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. $8$
• B. $-8$
• C. $3$
• D. $-3$

Answer 1

Answer: (C)

Since, it is in $AP$.
$\therefore log_{3}\left(2^{x} - 5\right) = \frac{log_{3} 2 + log_{3} \left(2^{x} - \frac{7}{2}\right)}{2}$
$ \Rightarrow 2log_{3} \left(2^{x} -5\right) = log_{3} 2\left(2^{x}-\frac{7}{2}\right) $
$ \Rightarrow \left(2^{x} - 5\right)^{2} = 2\cdot2^{x} -7 $
Put $x^{2 } = t $
$ t^{2} + 25 - 10 t = 2t - 7 $
$\Rightarrow t^{2} -12t + 32 = 0 $
$ \Rightarrow \left(t-8\right)\left(t-4\right) = 0 $
$\Rightarrow t = 4, 8$
$ \Rightarrow 2^{x} = 4, 2^{x} = 8$
$\Rightarrow x = 2, x = 3 $
$ \Rightarrow x = 3 $
$ \because x = 2$ does not exist in $log_{3}\left(2^{x} -5\right)$