Question

If ${\log }_{3}2$, ${\log }_3\left(2^x-5\right)$ and ${\log }_3\left(2^x-\frac{7}{2}\right)$ are in AP, then $x$ is equal to

• A. $2$
• B. $3$
• C. $4$
• D. $2,3$

Answer 1

The correct option is B

$3$

Explanation for the correct option

Given that, ${\log }_{3}2$, ${\log }_3\left(2^x-5\right)$ and ${\log }_3\left(2^x-\frac{7}{2}\right)$ are in an arithmetic progression

We know that if $a,b,c$ are in A.P. then $2b=a+c$

$\Rightarrow$ $2\times {\log }_3\left(2^x-5\right)=$${\log }_{3}2+{\log }_3\left(2^x-\frac{7}{2}\right)$

$\Rightarrow$ ${\log }_3{\left(2^x-5\right)}^2=$${\log }_3\left(2\times \left(2^x-\frac{7}{2}\right)\right)$ $\left[\because b{\log }_{a}m={\log }_a{m}^b,{\log }_{a}m+{\log }_{a}n={\log }_{a}mn\right]$

$\Rightarrow$ ${\log }_3{\left(2^x-5\right)}^2=$${\log }_3\left(2^{x+1}-7\right)$ $\left[\because {\log }_{a}b=l{\mathrm{og}}_{a}c\Rightarrow b=c\right]$

$\Rightarrow$ ${\left(2^x-5\right)}^2=$$\left(2^{x+1}-7\right)$

$\Rightarrow$ $2^{2x}-5\times 2^{x+1}+25=2^{x+1}-7$

$\Rightarrow$ $2^{2x}-6\times 2^{x+1}+32=0$

$\Rightarrow$ $2^{2x}-12\times 2^x+32=0$

$\Rightarrow$$2^{2x}-8\times 2^x-4\times 2^x+32=0$

$\Rightarrow$ $\left(2^x-4\right)\left(2^x-8\right)=0$

$\Rightarrow$ $2^x=4$ or $2^x=8$

$\Rightarrow$ $x=2$ or $x=3$

When $x=2$ , $\left(2^x-5\right)=4-5=-1$

Logarithm is not defined for negative numbers Hence, $x\ne 2$

Thus $x=3$

Hence option(B) i.e. $3$ is the correct answer.