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. 2
• B. 3
• C. 4
• D. 8

Answer 1

The correct option is B 3

We have 2 $lo{g}_3(2^x-5)=lo{g}_3(2^x-\frac{7}{2})+lo{g}_{3}2$
$\Rightarrow lo{g}_3(2^x-5{)}^2=lo{g}_{3}2(2^x-\frac{7}{2})\Rightarrow (2^x-5{)}^2=2^{x+1}-7\Rightarrow 2^{2x}-{10.2}^x+25=2\times 2^x-7\Rightarrow (2^x{)}^2-{12.2}^x+32=0\Rightarrow (2^x-8\left)\right(2^x-4)=0$
$\Rightarrow 2^x=8,2^x=4$
But $lo{g}_3(2^x-5)$ is not defined if $2^x<5.$

Thus $2^x=8\Rightarrow x=3.$