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$$(2x)^{\log 2} = (3y)^{\log 3} \\ 3^{\log x} = 2^{\log y}$$

Solve for $x$ and $y$.

My intuition for solving such problems is taking the logarithm on both sides but it does not work. I also tried using the swap rule by taking the base of the exponent and swapping it in the logarithm, but I am not able to do it.

(P.S. This is not a homework question. It is from the previous question papers of a math contest.)

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  • $\begingroup$ Taking log on both sides is the right approach. Where are you stuck? $\endgroup$ – user202729 Apr 28 '16 at 9:04
  • $\begingroup$ So i come to a stage where log2 log2x=log3log3y;log3logx =logylog2 $\endgroup$ – T Sidharth Apr 28 '16 at 9:04
  • $\begingroup$ After this i dont know how to manipulate the equations $\endgroup$ – T Sidharth Apr 28 '16 at 9:06
  • $\begingroup$ Do you know $\log(3y)=\log(3)+\log(y)$? $\endgroup$ – user202729 Apr 28 '16 at 9:09
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    $\begingroup$ So you have a system of linear equation of the form $ax+by=c,dx+ey=f$. Solve this is not hard. Then exponent log x and log y to get x and y. $\endgroup$ – user202729 Apr 28 '16 at 9:16
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DISCLAIMER: I posted this answer when the first equation read $2x^{\log 2}=3y^{\log 3}$ without parenthesis, but I see that that has been changed to $(2x)^{\log 2}=(3y)^{\log 3}$ in which case my answer below does not fit the problem. For this new equation $(1)$ it becomes $$ as-bt=b^2-a^2 $$ leading through the same steps as below to $$ s=-a\quad t=-b $$ so that $x=\frac12$ and $y=\frac13$.


Define $a=\log2,b=\log3,s=\log x$, and $t=\log y$. Then the first equation becomes $$ a+as=b+bt\\ \iff\\ as-bt=b-a\tag 1 $$ and the second equation becomes $$ bs=at\tag 2 $$ Multiplying $(1)$ by $b$ we get $$ abs-b^2t=b(b-a) $$ and using $(2)$ on the LHS then yields $$ a^2t-b^2t=b(b-a) $$ and since $a^2-b^2=(a+b)(a-b)$ we can divide by this on both sides to have $$ t=-\frac b{a+b} $$ so that $$ \log y=-\frac{\log 2}{\log 6} $$


A similar process started by multiplying $(1)$ by $a$ leads to $$ s=-\frac{a}{a+b}\iff\log x=-\frac{\log 3}{\log 6} $$ which after applying anti-logs should give you $$ x=\frac1{\sqrt[\log 6]3}=3^{-\frac1{\log 6}}\quad\text{ and }\quad y=\frac1{\sqrt[\log 6]2}=2^{-\frac1{\log 6}} $$ or however you want to format those results.

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