Solving inequlity with $e^x$

Question

I'm studying differential calculus, but one of the questions involves solving an inequality:

$$(x-2)e^x < 0$$

I intend to go deeper in solving inequalities later, but I just want to understand how the teacher got the following solution in order to advance in these lectures: $$x-2 < 0$$ $$x < 2$$

Where did the $e^x$ go? There's some rule to solve these inequalities involving $e$?

Answer 1

For all values of $x$, $e^x>0$ is true. This means that $a\cdot e^x > 0$ is true if and only if $a>0$.

Answer 2

Recall that for $a > 0$

$$ab < ac \iff b < c$$

Note that $e^x > 0 \ \forall x \in \mathbb R$

Answer 3

Divide by $e^x$

$$(x-2)e^x < 0$$ $$\iff x - 2 < 0$$

This is valid since $e^x > 0$ $\forall x \in \mathbb{R}$