Yajat Special (Problem $1$ )

Let $u = \log_{2}x$ so that we obtain the quadratic equation:

$u - \frac{6}{u} = 5 \Rightarrow u^2 - 5u - 6 = (u-6)(u+1) = 0 \Rightarrow u = -1, 6.$

which gives back $\log_{2}x = -1, 6 \Rightarrow x =\frac{1}{2}, 64$ . Since $x \in \mathbb{R^{+}}$ , the answer is YES.

Given that

$\begin{aligned} \log_2 x - 6 \log_x 2 & = 5 \\ \log_2 x - 6 \cdot \frac {\log_2 2}{\log_2 x} & = 5 \\ \log_2 x - 6 \cdot \frac 1{\log_2 x} & = 5 \\ \log_2^2 x - 5 \log_2 x - 6 & = 0 \\ (\log_2 x - 6)(\log_2 x + 1) & = 0 \end{aligned}$

$\implies \log_2 x = \begin{cases} 6 & \implies x = 2^6 = 64 \\ -1 & \implies x = 2^{-1} = \dfrac 12 \end{cases}$

Yes , it can be solved.