Question 240520
{{{log(x, (16)) = -2}}}
All of this...
I have {{{x^-2=16}}}

{{{x^-2^-1=16^-1}}}

{{{x^2=1/16}}}

{{{x = sqrt(1/16)}}} (Rejecting -sqrt(1/16) because x is the base of a logarithm and we don't have logarithms with negative bases.)

... is great. But you should simplify your answer:

{{{x = sqrt(1/16) = sqrt(1)/sqrt(16) = 1/4}}}