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

Aug 7, 2016

$- \frac{1}{5}$

#### Explanation:

route 1 use the definition of log
let ${\log}_{32} \left(\frac{1}{2}\right) = y$

$\implies {32}^{y} = \frac{1}{2}$

and $32 = {2}^{5}$

$\implies {\left({2}^{5}\right)}^{y} = \frac{1}{2}$

$\implies {2}^{5 y} = \frac{1}{2}$

invert the LHS
$\implies \frac{1}{{2}^{- 5 y}} = \frac{1}{2} ^ 1$

so $- 5 y = 1 , y = - \frac{1}{5}$

route 2 flip the base

${\log}_{32} \left(\frac{1}{2}\right)$

$= \frac{{\log}_{x} \left(\frac{1}{2}\right)}{{\log}_{x} \left(32\right)}$
$= \frac{- {\log}_{x} \left(2\right)}{{\log}_{x} \left(32\right)}$

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

$= - \frac{{\log}_{2} \left(2\right)}{{\log}_{2} \left(32\right)}$

$= - \frac{{\log}_{2} \left(2\right)}{{\log}_{2} \left({2}^{5}\right)}$

$= - \frac{{\log}_{2} \left(2\right)}{5 {\log}_{2} \left(2\right)} = - \frac{1}{5}$