Question 1080330
<pre>
 {{{2log(5,(2x-y))}}}{{{""=""}}}{{{log(5,(xy^2)) - log(5,(y))}}}

 {{{log(5,(2x-y)^2)}}}{{{""=""}}}{{{log(5,((xy^2)/y^""))}}}

 {{{(2x-y)^2}}}{{{""=""}}}{{{(xy^2)/y^""}}}

 {{{(2x-y)^2}}}{{{""=""}}}{{{(xy^cross(2))/cross(y^"")}}}

 {{{(2x-y)^2}}}{{{""=""}}}{{{xy}}}

 {{{4x^2-4xy+y^2}}}{{{""=""}}}{{{xy}}}

 {{{4x^2-5xy+y^2}}}{{{""=""}}}{{{0}}}

 {{{4x-y)(x-y)}}}{{{""=""}}}{{{0}}}

{{{matrix(5,4,       

4x-y=0, ";", "", x-y=0,
 4x=y,  ";", "",   y=x,
4x/y=y/y,";", "",  y/y=x/y,
4x/y=1,  ";", "",    1=x/y,
x/y=1/4, ";", "",   x/y=1)}}}

However, such equations often have extraneous 
answers, so we must check.  Logarithms may
be taken only of positive numbers.

Checking
{{{x/y}}}{{{""=""}}}{{{1/4}}}
{{{y}}}{{{""=""}}}{{{4x}}}
{{{2log(5,(2x-y))}}}{{{""=""}}}{{{log(5,(xy^2)) - log(5,(y))}}}
{{{2log(5,(2x-4x))}}}{{{""=""}}}{{{log(5,(x(4x)^2)) - log(5,(4x))}}}
{{{2log(5,(-2x))}}}{{{""=""}}}{{{log(5,(x(16x^2)^"")) - log(5,(4x))}}}
{{{2log(5,(-2x))}}}{{{""=""}}}{{{log(5,(16x^3)) - log(5,(4x))}}}
We can see that on the left side, x must be negative 
to insure that a logarithm is being taken only of a 
positive number.  However that causes both terms on 
the right to be logarithms of negative numbers. 

So we must discard {{{x/y}}}{{{""=""}}}{{{1/4}}}

Checking 
{{{x/y}}}{{{""=""}}}{{{1}}}
{{{y}}}{{{""=""}}}{{{x}}}
{{{2log(5,(2x-y))}}}{{{""=""}}}{{{log(5,(xy^2)) - log(5,(y))}}}
{{{2log(5,(2x-x))}}}{{{""=""}}}{{{log(5,(x(x)^2)) - log(5,(x))}}}
{{{2log(5,(x))}}}{{{""=""}}}{{{log(5,(x^3)) - log(5,(x))}}}
{{{2log(5,(x))}}}{{{""=""}}}{{{log(5,(x^3/x))}}}
{{{2log(5,(x))}}}{{{""=""}}}{{{log(5,(x^2))}}}
{{{2log(5,(x))}}}{{{""=""}}}{{{2log(5,(x))}}}

Which checks.

The only solution is {{{x/y}}}{{{""=""}}}{{{1}}}

Edwin</pre>