# Logarithm in the exponent

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

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.)

• Taking log on both sides is the right approach. Where are you stuck? – user202729 Apr 28 '16 at 9:04
• So i come to a stage where log2 log2x=log3log3y;log3logx =logylog2 – T Sidharth Apr 28 '16 at 9:04
• After this i dont know how to manipulate the equations – T Sidharth Apr 28 '16 at 9:06
• Do you know $\log(3y)=\log(3)+\log(y)$? – user202729 Apr 28 '16 at 9:09
• 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. – user202729 Apr 28 '16 at 9:16

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.