Question 896967
{{{16^(x+3)=64^(2x-5)}}}

Take log to the base 4.

{{{log(16^(x+3))=log(64^(2x-5))}}}

{{{(x+3)*log16= (2x-5)log64}}}

log is to the base 4. So you can write..

{{{(x+3)*2= (2x-5)3}}}

{{{2x +6 = 6x -15}}}

{{{21=4x}}}

{{{x=21/4}}}