Question

If ${\log }_{10}(2^x+1)+2x={\log }_{10}6+x{\log }_{10}50$, then the value of $x$ is

Answer 1

${\log }_{10}(2^x+1)+2x={\log }_{10}6+x{\log }_{10}50$
$\Rightarrow {\log }_{10}(2^x+1)+2x-x{\log }_{10}\left(\frac{100}{2}\right)={\log }_{10}6$
$\Rightarrow {\log }_{10}(2^x+1)+2x-x({\log }_{10}100-{\log }_{10}2)={\log }_{10}6$
$\Rightarrow {\log }_{10}(2^x+1)+2x-2x+{\log }_{10}{2}^x={\log }_{10}6$
$\Rightarrow {\log }_{10}\left[\right(2^x+1\left)2^x\right]={\log }_{10}6$
$\Rightarrow 2^x(2^x+1)=6$
$\Rightarrow (2^x{)}^2+2^x-6=0$
$\Rightarrow (2^x+3)(2^x-2)=0$
$\Rightarrow 2^x-2=0$ as $2^x+3\ne 0$
$\Rightarrow x=1$