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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Basics of Mathematics

JEE Main · Mathematics · Basics of Mathematics

If _2 x+ _4 x+ _y x+ _ 16 x-(25)/(36) and x=2^k , then k is

If log⁡2x+log⁡4x+log⁡yx+log⁡16x−2536\log _2 x+\log _4 x+\log _y x+\log _{16} x-\frac{25}{36} and x=2kx=2^k, then kk is

Options

  1. A1
  2. B12\frac{1}{2}
  3. C13\frac{1}{3}
  4. D18\frac{1}{8}

Correct answer

C. 13\frac{1}{3}

Step-by-step solution

log⁡2x+log⁡4x+log⁡8x+log⁡16x=2536log⁡xlog⁡2+log⁡xlog⁡4+log⁡xlog⁡8+log⁡xlog⁡16=2536log⁡xlog⁡2+log⁡x2log⁡2+log⁡x3log⁡2+log⁡x4log⁡2=2536log⁡xlog⁡2[1+12+13+14]=2536log⁡xlog⁡2[2512]=25362512log⁡2x=2536log⁡2x=13∴x=213\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=2kx=2^{\mathrm{k}} ... [Given] ∴k=13\therefore \quad \mathrm{k}=\frac{1}{3}
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