Question

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

Answer 1

Given equation:
$1+{\log }_x\left(\frac{4-x}{10}\right)=({\log }_{10}{\log }_{10}n-1){\log }_{x}10$
For the $\log$ to be defined,
$\frac{4-x}{10}>0;x>0;x\ne 1\Rightarrow x<4\Rightarrow x\in (0,4)-\left\{1\right\}\cdots \left(1\right)$

Now,
$1+{\log }_x\left(\frac{4-x}{10}\right)=({\log }_{10}{\log }_{10}n-1){\log }_{x}10\Rightarrow 1+{\log }_x(4-x)-{\log }_{x}10={\log }_{x}10\left({\log }_{10}{\log }_{10}n\right)-{\log }_{x}10\Rightarrow {\log }_x\left(x\right(4-x\left)\right)={\log }_{x}10\left({\log }_{10}{\log }_{10}n\right)\Rightarrow {\log }_{10}(4x-x^2)={\log }_{10}{\log }_{10}n\Rightarrow 4x-x^2={\log }_{10}n\Rightarrow x^2-4x=-{\log }_{10}n\Rightarrow (x-2{)}^2=4-{\log }_{10}n\Rightarrow x=2\pm \sqrt{4-{\log }_{10}n}$

When $n\in (1,{10}^4)-\left\{{10}^3\right\}$, so
${\log }_{10}n\in (0,4)-\left\{3\right\}\Rightarrow 4-{\log }_{10}n\in (0,4)-\left\{1\right\}\Rightarrow \sqrt{4-{\log }_{10}n}\in (0,2)-\left\{1\right\}$

For a given value of $n\in (1,{10}^4)-\left\{{10}^3\right\}$, we get two solutions as
$x_1=2+\sqrt{4-{\log }_{10}n}{x}_2=2-\sqrt{4-{\log }_{10}n}$