Question

Find the value of $x$ for which ${\log }_2({5.2}^x+1),{\log }_4(2^{1-x}+1)$ and $1$ form an arithmetic progression.

Answer 1

$lo{g}_2({5.2}^x+1),{\log }_4(2^{1-x}+1)and1$ are in AP.

Common difference $d=a_2-a_1=a_3-a_2$

$\therefore {\log }_4(2^{1-x}+1)-{\log }_2({5.2}^x+1)=1-{\log }_4(2^{1-x}+1)$

$\Rightarrow {\log }_4(2^{1-x}+1)+{\log }_4(2^{1-x}+1)={\log }_{2}2+{\log }_2({5.2}^x+1)[{\log }_{a}a=1]$

$\Rightarrow 2{\log }_4(2^{1-x}+1)={\log }_{2}2+{\log }_2({5.2}^x+1)$

$\Rightarrow \frac{2}{2}\cdot {\log }_2(2^{1-x}+1)={\log }_2\left(2\right({5.2}^x+1\left)\right)$

$[Since,{\log }_b^{n}a=n{\log }_{b}a$ & ${\log }_{c}a+{\log }_{c}b={\log }_c\left(ab\right)]$

$\therefore 2^{1-x}+1={5.2}^{x+1}+2$

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

Let $2^x=y$

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

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

$\Rightarrow 10y^2+5y-4y-2=0$

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

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

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

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

$\therefore 2^x=\frac{2}{5}$(Since exponent of any real number can never be a negative value).

Applying ${\log }_2$ on both sides,

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

$\Rightarrow x={\log }_{2}2-{\log }_{2}5$

$\Rightarrow x=1-{\log }_{2}5$