Q78 / 150 · MHT CET 2023 (09 May Shift 2) · Mathematics
If $\log _2 x+\log _4 x+\log _y x+\log _{16} x-\frac{25}{36}$ and $x=2^k$, then $k$ is
- A1
- B$\frac{1}{2}$
- Correct answerC$\frac{1}{3}$
- D$\frac{1}{8}$
Solution
$\begin{aligned}
& \log _2 x+\log _4 x+\log _8 x+\log _{16} x=\frac{25}{36} \\
& \frac{\log x}{\log 2}+\frac{\log x}{\log 4}+\frac{\log x}{\log 8}+\frac{\log x}{\log 16}=\frac{25}{36} \\
& \frac{\log x}{\log 2}+\frac{\log x}{2 \log 2}+\frac{\log x}{3 \log 2}+\frac{\log x}{4 \log 2}=\frac{25}{36} \\
& \frac{\log x}{\log 2}\left[1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right]=\frac{25}{36} \\
& \frac{\log x}{\log 2}\left[\frac{25}{12}\right]=\frac{25}{36} \\
& \frac{25}{12} \log _2 x=\frac{25}{36} \\
& \log _2 x=\frac{1}{3} \\
& \therefore \quad x=2^{\frac{1}{3}} \\
&
\end{aligned}$
But $x=2^{\mathrm{k}}$ ... [Given]
$\therefore \quad \mathrm{k}=\frac{1}{3}$
Hence, the correct answer is option: 'C'.