<s>[INST] If $x^{2a-3} y^2 = x^{6-a} y^{5a}$ then $a\log(\frac{x}{y}) = ?$ If $x^{2a-3} y^2 = x^{6-a} y^{5a}$ then $a\log(\frac{x}{y}) = ?$
I have Tried this and stucked here.
 [/INST] You are given
$x^{2a-3} y^2=x^{6-a} y^{5a}
$.
Multiplying by
$x^{a-6}y^{-2}
$,
this becomes
$x^{3a-9}=y^{5a-2}
$.
Looking at this,
I notice that,
if the
$y^2$
were
$y^{2a}$,
the right side would be
$y^{3a}$.
This indicates to me that
there might be
an error in
the statement of the problem.
However,
working with the problem
as written,
$(3a-9)\log x=(5a-2)\log y
$
so that
$\log y
=\frac{3a-9}{5a-2}\log x
$.
Therefore
$\begin{array}\\
a\log(x/y)
&=a(\log x-\log y)\\
&=a(\log x-\frac{3a-9}{5a-2}\log x)\\
&=\log x(a-\frac{3a-9}{5a-2})\\
&=a\log x(1-\frac{3a-9}{5a-2})\\
&=a\log x\frac{(5a-2)-(3a-9)}{5a-2}\\
&=a\log x\frac{2a+7}{5a-2}\\
\end{array}
$
</s>