Originally Posted by

**Soroban** Hello, cupid!

I don't understand their answers.

First, note that: .$\displaystyle x\,>\,1$

The radicand must be positive.

. . [1] The numerator and denominator are both positive, or

. . [2] The numerator and denominator are both negative.

Case [1]

$\displaystyle \log_{0.3}(x-1) \:>\:0 \quad\Rightarrow\quad x-1 \:>\:0.3^0 \quad\Rightarrow\quad x \:>\:2$

$\displaystyle (x+2)(x-4) \:>\:0 \quad\Rightarrow\quad x \,<\,\text{-}2\,\text{ or }\,x \,>\,4$

. . Hence: .$\displaystyle x\,>\,4$

Case [2]

$\displaystyle \log_{0.3}(x-1)\,\le\,0 \quad\Rightarrow\quad x-1\,\le\,0.3^0 \quad\Rightarrow\quad 1\,<\,x \,\le\,2$

$\displaystyle (x+2)(x-4)\:<\:0 \quad\Rightarrow\quad -2 \,<\,x\,<\,4$

. . Hence: .$\displaystyle 1\,<\,x\,\le\,2$

$\displaystyle \text{The domain is: }\;\left(1,2\,\right] \cup (4,\infty)$