Question

If $lo{g}_{3}2,lo{g}_3\left(2^x-5\right)$ and $lo{g}_3\left(2^x-\frac{7}{2}\right)\in$ 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 $2b=a+c$
$\therefore 2lo{g}_3(2^x-5)=lo{g}_3(2^{x+1}-7)$
$(\because log(ab)=loga+logb)$
$\Rightarrow (2^x-5{)}^2=2^{x+1}-7$
$\Rightarrow 2^{2x}-10\cdot 2^x+25=2^{x+1}-7$
$\Rightarrow 2^{2x}-12\cdot 2^x+32=0$
$\Rightarrow (2^x-8)(2^x-4)=0$
$\Rightarrow$ 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
$\therefore x=3$