If \(\log _2 x+\log _4 x+\log _8 x+\log _{16} x=\frac{25}{36}\) and \(x=2^k\), then \(k\) is

\(\begin{aligned}
& \log _2 x+\log _4 x+\log _8 x+\log _{16} x=\frac{25}{36} \\
& \frac{\log x}{\log 2}+\frac{\log x}{\log 4}+\frac{\log x}{\log 8}+\frac{\log x}{\log 16}=\frac{25}{36} \\
& \frac{\log x}{\log 2}+\frac{\log x}{2 \log 2}+\frac{\log x}{3 \log 2}+\frac{\log x}{4 \log 2}=\frac{25}{36} \\
& \frac{\log x}{\log 2}\left[1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right]=\frac{25}{36} \\
& \frac{\log x}{\log 2}\left[\frac{25}{12}\right]=\frac{25}{36} \\
& \frac{25}{12} \log _2 x=\frac{25}{36} \\
& \log _2 x=\frac{1}{3} \\
& \therefore \quad x=2^{\frac{1}{3}} \\
&
\end{aligned}\)
But \(x=2^{\mathrm{k}}\) ... [Given]
\(\therefore \quad \mathrm{k}=\frac{1}{3}\)