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

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 $\alpha , \beta \in m a t h \boldsymbol{R}$, then:

• ${h}^{0} = 1$
• ${h}^{- \alpha} = \frac{1}{h} ^ \alpha$
• ${h}^{\alpha} {h}^{\beta} = {h}^{\alpha + \beta}$
• ${h}^{\alpha} / {h}^{\beta} = {h}^{\alpha - \beta}$
• ${h}^{\alpha} {k}^{\alpha} = {\left(h k\right)}^{\alpha}$
• ${h}^{\alpha} / {k}^{\alpha} = {\left(\frac{h}{k}\right)}^{\alpha}$
• ${\left({h}^{\alpha}\right)}^{\beta} = {h}^{\alpha \beta}$

In case $\alpha , \beta \in m a t h \boldsymbol{Z}$, then the properties above are valid for all $h , k \in m a t h \boldsymbol{R}$ 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 ${\left(- 8 x\right)}^{3} = {\left(- 8\right)}^{3} {x}^{3} = \left[{\left(- 1\right)}^{3} {8}^{3}\right] {x}^{3} = - {8}^{3} {x}^{3}$ and ${\left(5 x\right)}^{2} = {5}^{2} {x}^{2}$. If we put the two pieces together we get ${\left(- 8 x\right)}^{3} {\left(5 x\right)}^{2} = \left(- {8}^{3} {x}^{3}\right) \left({5}^{2} {x}^{2}\right) = - {8}^{3} {5}^{2} \left({x}^{3} {x}^{2}\right)$.
By the third property ${x}^{3} {x}^{2} = {x}^{3 + 2} = {x}^{5}$. So in the end we get
${\left(- 8 x\right)}^{3} {\left(5 x\right)}^{2} = - {8}^{3} {5}^{2} {x}^{5}$.

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:
$- {8}^{3} {5}^{2} {x}^{5} = - {\left({2}^{3}\right)}^{3} {5}^{2} {x}^{5} = - {2}^{3 \cdot 3} {5}^{2} {x}^{5} = - {2}^{9} {5}^{2} {x}^{5}$.
Notice that the second equality is kindly provided by the last property of the list above.