Question

Solve the following inequality:
${\log }_{1/5}(x^2-6x+18)+2{\log }_5(x-4)<0$

Answer 1

Given expression: $-{\log }_5(x^2-6x+18)+{\log }_5(x-4{)}^2<0$

${\log }_5\frac{x^2-8x+16}{x^2-6x+18}<0$

$\frac{x^2-8x+16}{x^2-6x+18}<1x^2-8x+16<x^2-6x+18-2<2x-1<xx\in (-1,\infty )$ and $x^2-6x+18>0x\in R(x-4)>0x>4x\in (4,\infty )$

Taking intersection, we get $x\in (4,\infty )$