Question

Solve: $\left(0.5\right)$ ${\log }_{1/3}\frac{x+5}{x^2+3}$ $>$ 1

Answer 1

Given:$-{\log }_3\left(\frac{x+5}{x^2+3}\right)>2$

${\log }_3\left(\frac{x+5}{x^2+3}\right)<-2$

$0<\frac{x+5}{x^2+3}<3^{-2}$

$\frac{x+5}{x^2+3}>0$

$x>-5$

$\frac{x+5}{x^2+3}<\frac{1}{9}$

$\frac{9x+5-x^2-3}{x^2+3}<0$

$-x^2+9x+2<0$

$x^2-9x-2>0$

$x=\frac{9\pm \sqrt{81+8}}{2}=\frac{9\pm \sqrt{89}}{2}$

$x\epsilon (\frac{9-\sqrt{89}}{2},\frac{9+\sqrt{89}}{2})andx>-5$

$So,$

$x\epsilon (\frac{9-\sqrt{89}}{2},\frac{9+\sqrt{89}}{2})$