Solve for $x,y$: $(3x)^{\log3} = (4y)^{\log4}$ and $4^{\log x} = 3^{\log y}$.

Question

if $(3x)^{\log3} = (4y)^{\log4}$ and $4^{\log x} = 3^{\log y}$,

then how do I solve for x?

I tried taking log on both sides but after few steps I got stuck

Answer 1

When you take the logarithm on each side of each equation, the two equations become

$$(\log3)(\log3+\log x)=(\log4)(\log4+\log y)\quad\text{and}\quad(\log x)(\log4)=(\log y)(\log3)$$

Solving the second equation for $\log y$ gives

$$\log y={\log4\over\log3}\log x$$

Plugging this into the first equation gives

$$(\log3)(\log3+\log x)=(\log4)\left(\log4+{\log4\over\log3}\log x\right)$$

which, on moving stuff around, becomes

$$\left(\log3-{(\log4)^2\over\log3} \right)\log x=(\log4)^2-(\log3)^2$$

It should be easy from here to see that

$${\log x\over\log3}=-1$$

which implies $\log x=-\log3=\log(1/3)$, so that $x=1/3$.

Answer 2

I believe a hint is in order $$ 4^{\log x} = 3^{\log y}\implies \log 4 \log x = \log 3\log y\implies y = 10^{\frac{\log 4}{\log 3}\log x}=x^{\frac{\log 4}{\log 3}} $$ Insert into the other one. Re-arrange the first equation to obtain a similar exponent which should be $$x^{\lambda\left(\log^24-\log^2 3\right)} = 10^{\log^24-\log^2 3}$$ it should become clear.