<s>[INST] How to solve this logarithm problem If $x$ and $y$ are positive real numbers and $x^{\log _yx}=2$ And $y^{\log _xy}=16$ Find $\log _yx$ and find $x$.
The way I tried to solve it was to take $\log _x$ on the base $x$ one and $\log _y$ for the $y$ one. Then something really messy came up. I got that $\log _yx=\sqrt[3]{\dfrac{1}{4}}$ but I don't know if it's correct or not.
 [/INST] Note the following:
$$x^{\log_y x} = 2 \implies x^{\frac{\log_2 x}{\log_2 y}} = 2$$
$$\implies 2^{\frac{\log^2_2 x}{\log_2 y}}=2 \implies \frac{\log^2 x}{\log_2 y} = \log_2 2 = 1,$$ or in particular,
$$\frac{\log^2_2 x}{\log_2 y} = 1.$$
Similarly,
$$y^{\log_x y} = 16 \implies \frac{\log^2_2 y}{\log_2 x} = \log_2 16 = 4,$$
or in particular, $$\frac{\log^2_2 y}{\log_2 x} = 4.$$
So write $a = \log_2 x$ and $b=\log_2 y$. Then on the one hand, the above derived equations become $$\frac{a^2}{b} = 1,$$ and $$\frac{b^2}{a} = 4.$$ On the other hand, $\log_y x = \frac{a}{b}$. So let us now calculate $\frac{a}{b}$. Note that
$$\frac{a^2}{b} = 1 \implies a^2 = b.$$ Plugging $a^2=b$ into $\frac{b^2}{a} = 4$ gives $\frac{a^4}{a}=a^{3} = 4$, and so plugging this into $\frac{a^2}{b}=1$ [and dividing both sides by $a$ to get $\frac{a}{b}=\frac{1}{a}$] gives
$$\frac{a}{b}=\frac{1}{a} = a^{-1} = (4)^{-1/3}.$$ So you arrived at the correct answer.
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