Question

Find the value of $x$, if
${\log }_2(5.2^x+1),{\log }_4(2^{1-x}+1)$ and $1$ are in A.P.

• A. $\frac{\log 2-\log 5}{\log 2}$
• B. $\frac{\log 2-\log 3}{\log 2}$
• C. $\frac{\log 2-\log 5}{\log 5}$
• D. $\frac{\log 3-\log 5}{\log 2}$

Answer 1

The correct option is A $\frac{\log 2-\log 5}{\log 2}$

$lo{g}_2({5.2}^x+1),lo{g}_4(2^{1-x}+1),1$ are in AP.

Common difference 'd' is given by $d=a_2-a_1=a_3-a_2$

$lo{g}_4(2^{1-x}+1)-lo{g}_2({5.2}^x+1)=1-lo{g}_4(2^{1-x}+1)$

$2lo{g}_4(2^{1-x}+1)=lo{g}_{2}2+lo{g}_2({5.2}^x+1)$

$\frac{2}{2}lo{g}_2(2^{1-x}+1)=lo{g}_2(10\cdot 2^x+2)$

$2^{1-x}+1=10\cdot 2^x+2$

Let $2^x=y$

$\frac{2}{y}=10y+1$

$10y^2+y-2=0$

$(5y-2)(2y+1)=0$

$y=\frac{2}{5}ory=\frac{-1}{2}$

$2^x=\frac{2}{5}or{2}^x=\frac{-1}{2}$

$2^x=\frac{2}{5}$

Applying log on both sides

$lo{g}_2\left(2x\right)=lo{g}_2\left(\frac{2}{5}\right)$

$x=lo{g}_{2}2-lo{g}_{2}5$
$\therefore x=\frac{\log 2-\log 5}{\log 2}$