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If $x^{2a-3} y^2 = x^{6-a} y^{5a}$ then $a\log(\frac{x}{y}) = ?$ I have Tried this and stucked here.

Akash Saxena
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    OP, please tell us the context for this question, and show us what work you've done to try to solve it. – Chris Jun 21 '17 at 03:40
  • Have you tried applying a logarithm to both sides of the equality that you were given? – Tucker Jun 21 '17 at 03:48
  • Yeah i have applied Logarithm to both sides but i unable to get required answer. – Akash Saxena Jun 21 '17 at 03:52
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    @AkashSaxena i have applied ... You should edit your question and add what you tried and where you got stuck. That said, what makes you expect that there would be a unique answer? – dxiv Jun 21 '17 at 04:22

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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} $

marty cohen
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