Question

Solve the following inequality:
${\left(\frac{1}{3}\right)}^{{\log }_{1/9}(x^8-\frac{10}{3}x+1)}⩽1.$

Answer 1

$3^{\frac{-\log (x^8-\frac{10x}{3}+1)}{-2\log 3}}\le 13^{\frac{\log (x^8-\frac{10x}{3}+1)}{2\log 3}}\le 1\frac{\log (x^8-\frac{10x}{3}+1)}{2\log 3}\le 0x^8-\frac{10x}{3}+1\le 1(x^7-\frac{10}{3})x\le 0x\in [0,{\left(\frac{10}{3}\right)}^{1/7}]$