Mathematics - Sequence & Series Question with Solution | TestHub

As ${\log }_{3}2,{\log }_3\left(2^x-5\right),{\log }_3\left(2^x-\frac{7}{2}\right)$ are in arithmetic progression, $2{\log }_3\left(2^x-5\right)={\log }_{3}2+{\log }_3\left(2^x-\frac{7}{2}\right)$

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

Let $2^x=t$

So, ${\log }_3{\left(t-5\right)}^2={\log }_{3}2\left(t-\frac{7}{2}\right)$

$\Rightarrow {\left(t-5\right)}^2=2t-7$

$\Rightarrow t^2-12t+32=0$

$\Rightarrow \left(t-4\right)\left(t-8\right)=0$

$\Rightarrow 2^x=4$ or $2^x=8$

Here, $x\ne 2,$ as ${\log }_3\left(4-5\right)$ does not exist. 

So, $x=3$