Question

The sum of all the possible value(s) of $x$ for which ${\log }_{3}2,{\log }_3(2^x-5),{\log }_3(2^x-\frac{7}{2})$ are in A.P., is

Answer 1

Given that ${\log }_{3}2,{\log }_3(2^x-5),{\log }_3(2^x-\frac{7}{2})$ are in A.P.
For the $\log$ function to be defined,
$2^x-5>0\Rightarrow 2^x>5\cdots \left(1\right)2^x-\frac{7}{2}>0\Rightarrow 2^x>\frac{7}{2}\cdots \left(2\right)$
From equation $\left(1\right)$ and $\left(2\right)$,
$2^x>5$

Using the property of A.P.,
${\log }_{3}2+{\log }_3(2^x-\frac{7}{2})=2{\log }_3(2^x-5)\Rightarrow {\log }_{3}2\times (2^x-\frac{7}{2})={\log }_3(2^x-5{)}^2\Rightarrow 2\times (2^x-\frac{7}{2})=(2^x-5{)}^2$
Assuming $2^x=t$
$\Rightarrow 2\times (t-\frac{7}{2})=(t-5{)}^2\Rightarrow 2t-7=t^2-10t+25\Rightarrow t^2-12t+32=0\Rightarrow (t-4\left)\right(t-8)=0\Rightarrow t=4,8\Rightarrow 2^x=4,8$
But $2^x>5$
So, $2^x=8$
$\Rightarrow x=3$

Hence, the sum of all possible values of $x$ is $3$.