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

2 Answers
Nov 9, 2015

Let's start with the negative exponent.

We can appy the following power rule: #x^(-1) = 1/x# or, more general: #x^(-a) = 1 / x^a#.

So,
#-4^(-5/2) = - 1/4^(5/2)#.

As next, let's take care of the #5/2#.
First of all, the fraction can be split into #5/2 = 5 * 1/2#.

Another power rule states: #x^(a*b) = (x^a)^b#.
Here, this means that you can either compute #(4^5)^(1/2)# or #(4^(1/2))^5# instead of #4^(5/2)#.

Let's stick with #(4^(1/2))^5#.

What does #4^(1/2)# mean? #x^(1/2) = sqrt(x)#, so #4^(1/2) = sqrt(4) = 2#.
So you get #4^(5/2) = (4^(1/2))^5 = (4^(1/2))^5 = (sqrt(4))^5 = 2^5 = 32#.

In total, your solution is:

#-4^(-5/2) = - 1/4^(5/2) = - 1 / ((4^(1/2))^5) = - 1/((sqrt(4))^5) = - 1 / 2^5 = - 1 / 32#.

Nov 9, 2015

Answer:

  • 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:
#-(4^(-5/2))#
When we have to deal with negative exponents, we have a set of rules that simplyfies them. One of them is:
#a^(-n) = 1/(a^n)#
Another one, that we will use to convert exponents into square roots is:
#a^(m/n) = root(n)a^m#

So let's begin!
We already have #-(4^(-5/2))#, so we will just star by using the first rule (the one about making exponents positive).

#-(4^(-5/2)) = -1/(4^(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!

#-1/(sqrt(4^5)) = -1/(2^5) = -1/32#