Question

If the product of all solutions of the equation $\frac{\left(2019\right)x}{2020}=(2019{)}^{{\log }_x\left(2020\right)}$ can be expressed in the lowest form as$\frac{m}{n}$, then the value of $m-n$ is

• A. $0$
• B. $1$
• C. $2$
• D. $5$

Answer 1

The correct option is B $1$

$\frac{\left(2019\right)x}{2020}=(2019{)}^{{\log }_x\left(2020\right)}$
$\Rightarrow {\log }_x\left(2019\right)+1-{\log }_{x}2020={\log }_{x}2020\cdot {\log }_{x}2019$
$\Rightarrow {\log }_x\left(2019\right)[1-{\log }_{x}2020]+(1-{\log }_{x}2020)=0$
$\Rightarrow ({\log }_{x}2019+1)(1-{\log }_{x}2020)=0\Rightarrow x=\frac{1}{2019}\text{(or)}x=2020\Rightarrow \text{Product}=\frac{2020}{2019}\Rightarrow m-n=1$