# Question #11be7

Mar 19, 2017

When you are sure of the sign, eg $| {x}^{2} |$

Mar 19, 2017

This can be done when the argument of the absolute value is non-negative.

#### Explanation:

Examples:

$5 \left\mid {x}^{2} \right\mid = 5 \left({x}^{2}\right)$ $\text{ }$ (Of course, then parentheses are not needed here.)

$\log \left\mid {x}^{2} + 3 \right\mid = \log \left({x}^{2} + 3\right)$

$\sin \left\mid \sqrt{x} \right\mid = \sin \left(\sqrt{x}\right)$

In the following, the absolute value cannot be replaced:

$\left\mid {x}^{3} + 5 \right\mid$

$\left\mid \sqrt[5]{x} + 7 \right\mid$

Also

If the argument $u$ is non-positive, then $\left\mid u \right\mid = - \left(u\right)$

So $\log \left(\left\mid - {x}^{2} + 2 x - 1 \right\mid\right) = \log \left(- \left(- {x}^{2} + 2 x - 1\right)\right) = \log \left({x}^{2} - 2 x + 1\right)$