In the Middle of the Tower

$\begin{aligned} 2^{x^{y^4}} & = 2^{(x^y)^{y^3}} = 2^{3^{y^3}} = 5 \\ 3^{y^3} \ln 2 & = \ln 5 \\ 3^{y^3} & = \frac {\ln 5}{\ln 2} \\ y^3 \ln 3 & = \ln \left(\frac {\ln 5}{\ln 2} \right) \\ y^3 & = \frac {\ln \left(\frac {\ln 5}{\ln 2} \right)}{\ln 3} \\ \implies y & = \sqrt[3]{\frac {\ln \left(\frac {\ln 5}{\ln 2} \right)}{\ln 3}} \approx 0.915287667 \end{aligned}$

Since $x^y = 3$ ,

$\begin{aligned} \implies y \ln x & = \ln 3 \\ \ln x & = \frac {\ln 3}y \\ x = e^{\frac {\ln 3}y} \\ \implies xy & = ye^{\frac {\ln 3}3} \\ & \approx 0.915287667 e^{\frac {\ln 3}{0.915287667}} \\ & \approx \boxed{3.04} \end{aligned}$

Finding the value of $y$ .

$\begin{array}{cc} \begin{aligned} x^{y} = \space & 3 \\ y \ln x = \space & \ln 3 \\ \implies \color{#3D99F6} \ln x \color{#333333} = \space & \color{#3D99F6} \dfrac{\ln 3}{y} \end{aligned} & \; \end{array}$

$\begin{aligned} 2^{x^{y^{4}}} = \space & 5 \\ x^{y^{4}} \ln 2 = \space & \ln 5 \\ x^{y^{4}} = \space & \dfrac{\ln 5}{\ln 2} \\ y^{4} \ln x = \space & \ln \left( \dfrac{\ln 5}{\ln 2} \right) \\[0.75em] y^{4} \color{#3D99F6} \ln x \color{#333333} = \space & \ln ( \ln 5) - \ln ( \ln 2) \\ y^{4} \color{#3D99F6} \dfrac{\ln 3}{y} \color{#333333} = \space & \ln ( \ln 5) - \ln ( \ln 2) \\ y^{3} \ln 3 = \space & \ln ( \ln 5) - \ln ( \ln 2) \\[0.3em] y^{3} = \space & \dfrac{ \ln ( \ln 5) - \ln ( \ln 2)}{\ln 3} \\[0.6em] y = \space & \sqrt[3]{\dfrac{ \ln ( \ln 5) - \ln ( \ln 2)}{\ln 3}} \end{aligned}$

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Finding the value of $x$ .

$\begin{aligned} \color{#3D99F6} \ln x \color{#333333} = \space & \color{#3D99F6} \dfrac{\ln 3}{y} \\ \implies x = \space & e \space \text{raised to} \space \dfrac{\ln 3}{y} \\ x = \space & e \space \text{raised to} \space \dfrac{ \ln 3}{ \sqrt[3]{\frac{ \ln ( \ln 5) - \ln ( \ln 2)}{\ln 3}}} \end{aligned}$

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$\begin{cases} x = 3.3210859... \\ y = 0.9152876... \\ xy = \boxed{3.039748...} \end{cases}$