Q93 / 170 · Manipal (MET) 2021 · Mathematics

If $\log _3 2, \log _3\left(2^x-5\right)$ and $\log _3\left(2^x-\frac{7}{2}\right)$ are in $A P$, the value of $x$ is

  • A
    2
  • Correct answer
    B
    3
  • C
    0
  • D
    $\frac{1}{3}$

Solution

$\begin{aligned} & \log _3 2, \log _3\left(2^x-5\right), \log _3\left(2^x-\frac{7}{2}\right) \text { are in AP. } \\ & \Rightarrow \log _3\left(2^x-5\right)=\frac{\log _3\left(2^x-\frac{7}{2}\right)+\log _3 2}{2} \\ & \Rightarrow \quad\left(2^x-5\right)^2=2 \cdot\left(2^x-\frac{7}{2}\right) \\ & \Rightarrow \quad\left(2^x\right)^2+25-10 \cdot 2^x=2 \cdot 2^x-7 \\ & \Rightarrow\left(2^x\right)^2-12 \cdot 2^x+25+7=0 \\ & \Rightarrow \quad\left(2^x\right)^2-12 \cdot 2^x+32=0 \\ & \Rightarrow \quad 2^x=8 \text { or } 2^x=4 \\ & \Rightarrow x=3 \text { or } 2\end{aligned}$ $\Rightarrow x=3(\because x=2$ does not satisfy the given series $)$

Hence, the correct answer is option: 'B'.

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