Question

Number of solution(s) of the equation $1+{\log }_x\left(\frac{4-x}{10}\right)=({\log }_{10}{\log }_{10}n-1){\log }_{x}10$ for a given value of $n\in (1,{10}^4)-\left\{{10}^3\right\}$ is

Answer 1

The domain: $x\in (0,4)-\left\{1\right\}$
$1+\frac{{\log }_{10}\left(\frac{4-x}{10}\right)}{{\log }_{10}x}={\log }_{10}\left(\frac{{\log }_{10}n}{10}\right).\frac{1}{{\log }_{10}x}$
${\log }_{10}(x.\frac{4-x}{10})={\log }_{10}\left(\frac{{\log }_{10}n}{10}\right)$
$x^2-4x+{\log }_{10}n=0$
$x=2\pm \sqrt{4-{\log }_{10}n}$
So if $1<n<{10}^4,n\ne {10}^3$
equation has two different roots
$x_1=2+\sqrt{4-{\log }_{10}n},x_2=2-\sqrt{4-{\log }_{10}n}$ for all the values of $x$ in the domain.