Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. If $ log2, log\left(2^{x} -1\right) $and $log\left(2^{x}+3\right)$ are in $A.P$., then the value of $x$ is

Sequences and Series

Solution:

$2\, log \left(2^{x} -1\right) = log\, 2 +log \left(2^{x} +3\right)$
$ \Rightarrow \left(2^{x}-1\right)^{2} = 2\cdot\left(2^{x} +3\right)$
$\Rightarrow \left(2^{x}\right)^{2} -4\cdot2^{x} -5 =0$
$\Rightarrow \left(2^{x}-5\right)\left(2^{x}+1\right) = 0$
$\Rightarrow x= log_{2}\,5$, as $2^{x} +1\ne0$