Question 1000716

{{{8^(x+3)=9^x}}}

{{{(2^3)^(x+3)=(3^2)^x}}}

{{{2^(3x+9)=3^(2x)}}}

{{{log(2^(3x+9))=log(3^(2x))}}}

{{{(3x+9)log(2)=(2x)log(3)}}}...........since {{{log(2)=0.301029995663981}}}and {{{log(3)=0.47712125471}}}, we have

{{{(3x+9)0.301029995663981=(2x)0.47712125471}}}

{{{0.903089986991943x+2.709269960975829=0.95424250942x}}}

{{{2.709269960975829=0.95424250942x-0.903089986991943x}}}

{{{2.709269960975829=0.05115252243x}}}

{{{x=2.709269960975829/0.05115252243}}}

{{{x = 52.9645}}}

or, rounded to three decimals

{{{x=52.965}}}