Question

Solve: ${\log }_{0.2}{\log }_6\left(\frac{x^2-1}{x^2+1}\right)=0$.

Answer 1

$lo{g}_{0.2}lo{g}_6\left(\frac{x^2-1}{x^2+1}\right)=0$

$lo{g}_6\left(\frac{x^2-1}{x^2+1}\right)=(0.2{)}^0=1$

$\Rightarrow \frac{x^2-1}{x^2+1}=6^1=6$

$\Rightarrow x^2-1=6x^2+6$

$\Rightarrow 6x^2-x=-1-6$

$\Rightarrow 5x^2=-7$

$\Rightarrow x^2=-7/5$

substitute $x^2=-7/5$ in $lo{g}_6\left(\frac{x^2-1}{x^2+1}\right)$

$=lo{g}_6\left(\frac{-7/5-1}{-7/5+1}\right)$

$=lo{g}_6\left(\frac{\frac{+12}{5}}{12/5}\right)=6$

=1

$\therefore x^2=-7/5$ is the solution of the above equation