# How do you apply the exponential properties to simplify #(-8x)^3(5x)^2#?

##### 1 Answer

I don't like the attribution "exponential" that is often used to denote these properties. I find it misleading, because there are cases in which these are applied but... there's no exponential involved! For example... your case (just integer powers, not variable exponents)! Anyway, that's just a personal thought. Let's face the question!

First of all, let's recall the properties. Let

#h^0=1# #h^{-alpha}=1/h^alpha# #h^alpha h^beta=h^{alpha+beta}# #h^alpha/h^beta=h^{alpha-beta}# #h^alpha k^alpha=(hk)^alpha# #h^alpha/k^alpha=(h/k)^alpha# #(h^alpha)^beta=h^{alpha beta}#

In case

By the fifth property of the list above, we have that

By the third property

If you like "big" numbers, you can compute the numeric part:

Mathematicians often *love* prime factorizations, so another (more elegant) possibility is to express the result in the following way:

Notice that the second equality is *kindly* provided by the last property of the list above.