Question

If ${log}_{3}2,{log}_3(2^x-5)$, and ${log}_3(2^x-7/2)$ are in A.P, then the value of x is

Answer 1

$2b=a+c$

where a, b, c are in AP

$\Rightarrow 2lo{g}_3(2^x-5)=lo{g}_3^2+lo{g}_3(2^x-\frac{7}{2})$

$\Rightarrow lo{g}_3(2^x-5{)}^2=lo{g}_3(\left(2\right)(2^x-\frac{7}{2}))$

$\Rightarrow (2^x-5{)}^2={2.2}^x-7$ Let $2^x=a$

$\Rightarrow a^2+25-10a=2a-7$

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

$\Rightarrow (a-4)(a-8)=0$

So $2^x=4$ or $2^x=8$

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