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. $\frac{log5}{log2}$
• B. $lo{g}_{2}0.6$
• C. $1-\frac{log5}{log2}$
• D. $\frac{log2}{log5}$

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, $2^x=y$
$\Rightarrow \frac{2}{y}+1=(5y+1)2$
$\Rightarrow 2+y=10y^2+2y$
$\Rightarrow 10y^2+y-2=0\Rightarrow y=\frac{2}{5},\frac{-1}{2}$
$\Rightarrow 2^x=\frac{2}{5}$ or $\frac{-1}{2}$ (not possible)
$\Rightarrow 2^x=\frac{2}{5}$
$\Rightarrow x={\log }_2\left(\frac{2}{5}\right)=1-{\log }_{2}5=1-\frac{\log 5}{\log 2}$