# How do you simplify (a^4)^-5 * a^13?

May 17, 2015

By definition, ${x}^{- N} = \frac{1}{{x}^{N}}$.
Therefore,
${\left({a}^{4}\right)}^{- 5} = \frac{1}{{\left({a}^{4}\right)}^{5}}$

By definition. ${\left({x}^{M}\right)}^{N} = \left({x}^{M}\right) \cdot \left({x}^{M}\right) \cdot \left({x}^{M}\right) \ldots$
(where multiplication is performed $N$ times)
But each ${x}^{M} = x \cdot x \cdot x \ldots$
(where multiplication is performed $M$ times)
Therefore,
${\left({x}^{M}\right)}^{N} = x \cdot x \cdot x \cdot \ldots$
(where multiplication is performed $M \cdot N$ times).
Hence, ${\left({x}^{M}\right)}^{N} = {x}^{M \cdot N}$

Using the above, we can write:
${\left({a}^{4}\right)}^{- 5} = \frac{1}{{\left({a}^{4}\right)}^{5}} = \frac{1}{a} ^ \left(4 \cdot 5\right) = \frac{1}{a} ^ 20$

The original expression, therefore, equals to
${\left({a}^{4}\right)}^{- 5} \cdot {a}^{13} = {a}^{13} / {a}^{20} = \frac{1}{a} ^ 7 = {a}^{- 7}$