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 $2x$ is equal to

• A. 6
• B. 9
• C. 3
• D. 1

Answer 1

Answer: (A)

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