Question

If $\log _5 2, \log _5\left(2^x-5\right)$ and $\log _5\left(2^x-7 / 2\right)$ are in A.P., then value of $2 x$ is equal to

• A. 6
• B. 9
• C. 3
• D. 1

Answer 1

Answer: (A)

$ 2 \log _5\left(2^x-5\right)=\log _5 2+\log _5\left(2^x-7 / 2\right)$
$ \Rightarrow \left(2^x-5\right)^2=2\left(2^x-\frac{7}{2}\right)$
$ \left.\Rightarrow t^2-10 t+25=2 t-7 \text { \{put } 2^x=t\right\} $
$\Rightarrow t^2-12 t+32=0 $
$\Rightarrow t=8,4 $
$ \therefore 2^x=4 \text { or } 2^x=8 $
$ \therefore x=2,3 (\because 2^x-5>0$
$ \Rightarrow 2^x>5)$
so only solution $x=3$
$\therefore 2 x=6$