Question

If ${\log }_2(5\cdot 2^x+1),{\log }_4(2^{1-x}+1)$ and $1$ are in AP, then $x$ equals to

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

Answer 1

Answer: (B)

$1-{\log }_{2}5$

The given numbers are in AP.
Therefore, $2{\log }_4(2^{1-x}+1)={\log }_2(5.2^x+1)+1$
$\Rightarrow 2{\log }_{2^2}(\frac{2}{2^x}+1)={\log }_2(5.2^x+1)+{\log }_{2}2$
$\Rightarrow {\log }_{2^2}(\frac{2}{2^x}+1)={\log }_2(10.2^x+2)$
$\Rightarrow \frac{2}{2^x}+1=10.2^x+2$
$\Rightarrow \frac{2}{y}+1=10y+2$ $(2^x=y)$
$\Rightarrow 10y^2+y-2=0$
$\Rightarrow y=\frac{2}{5}ory=-\frac{1}{2}$
$\Rightarrow 2^x=\frac{2}{5};2^x=-\frac{1}{2}$
$\Rightarrow x={\log }_2\left(\frac{2}{5}\right)$ ....($\because$ $2^x$ cannot be negative)
$\Rightarrow x={\log }_{2}2-{\log }_{2}5$
$\Rightarrow x=1-{\log }_{2}5$