Question

Find the number of positive integers satisfying the equation $x+{\log }_{10}(2^x+1)=x{\log }_{10}5+{\log }_{10}6$.

Answer 1

$x+{\log }_{10}(2^x+1)=x{\log }_{10}5+{\log }_{10}6$
$\Rightarrow x{\log }_{10}10+{\log }_{10}(2^x+1)=x{\log }_{10}5+{\log }_{10}6[\because {\log }_{a}a=1]$
$\Rightarrow {\log }_{10}{10}^x+{\log }_{10}(2^x+1)={\log }_{10}{5}^x+{\log }_{10}6[\because m\log a=\log a^m]$
$\Rightarrow {\log }_{10}{10}^x(2^x+1)={\log }_{10}{5}^{x}6[\because \log a+\log b=\log (ab\left)\right]$
$\Rightarrow 2^x{5}^x(2^x+1)=6\cdot 5^x$
$\Rightarrow {\left(2^x\right)}^2+2^x-6=0$
$\Rightarrow 2^x=2as{2}^x=-3$ is not possible
$\Rightarrow x=1$
Ans: 1