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

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Aicharya · Accepted Answer

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

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

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