Question 356163
Solve the equation 2^X=3^4-x. Write your solution exactly in terms of logarithms. Do not give a decimal approximation. 

{{{2^x}}}{{{""=""}}}{{{3^(4-x)}}}

Take the log of both sides:

{{{log(2^x)}}}{{{""=""}}}{{{log(3^(4-x))}}}

Move the exponents in front of the logs:

{{{x*log(2)}}}{{{""=""}}}{{{(4-x)log(3)}}}

Distribute on the right

{{{x*log(2)}}}{{{""=""}}}{{{4log(3)-x*log(3)}}}

Get the terms that contain x on the left and other one on the right:


{{{x*log(2)+x*log(3)}}}{{{""=""}}}{{{4log(3)}}}

Factor out x on the left:

{{{x(log(2)+log(3))}}}{{{""=""}}}{{{4log(3)}}}

Divide both sides by {{{(log(2)+log(3))}}}

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

{{{x(cross(log(2)+log(3)))/(cross((log(2)+log(3))))}}}{{{""=""}}}{{{4log(3)/(log(2)+log(3))}}}

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