Question 903215
<pre>
{{{system(matrix(2,1,"", 3^y*64^(1/x)=36),matrix(2,1,"", 5^y*512^(1/x)=200)) }}}

The {{{1/x}}} is cumbersome, so let's substitute  {{{1/x=u}}}

{{{system(3^y*64^u=36,5^y*512^u=200)}}}

Write 64 as {{{2^6}}},
Write 36 as {{{2^2*3^2}}}
Write 512 as {{{2^9}}}
Write 200 as {{{2^3*5^2}}}

{{{system(3^y*(2^6)^u=2^2*3^2,5^y*(2^9)^u=2^3*5^2)}}}

{{{system(3^y*2^(6u)=2^2*3^2,5^y*2^(9u)=2^3*5^2)}}}

Divide the first equation through by {{{3^2*2^(6u)}}}
Divide the second equation through hy {{{5^2*2^(9u)}}}

{{{system((3^y*2^(6u))/(3^2*2^(6u))=(2^2*3^2)/(3^2*2^(6u)),(5^y*2^(9u))/(5^2*2^(9u))=(2^3*5^2)/(5^2*2^(9u)))}}}, {{{system((3^y*cross(2^(6u)))/(3^2*cross(2^(6u)))=(2^2*cross(3^2))/(cross(3^2)*2^(6u)),(5^y*cross(2^(9u)))/(5^2*cross(2^(9u)))=(2^3*cross(5^2))/(cross(5^2)*2^(9u)))}}}

Subtracting exponents of like bases:

{{{system(3^(y-2)=2^(2-6u),5^(y-2)=2^(3-9u))}}}

Take out common factors in exponents on the right:

{{{system(3^(y-2)=2^(2(1-3u)),5^(y-2)=2^(3(1-3u)))}}}

Write powers of 2 as 2nd and 3rd powers of {{{2^(1-3u)}}}

{{{system(3^(y-2)=(2^(1-3u))^2,5^(y-2)=(2^(1-3u))^3)}}}

To make the right sides equal raise both sides of the first
equation to the 3rd power and raise both sides of the second
equation to the second power:

{{{system((3^(y-2))^3=((2^(1-3u))^2)^3,(5^(y-2))^2=((2^(1-3u))^3)^2)}}}


{{{system((3^(y-2))^3=(2^(1-3u))^6,(5^(y-2))^2=(2^(1-3u))^6)}}}

{{{system(3^(3(y-2))=2^(6(1-3u)),5^(2(y-2))=2^(6(1-3u)))}}}

Divide equals by equals.  Since the right sides are equal, the
right side will be just 1.

{{{(3^(3(y-2)))/(5^(2(y-2)))=1}}}

The left side can be written as

{{{(3^3/5^2)^(y-2)=1}}}

The only exponent of a base (other than 1) that equals 1 is 0,
therefore

{{{y-2=0}}}

{{{y=2}}}

Substituting in

{{{3^y*2^(6u)=2^2*3^2}}}

{{{3^2*2^(6u)=2^2*3^2}}}

{{{2^(6u)=2^2}}}

Equating the exponents of 2

{{{6u = 2}}}

{{{u=2/6=1/3}}}

{{{1/x=u=1/3}}}

{{{x=3}}}

Solution: (x,y) = (3,2) 

Edwin</pre>