1}{4}}{\ln 8}=\frac{\ln 4^{-1}}{\ln 8}=\frac{-\ln 4}{\ln 8}=\frac{-\ln 2^2}{\ln 2^3}=\frac{- 2\ln 2}{3\ln 2}=-\frac{2}{3}$. The $\ln 2$ terms cancel out (since $\ln 2
eq0$).

Answer:

$-\frac{2}{3}$