Question

Solve $\frac{1+\log e^x}{1-\log e^x}$

Answer 1

$\frac{1+{\log }_e{e}^x}{1-{\log }_e{e}^x}=\frac{1+x{\log }_{e}e}{1-x{\log }_{e}e}$

As we know that ${\log }_{e}e=1$

so, $\frac{1+{\log }_e{e}^x}{1-{\log }_e{e}^x}$ = $\frac{1+x}{1-x}$