How do you evaluate $log_32 (1/2)$ ?

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Answer 1 · 2016-08-07T13:55:33

$- 1/5$

route 1 use the definition of log
let $log_(32) (1/2) = y$

$implies 32^y = 1/2$

and $32 = 2^5$

$implies (2^5)^y = 1/2$

$implies 2^(5y) = 1/2$

invert the LHS
$implies 1/(2^(-5y)) = 1/2^1$

so $-5y = 1, y = - 1/5$

route 2 flip the base

$log_(32) (1/2)$

$= (log_(x) (1/2))/(log_(x) (32))$
$= (-log_(x) (2))/(log_(x) (32))$

we can choose x to be what we want and here it seems that 2 must be a good candidate

$= -(log_(2) (2))/(log_(2) (32))$

$=- (log_(2) (2))/(log_(2) (2^5))$

$= -(log_(2) (2))/(5 log_(2) (2)) = - 1/5$

How do you evaluate log_32 (1/2)log_32 (1/2)?