Question

If $lo{g}_{3}2,$ $lo{g}_3(2^x-5)$ and $lo{g}_3(2^x-\frac{7}{2})$ 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 $AP$.
$\therefore lo{g}_3\left(2^x-5\right)=\frac{lo{g}_{3}2+lo{g}_3\left(2^x-\frac{7}{2}\right)}{2}$
$\Rightarrow 2lo{g}_3\left(2^x-5\right)=lo{g}_{3}2\left(2^x-\frac{7}{2}\right)$
$\Rightarrow {\left(2^x-5\right)}^2=2\cdot 2^x-7$
Put $x^2=t$
$t^2+25-10t=2t-7$
$\Rightarrow t^2-12t+32=0$
$\Rightarrow \left(t-8\right)\left(t-4\right)=0$
$\Rightarrow t=4,8$
$\Rightarrow 2^x=4,2^x=8$
$\Rightarrow x=2,x=3$
$\Rightarrow x=3$
$\because x=2$ does not exist in $lo{g}_3\left(2^x-5\right)$