Question

If ${\log }_2({5.2}^x+1)$, ${\log }_4(2^{1-x}+1)$ and $1$ are in A.P, then the value of $x$ $(x>0)$ is equal to

• A. ${\log }_5\frac{5}{2}$
• B. ${\log }_2\frac{5}{2}$
• C. ${\log }_5\frac{2}{5}$
• D. ${\log }_2\frac{2}{5}$

Answer 1

Answer: (D)

${\log }_2\frac{2}{5}$

${\log }_2({5.2}^x+1),{\log }_4(2^{1-x}+1),1\text{are in A.P}{\log }_2({5.2}^x+1)+{\log }_{2}2=2{\log }_4(2^{1-x}+1)\Rightarrow {\log }_2({5.2}^x+1)\times 2={\log }_2(2^{1-x}+1){10.2}^x+2=\frac{2}{2^x}+1\text{Let}{2}^x=a,\text{So we get}10a+1=\frac{2}{a}\Rightarrow 10a^2+a-2=0\Rightarrow (5a-2)(2a+1)=0\Rightarrow a=\frac{2}{5},\frac{-1}{2}\therefore 2^x=\frac{2}{5}...(\because x>0)x={\log }_2\frac{2}{5}$