Question

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

Answer 1

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,k>0` and `alpha,beta in mathbb{R}`, then:

• `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 `alpha,beta in mathbb{Z}`, then the properties above are valid for all `h,k in mathbb{R}` and not only for the positive ones. This is actually your case: you have powers of an unknown value `x`, which might be negative (there are no restrictions on `x`).

By the fifth property of the list above, we have that `(-8x)^3=(-8)^3 x^3=[(-1)^3 8^3] x^3=-8^3 x^3` and `(5x)^2=5^2 x^2`. If we put the two pieces together we get `(-8x)^3 (5x)^2=(-8^3 x^3) (5^2 x^2)=-8^3 5^2 (x^3 x^2)`.
By the third property `x^3 x^2=x^{3+2}=x^5`. So in the end we get
`(-8x)^3 (5x)^2=-8^3 5^2 x^5`.

If you like "big" numbers, you can compute the numeric part:
`-8^3 5^2 x^5=-12800 x^5`

Mathematicians often love prime factorizations, so another (more elegant) possibility is to express the result in the following way:
`-8^3 5^2 x^5=-(2^3)^3 5^2 x^5=-2^{3*3} 5^2 x^5=-2^9 5^2 x^5`.
Notice that the second equality is kindly provided by the last property of the list above.