If log_32, log_3(2^x-5), log_3(2^x- 7 2) are in an arithmetic progression, then the value of x is equal to .

As ${\mathrm{log}}_{3}2,{\mathrm{log}}_{3}({2}^{x}-5),{\mathrm{log}}_{3}({2}^{x}-\frac{7}{2})$log3​2,log3​(2x−5),log3​(2x−27​) are in arithmetic progression, $2{\mathrm{log}}_{3}({2}^{x}-5)={\mathrm{log}}_{3}2+{\mathrm{log}}_{3}({2}^{x}-\frac{7}{2})$2log3​(2x−5)=log3​2+log3​(2x−27​)

$$\Rightarrow {\mathrm{log}}_{3}{({2}^{x}-5)}^{2}={\mathrm{log}}_{3}2({2}^{x}-\frac{7}{2})$$⇒log3​(2x−5)2=log3​2(2x−27​)

Let ${2}^{x}=t$2x=t

So, ${\mathrm{log}}_{3}{(t-5)}^{2}={\mathrm{log}}_{3}2(t-\frac{7}{2})$log3​(t−5)2=log3​2(t−27​)

$$\Rightarrow {(t-5)}^{2}=2t-7$$⇒(t−5)2=2t−7

$$\Rightarrow {t}^{2}-12t+32=0$$⇒t2−12t+32=0

$$\Rightarrow (t-4)(t-8)=0$$⇒(t−4)(t−8)=0

$\Rightarrow {2}^{x}=4$⇒2x=4 or ${2}^{x}=8$2x=8

Here, $x\neq 2,$x=2, as ${\mathrm{log}}_{3}(4-5)$log3​(4−5) does not exist.

So, $x=3$x=3