interesting question

\(x^{log_y{x}}=2  \qquad  \qquad \qquad y^{log_x{y}}=2^4\\ x^{\frac{log x}{log y}}=2  \qquad  \qquad \qquad y^{\frac{log y}{log x}}=2^4\\ log\;{x^{\frac{log x}{log y}}}=log\;{2}  \qquad  \quad \log{y^{\frac{log y}{log x}}}=\log{2^4}\\ {\frac{log x}{log y}}*log\;x=log\;{2}  \qquad  \quad {\frac{log y}{log x}}*\log\;y=4\log\;2\\ \frac{({log x})^2}{log y}=log\;{2}  \qquad  \quad \qquad \frac{(log y)^2}{4log x}=\log\;2\\~\\ so\\ 4{{(log x})^3}=(log y)^3\\ \sqrt[3]{4}=\frac{logy}{logx}\\ log_xy = \sqrt[3]{4}\)

Oh I have to solve for x...

\(\frac{logx}{log y}=\frac{1}{\sqrt[3]{4}}\\ log_y x=4^{(-1/3)}\\ y^{log_y x}=y^{4^{(-1/3)}}\\ x=y^{4^{(-1/3)}}\\ \)

Here's my solution to this:

my attempt:

Take logx of both sides

\(log_x{x^{log_y{x}}}=log_x2\)

(1) \(log_yx=log_x2\)

Take logy of both sides

\(log_y{y^{log_xy}}=log_y16\)

\(log_xy=log_y16\)

for (2) we isolate \(log_y2\) on one side

\(log_xy=4*log_y2\)

(2) \(\frac{log_xy}{4}=log_y2\)

For (1) we invert the logs to substitute into 2

\(log_yx=log_x2\) -> \(log_xy=log_2x\)

substituting:

\(\frac{log_2x}{4}=log_y2\)

Then we isolate y which is

\(y=2^{log_x16}\)

Substituing back into (1)
\(log_{2^{log_x16}}x=log_x2\)

\(\frac{log_2x}{log_x16}=log_x2\)

\(\frac{log_2x}{4*log_x2}=log_x2\)

\(log2_x=4*(log_x2)^2\)

We now make \(log_x2=a\)

\(\frac{1}{a}=4a^2\)

\(a=\frac{1}{4^{\frac{1}{3}}}\)

\(log_x2=\frac{1}{4^{\frac{1}{3}}}\)

solving for x we get x = 2^(cbrt(4))