Question

If ${\log }_2({5.2}^x+1),{\log }_4(2^{1-x}+1)$ and 1 are in AP, then $x$ equals

• A. $lo{g}_{2}5$
• B. $1-lo{g}_{2}5$
• C. $lo{g}_{5}2$
• D. None of these

Answer 1

Answer: (B)

The given numbers are in AP. $\therefore$ $2{\log }_4(2^{1-x}+1)={\log }_2({5.2}^x+1)+1$ $\Rightarrow$ $2{\log }_2\left(\frac{2}{2^x}+1\right)={\log }_2({5.2}^x+1)+{\log }_{2}2$ $\Rightarrow$ $\frac{2}{2}{\log }_2\left(\frac{2}{2^x}+1\right)={\log }_2({5.2}^x+1)2$ $\Rightarrow$ ${\log }_2\left(\frac{2}{2^x}+1\right)={\log }_2({10.2}^x+2)$ $\Rightarrow$ $\frac{2}{2^x}+1={10.2}^x+2$ $\Rightarrow$ $\frac{2}{y}+1=10y+2,$ where $2^x=y$ $\Rightarrow$ $10y^2+y-2=0\Rightarrow (5y-2)(2y+1)=0$ $\Rightarrow$ $y=\frac{2}{5}$ or $y=\frac{-1}{2}\Rightarrow 2^x=\frac{2}{5}$ or $2^x=\frac{-1}{2}$ $\Rightarrow$ $x={\log }_2\left(\frac{2}{5}\right)$ $\Rightarrow$ $x={\log }_{2}2-{\log }_{2}5$ $\Rightarrow$ $x=1-{\log }_{2}5$