Question

Solve the following inequality:
$\log {10}^{\log (x^2+21)}>1+\log x$

Answer 1

$\log {\left(10\right)}^{\log (x^2+21)}>1+\log x$

$\log (x^2+21)>(1+\log x)$$\log (x^2+21)>\log \left(10x\right)$

$x^2+21>10x$

$x^2-10x+21>0$

$xϵ(-\infty ,3)\bigcup (7,\infty )$ but $x>0$

Taking intersection $xϵ(0,3)\bigcup (7,\infty )$