# How do you simplify  -4^(-5/2)?

Nov 9, 2015

We can appy the following power rule: ${x}^{- 1} = \frac{1}{x}$ or, more general: ${x}^{- a} = \frac{1}{x} ^ a$.

So,
$- {4}^{- \frac{5}{2}} = - \frac{1}{4} ^ \left(\frac{5}{2}\right)$.

As next, let's take care of the $\frac{5}{2}$.
First of all, the fraction can be split into $\frac{5}{2} = 5 \cdot \frac{1}{2}$.

Another power rule states: ${x}^{a \cdot b} = {\left({x}^{a}\right)}^{b}$.
Here, this means that you can either compute ${\left({4}^{5}\right)}^{\frac{1}{2}}$ or ${\left({4}^{\frac{1}{2}}\right)}^{5}$ instead of ${4}^{\frac{5}{2}}$.

Let's stick with ${\left({4}^{\frac{1}{2}}\right)}^{5}$.

What does ${4}^{\frac{1}{2}}$ mean? ${x}^{\frac{1}{2}} = \sqrt{x}$, so ${4}^{\frac{1}{2}} = \sqrt{4} = 2$.
So you get ${4}^{\frac{5}{2}} = {\left({4}^{\frac{1}{2}}\right)}^{5} = {\left({4}^{\frac{1}{2}}\right)}^{5} = {\left(\sqrt{4}\right)}^{5} = {2}^{5} = 32$.

$- {4}^{- \frac{5}{2}} = - \frac{1}{4} ^ \left(\frac{5}{2}\right) = - \frac{1}{{\left({4}^{\frac{1}{2}}\right)}^{5}} = - \frac{1}{{\left(\sqrt{4}\right)}^{5}} = - \frac{1}{2} ^ 5 = - \frac{1}{32}$.

Nov 9, 2015
• Make the exponent positive
• A bit of algebra

#### Explanation:

Let's start off by having a look at the expression. The minus sign is not a part of the base (which is 4), so therefore, we can write the expression like this:
$- \left({4}^{- \frac{5}{2}}\right)$
When we have to deal with negative exponents, we have a set of rules that simplyfies them. One of them is:
${a}^{- n} = \frac{1}{{a}^{n}}$
Another one, that we will use to convert exponents into square roots is:
${a}^{\frac{m}{n}} = {\sqrt[n]{a}}^{m}$

So let's begin!
We already have $- \left({4}^{- \frac{5}{2}}\right)$, so we will just star by using the first rule (the one about making exponents positive).

$- \left({4}^{- \frac{5}{2}}\right) = - \frac{1}{{4}^{\frac{5}{2}}}$

Followed by the rule about exponents into square roots:

-1/4^(5/2) = -1/(root(2)4^5) = -1/(sqrt(4^5)

Just some algebra that remains!

$- \frac{1}{\sqrt{{4}^{5}}} = - \frac{1}{{2}^{5}} = - \frac{1}{32}$