Question

If ${\log }_2(5\times 2^x+1),{\log }_4(2^{1-x}+1)$ and $1$ are in A.P., then $x$ equals

• A. $1-{\log }_{2}5$
• B. $1+{\log }_{5}2$
• C. ${\log }_{5}2$
• D. $\log 2$

Answer 1

The correct option is A $1-{\log }_{2}5$

$lo{g}_2(5\times 2^x+1)$, $lo{g}_4(2^{1-x}+1)$, $1$ are in AP

If a, b, c are in AP

Then, $2b=a+c$

$2{\log }_4(2^{1-x}+1)={\log }_2(5\times 2^x+1)+1$

$\frac{2}{2}{\log }_2(2^{1-x}+1)={\log }_2\left(2\right(5\times 2^x+1\left)\right)$

[Since, ${\log }_b^n=a=n{\log }_b^a$, ${\log }_c^a+{\log }_c^b={\log }_c^{ab}$]

${\log }_2(2^{1-x}+1)={\log }_2\left(2\right(5\times 2^x+1\left)\right)$

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

$\frac{2}{2^x}={10.2}^x+1$

Let $2^x=y$

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

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

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

$2^x=\frac{2}{5}$ (take $lo{g}_2$ both sides) and $2^x$ cannot be negative

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

$x=1-lo{g}_{2}5$.