Question 290793
Given that 'p' equals log16 to the base 'q' (p=logq 16),
express in terms of p:
.
{{{p=log(q,16)}}}
.
{{{q^p=q^log(q,16)}}}
.
{{{q^p=16}}}
.
{{{q^p=16}}}
.
{{{q=16^(1/p)}}}
That is, 
q equals 16 raised to the power of 1/p
or
q equals the pth root of 16.