Question 973207
if you have

{{{12^(2x)-8=15}}} then

{{{12^(2x)=15+8}}}....use log base ten

{{{log(12^(2x))=log(23)}}}

{{{(2x)log(12)=log(23)}}}

{{{2x=log(23)/log(12)}}}

{{{x=(log(23)/log(12))/2}}}

{{{x=(log(23)/2log(12))}}}

{{{x=(log(23)/2log(2^2*3))}}}

{{{x=x = log(23)/(2(2log(2)+log(3)))}}}

{{{x=0.6309}}} => approximately


but, if you have

{{{12^(2x-8)=15}}} then

{{{log(12^(2x-8))=log(15)}}}

{{{(2x-8)log(12)=log(15)}}}

{{{2x-8=log(15)/log(12)}}}

{{{2x-8=log(3*5)/log(3*2^2)}}}

{{{2x-8=(log(3)+log(5))/(log(3)+2log2))}}}

{{{2x=(log(3)+log(5))/(log(3)+2log2))+8}}}

{{{x=(log(3)+log(5))/(log(3)+2log2))/2+8/2}}}

{{{x=(log(3)+log(5))/(2(log(3)+2log2)))+4}}}

approximately  {{{x=4.5449}}}