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

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.)

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.