Question

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

• A. 2
• B. 3
• C. 4
• D. 2,3

Answer 1

Answer: (B)

3

Given ${\log }_{3}2,{\log }_3(2^x-5){\log }_3(2^x-\frac{7}{2})$ are in A.P
$\Rightarrow 2{\log }_3(2^x-5)={\log }_{3}2+{\log }_3(2^x-\frac{7}{2})$
$\Rightarrow {\log }_3{(2^x-5)}^2={\log }_{3}2\cdot (2^x-\frac{7}{2})$
$\Rightarrow {(2^x-5)}^2=2(2^x-\frac{7}{2})$
Let $2^x=t$
${\Rightarrow (t-5)}^2=2t-7$
$\Rightarrow t^2-10t+25=2t-7$
${\Rightarrow t}^2-12t+32=0$
${\Rightarrow t}^2-8t-4t+32=0$
$\Rightarrow (t-8\left)\right(t-4)=0$
$\Rightarrow (t-8)=0$ or $(t-4)=0$
$\Rightarrow t=8$ or $t=4$
$\Rightarrow 2^x=8$ or $2^x=4$
$\Rightarrow 2^x=2^3$ or $2^x=2^2$
Comparing the exponents, we get
$x=2$ or $x=3$
Therefore, $x$ cannot take 2 because $2^2-5=-1$
Hence,option 'A' is correct.