Question 1189533

{{{2^(x+1) = 3^(y+2) - 3^y}}}............note:{{{2^(x+1)=2*2^x}}}  and {{{3^(y+2) -3^y=3^y*3^2 -3^y=3^y*(3^2-1)=8*3^y}}}


{{{2*2^x= 8*3^y}}}.......simplify


{{{2^x= 4*3^y}}}.........take log of both sides


{{{log((2^x))= log((4*3^y))}}} 


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


{{{x*log((2))= log((2^2))+y*log((3))}}} 


{{{x*log((2))- 2log((2))=y*log((3))}}} 


{{{y=(x*log((2))- 2log((2)))/log((3))}}} 


solution: set {{{y}}} to zero


{{{0=(x*log((2))- 2log((2)))/log((3))}}}..........will be zero only if
 

{{{0=x*log((2))- 2log((2))}}}


{{{2log((2))=x*log((2))}}}.......simplify


{{{2=x}}}


so, solutions are

{{{x = 2}}}, {{{y = 0}}}