Question

The value of $x$ for which ${\log }_3(2^{1-x}+3),{\log }_{9}4$ and ${\log }_{27}{(2^x-1)}^3$ form an $A.P.$ is

• A. $\frac{11}{6}$
• B. $\frac{6}{11}$
• C. ${\log }_2\left(\frac{11}{6}\right)$
• D. $1$

Answer 1

Answer: (D)

$1$

Condition for $A.P.$ is: $2b=a+c$

$\Rightarrow 2{\log }_{9}4={\log }_3(2^{1-x}+3)+{\log }_{27}{(2^x+1)}^3$

$\Rightarrow 2\frac{\log 4}{\log 9}={\log }_3(2^{1-x}+3)+3\frac{\log (2^x+1)}{\log 27}$

$\Rightarrow 2\frac{{\log }_{3}4}{2{\log }_{3}3}={\log }_3(2^{1-x}+3)+3\frac{{\log }_3(2^x+1)}{3{\log }_{3}3}$

$\Rightarrow \frac{2}{2}{\log }_{3}4={\log }_3(2^{1-x}+3)+\frac{3}{3}{\log }_3(2^x-1)$

$\Rightarrow \frac{2}{2}\log 4=\log (2^{1-x}+3)+\frac{3}{3}\log (2^x-1)$

$\Rightarrow \log 4=\log \left[(2^{1-x}+3)(2^x-1)\right]$

$\therefore 4=(\frac{2}{y}+3)(y-1),$ where $y=2^x$

$\therefore 4y=(3y+2)(y-1)$ or $3y^2-5y-2=0$

$\therefore y=2,-\frac{1}{3}=2^x$

Since an exponential function cannot be $-ve,$

$\therefore 2^x=-\frac{1}{3}$ is rejected

$\therefore 2^x=2\Rightarrow x=1$