Question

Evaluate:

${\log }_{x^2}(x-5)\le 1$

Answer 1

${\log }_{x^2}(x-5)\le 1$

${\log }_{x^2}(x-5)\le {\log }_{x}x$

$\frac{1}{2}{\log }_x(x-5)\le {\log }_{x}x$

${\log }_x(x-5)\le 2{\log }_{x}x$

${\log }_x(x-5)\le {\log }_x{x}^2$

$x-5\le x^2$

$x^2-x+5\ge 0$

There is no real value of $x$ satisfying the property.