$[O{H}^{-}]={10}^{-\text{p}\text{O}\text{H}}$

In order to solve for pOH you must clear the exponent.

Step $1$

Take the logarithm of both sides of the equation.

$log[O{H}^{-}]=log({10}^{-\text{p}\text{O}\text{H}})$

Step $2$

Simplify the right side of this equation using the rules.

$log{a}^n=nloga,$ and $log10=1.$

$log({10}^{-\text{p}\text{O}\text{H}})=-(pOH)log10=-$pOH

The resulting equation is:

pOH$=-log[O{H}^{-}]$

$($b$)$ If $[O{H}^{-}]=1\mathrm{.}3{10}^{-8}$ then

pOH$=$