Question

If ${\log }_{10}2,{\log }_{10}(2^x-1)$ and ${\log }_{10}(2^x+3)$ be three consecutive terms of an A.P. then

• A. $x=0$
• B. $x=1$
• C. $x={\log }_{2}5$
• D. none of these

Answer 1

Answer: (C)

By the given condition
$2lo{g}_{10}\left(2^x-1\right)=lo{g}_{10}2+lo{g}_{10}\left(2^x+3\right)$
$\Rightarrow {\left(2^x-1\right)}^2=2\left(2^x+3\right)$
$\Rightarrow 2^{2x}-2.2^x+1=22^x+6$
$\Rightarrow 2^{2x}-4.2^x-5=0$
$\Rightarrow \left(2^x-5\right)\left(2^x+1\right)=0$
$2^x=5or{2}^x=-1$
But $2^x=-1$ is not possible.
$\therefore 2^x=5$
$\Rightarrow lo{g}_2{2}^x=lo{g}_{2}5$
$\Rightarrow xlo{g}_{2}2=lo{g}_{2}5$
$\Rightarrow x=lo{g}_{2}5\left[\because lo{g}_{2}2=1\right]$