$$3 ^ {\log{x}} = 4 ^ {\log{y}}$$ $$(4x)^ {\log{4}} = (3y)^{\log{3}}$$
I know the answer will be a pair of positive rational numbers, however I have no idea on how to start or approach this problem.
Could someone please help me solve this problem for solving for $x$ and $y$.
Thanks!
After taking logarithms, the first equation gives $$(\log3)\log x=(\log4)\log y\implies \log x=\frac{\log4}{\log3}\,\log y\tag{1}$$ and the second equation gives $$\log4(\log4+\log x)=\log3(\log3+\log x)$$ $$(\log4)\log x-(\log3)\log y=\log^23-\log^24\tag{2}$$ Now simply substitute $(1)$ into $(2)$. $$\left(\frac{\log^24}{\log3}-\log3\right)\log y=\log^23-\log^24$$ $$-(\log^23-\log^24)\log y=\log3(\log^23-\log^24)$$ $$\log y=-\log3\implies \boxed{y=\frac13}$$ Hence $$\log x = \frac{\log4}{\log3}(-\log3)=-\log4\implies\boxed{x=\frac14}$$