Question

If $\left(log\right)_{2} \left(5 \left(2^{x}\right) + 1\right), \, \left(log\right)_{4}⁡\left(2^{1 - x} + 1\right)$ and $1$ are in arithmetic progression, then $x$ is equal to

• A. $\frac{log 5}{log ⁡ 2}$
• B. $log_{2} 0.6$
• C. $1-\frac{log 5}{log ⁡ 2}$
• D. $\frac{log 2}{log ⁡ 5}$

Answer 1

Answer: (C)

$\log _{2}\left(5 \cdot 2^{x}+1\right), \frac{1}{2} \log _{2}\left(2^{1-x}+1\right), \log _{2} 2$ are in A.P.
$\Rightarrow 2 \times \frac{1}{2} \log _{2}\left(2^{1-x}+1\right)=\log _{2}\left(5 \cdot 2^{x}+1\right)+\log _{2} 2$
$\Rightarrow 2^{1-x}+1=\left(5 \cdot 2^{x}+1\right) 2$
Let, $2^{x}=y$
$\Rightarrow \frac{2}{y}+1=(5 y+1) 2$
$\Rightarrow 2+y=10 y^{2}+2 y$
$\Rightarrow 10 y^{2}+y-2=0 \Rightarrow y=\frac{2}{5}, \frac{-1}{2}$
$\Rightarrow 2^{x}=\frac{2}{5}$ or $\frac{-1}{2}$ (not possible)
$\Rightarrow 2^{x}=\frac{2}{5}$
$\Rightarrow x=\log _{2}\left(\frac{2}{5}\right)=1-\log _{2} 5=1-\frac{\log 5}{\log 2}$