Too many Logs!

$\large \begin{aligned} (\sqrt 2)^{\log_2 2^{\log_2 2^{\log_2 2^{\log_2 2^{\log_2 x}}}}} & = 5 & \small \color{#3D99F6} \text{Note that }\log_2 2^{\log_2 x} = \log_2 x \log_2 2 = \log_2 x \\ (\sqrt 2)^{\log_2 2^{\log_2 2^{\log_2 2^{\log_2 x}}}} & = 5 \\ (\sqrt 2)^{\log_2 2^{\log_2 2^{\log_2 x}}} & = 5 \\ (\sqrt 2)^{\log_2 2^{\log_2 x}} & = 5 \\ (\sqrt 2)^{\log_2 x} & = 5 \\ 2^{\frac 12 \log_2 x} & = 5 & \small \color{#3D99F6} \text{Taking }\log_2 \text{ on both sides} \\ \frac 12 \log_2 x & = \log_2 5 \\ \log_2 x & = 2\log_2 5 \\ & = \log_2 25 \\ \implies x & = \boxed{25} \end{aligned}$

Continuously using the this particular property of logarithms -> $\log_{a}b^{c} = c\cdot log_{a}b$ and $log_{a}a =1$ ,

The $\log_{2}x$ keep moving down until the equation becomes -> $\sqrt{2}^{log_{2}x}$ = 5

Squaring both sides => $2^{log_{2}x} = 25$

${log_{2}x}$ basically asks 2 to the power what equals ‘x’.

So, $2^{log_{2}x} = x$ .

Therefore the equation is true when $\textbf{x = 25}$

That's simple ,just use a^log(c) base a=C from the top corner