Question

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

• 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 $1-{\log }_{2}5$

Given : ${\log }_2(5\times 2^x+1),{\log }_4(2^{1-x}+1),1$ are in A.P.
For $\log$ to be defined,
$5\times 2^x+1>0,2^{1-x}+1>0\Rightarrow x\in R$

Now,
$2{\log }_4(2^{1-x}+1)={\log }_2(5\times 2^x+1)+1(\because 2b=a+c)\Rightarrow {\log }_2(2^{1-x}+1)={\log }_2\left[2\right(5\times 2^x+1\left)\right]\Rightarrow 2^{1-x}+1=10\times 2^x+2$
Assuming $2^x=y$, we get
$\Rightarrow \frac{2}{y}=10y+1\Rightarrow 10y^2+y-2=0\Rightarrow (5y-2)(2y+1)=0\Rightarrow y=\frac{2}{5}(\because y=2^x>0)\Rightarrow 2^x=\frac{2}{5}\Rightarrow x={\log }_2\left(\frac{2}{5}\right)\therefore x=1-{\log }_{2}5$