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.
 A: Define $\log _yx=z$ and $\log _xy=w$. Then $x=y^z$ and $y=x^w$. $(1)$
Then $x^z=2$ and hence $y^{z^2}=2$ and also $y^w=16$. So $y^{4z^2}=16$ and hence $w=4z^2$. $(2)$.
Also $y^z=x$ gives $y^{zw}=x^w=y$ hence $zw=1$. And by $(2)$, $\dfrac{1}{z}=4z^2$ and $z^3=\dfrac{1}{4}$ hence $z=\dfrac{1}{\sqrt[3]{4}}$ so $\log _yx=\dfrac{1}{\sqrt[3]{4}}$.
$x=2^{1/z}=2^{\sqrt[3]{4}}$. So everything is correct!
A: 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.
A: We have that  $x^{\log _yx}=2$, $y^{\log _xy}=16$ 
Taking log with base $2$ and converting all logs into base $2$: 
$\log _2x \cdot \log _yx=1 \implies (\log _2x)^2 = \log _2y$
$\log _2y \cdot \log _xy =4 \implies (\log _2y)^2 = 4 \log _2x $
Thus, $$(\log _2x)^2 =2 \sqrt{\log _2x} \implies \log _2x=2^{\frac{2}{3}} \implies x=2^{2^{\frac{2}{3}}}$$
$$\log _yx=\frac{\log _2x}{\log _2y}=2^{-\frac{2}{3}}=\sqrt[3]{\dfrac{1}{4}}$$
So your answer is correct.
