Question 1209565
<pre>

{{{system(2^x - 3^y = 5,
2^(x+2) + 3^(y+2) = 59)}}}

{{{system(2^x - 3^y = 5,
2^2*2^x + 3^2*3^y = 59)}}}

{{{system(2^x - 3^y = 5,
4*2^x + 9*3^y = 59)}}}

Multiply the 1st equation by -4

{{{system(-4*2^x +4*3^y = -20,
4*2^x + 9*3^y = 59)}}}

Adding them gives

{{{13*3^y=39}}}

{{{3^y=39/13}}}

{{{3^y=3}}}

{{{y=1}}}

Go back to 

{{{system(2^x - 3^y = 5,
4*2^x + 9*3^y = 59)}}}

Multiply the 1st equation by 9

{{{system(9*2^x - 9*3^y = 45,
4*2^x + 9*3^y = 59)}}}

Adding them gives

{{{13*2^x=104}}}
{{{2^x=104/13}}}
{{{2^x=8}}}
{{{x=3}}}

So xy = (3)(1)=3

Edwin</pre>