Question

If ${\log }_{3}2,{\log }_3(2^x-5),{\log }_3(2^x-\frac{7}{2})$ are in an arithmetic progression, then the value of $x$ is equal to

Answer 1

Since ${\log }_{3}2,{\log }_3(2^x-5),{\log }_3(2^x-\frac{7}{2})$ are in A.P.
$\therefore 2,2^x-5,2^x-\frac{7}{2}$ are in G.P.
$\Rightarrow (2^x-5{)}^2=2\cdot (2^x-\frac{7}{2})$
$\Rightarrow 2^{2x}-10\cdot 2^x+25={2.2}^x-7$
$\Rightarrow 2^{2x}-12\cdot 2^x+32=0$
$\Rightarrow (2^x-4)(2^x-8)=0$
$\Rightarrow x=2$ or $3$
But $x=2$ is not acceptable.
$\therefore x=3$