Question

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

• A. $2$
• B. $3$
• C. $4$
• D. $8$

Answer 1

The correct option is B $3$

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

$\Rightarrow {(2^x-5)}^2=2^{x+1}-7$

$\Rightarrow 2^{2x}-10\cdot 2^x+25=2^{x+1}-7$

$\Rightarrow {\left(2^x\right)}^2-12\cdot 2^x+32=0$

$\Rightarrow (2^x-8)(2^x-4)=0$

$\Rightarrow 2^x=8,2^x=4$

but ${\log }_3(2^x-5)$ is valid if $2^x>5$.

Thus $2^x=8\Rightarrow x=3$.
Ans: B