Question

Solve the following inequality:
$(0.5{)}^{{\log }_{\frac{1}{3}}\frac{x+5}{x^2+3}}>1$

Answer 1

$2^{-{\log }_{1/3}\frac{x+5}{x^2+3}}>1$

Taking ${\log }_2$ both sides

$-{\log }_{1/3}\frac{x+5}{x^2+3}>0\frac{-\left[\log \right(x+5)-\log (x^2+3\left)\right]}{-\log 3}>0\log (x+5)-\log (x^2+3)>0\log (x+5)>\log (x^2+3)x+5>x^2+3x^2-x-2<0(x-2)(x+1)<0x\in (-1,2)$

Also, $\frac{x+5}{x^2+3}>0x>-5\Rightarrow x\in (-5,\infty )$

Taking intersection, we get $x\in (-1,2)$