Question

If $log_{3}2, log_{3} \left(2^{x}-5\right)$ and $log_{3} \left(2^{x}-\frac{7}{2}\right) \in$ A.P., then $x$ equals

• A. $3$
• B. $4$
• C. $2$
• D. $0$

Answer 1

Answer: (A)

If $a, b, c \in$ A.P.
Then $2b= a + c$
$\therefore 2\,log_{3} \, (2^{x}-5) = log_{3} (2^{x+1}-7)$
$(\because log (ab) = log a +log b)$
$\Rightarrow (2^{x}-5)^{2} = 2^{x+1}-7$
$\Rightarrow 2^{2x}-10 \cdot 2^{x}+25=2^{x+1}-7$
$\Rightarrow 2^{2x}-12 \cdot 2^{x}+32=0$
$\Rightarrow (2^{x}-8) (2^{x}-4) =0$
$\Rightarrow $ either $2^{x} = 2^{3}$ or $2^{x}=2^{2}$
$\Rightarrow $ either $x=3$ or $x=2$
For $x=2, 2^{x}-5 <\,0$
$\therefore $ it is rejected
$\therefore x=3$