Question

If ${\log }_2(5\times 2^x+1),{\log }_4(2^{1-x}+1)$ and $1$ are in A.P., then $x$ equals

• A. ${\log }_{2}5$
• B. $1-{\log }_{5}2$
• C. ${\log }_{5}2$
• D. $1-{\log }_{2}5$

Answer 1

The correct option is D $1-{\log }_{2}5$

The given numbers are in A.P. so,
$2{\log }_4(2^{1-x}+1)=1+{\log }_2(5\times 2^x+1)\Rightarrow 2{\log }_{2^2}(\frac{2}{2^x}+1)={\log }_{2}2+{\log }_2(5\times 2^x+1)\Rightarrow \frac{2}{2}{\log }_2(\frac{2}{2^x}+1)={\log }_{2}2(5\times 2^x+1)\Rightarrow (\frac{2}{2^x}+1)=2(5\times 2^x+1)$
Assuming $2^x=y$
$\Rightarrow \frac{2}{y}+1=10y+2\Rightarrow 10y^2+y-2=0\Rightarrow (5y-2)(2y+1)=0\Rightarrow y=\frac{2}{5}\text{or}-\frac{1}{2}\Rightarrow 2^x=\frac{2}{5}\text{or}-\frac{1}{2}\Rightarrow 2^x=\frac{2}{5}[\because 2^x\text{cannot be negative}]$
$\Rightarrow x={\log }_2\left(2/5\right)=1-{\log }_{2}5$