laws of exponents

The so-called laws of exponents are the algebraic rules that arise naturally out of the definition of exponentiation of numbers. These rules are mostly very simple to derive, but there is great value in studying and practicing them until they are committed to memory, as this makes simplifying, combining, and expanding algebraic expressions much quicker and easier.

Recall that to exponentiate a number \(a\) to the power \(n\) is to multiply \(a\) times itself \(n\) times:

\[a^n = \underbrace{\,a \times a \times \cdots \times a\,\,}_{\text{$n$ factors}}\]

For reasons that will become clear, we define any number but zero, when raised to the power zero, to be equal to 1:

\[a^0 = 1,\,\,\,a \neq 0\]

We leave the expression \(0^0\) undefined.

The complete set of rules is as listed below. Click the show-me-why buttons to show/hide the corresponding derivation and explanation.

\({\Large a^na^m=a^{n+m}}\) show me why


Since \(a^n\) represents \(n\) factors of \(a\) and \(a^m\) represents \(m\) factors, their product must be \(n+m\) factors of \(a\):

\[\begin{eqnarray*}a^n a^m & = & \underbrace{\underbrace{\,a \times a \times \cdots \times a\,\,}_{\text{$n$ factors}}\times\underbrace{\,a \times a \times \cdots \times a\,\,}_{\text{$m$ factors}}\,\,}_{\text{$n+m$ factors}} \\ & = & a^{n+m}\end{eqnarray*}\]

\({\Large \left(a^n\right)^m=a^{nm}}\) show me why


Since \(a^n\) represents \(n\) factors of \(a\), raising this to the \(m\)th power must represent \(n\times m\) factors of \(a\):

\[\begin{eqnarray*}\left(a^n\right)^m & = & \underbrace{\underbrace{\,a \times a \times \cdots \times a\,\,}_{\text{$n$ factors}}\times\cdots\times\underbrace{\,a \times a \times \cdots \times a\,\,}_{\text{$n$ factors}}\,\,\,}_{\text{$m$ times}} \\ & = & a^{nm}\end{eqnarray*}\]

\({\Large a^{-n}=\displaystyle\frac{1}{a^n}}\) show me why


This is by definition, and is so defined because it is needed to make the other rules work properly. For instance, according to the rules already defined, \(a^na^{-n}=a^{n-n}=a^0=1\), but also \(a^na^{-n}=\displaystyle\frac{a^n}{a^n}=1\).

\({\Large (ab)^n=a^nb^n}\) show me why


This is by commutivity of multiplication:

\[\begin{eqnarray*}(ab)^n & = & \underbrace{\,ab \times ab \times \cdots \times ab\,\,\,}_{\text{$n$ factors}} \\ & = & \underbrace{\,a \times a \times \cdots \times a\,\,\,}_{\text{$n$ factors}}\,\times\underbrace{\,b \times b \times \cdots \times b\,\,}_{\text{$n$ factors}} \\ & = & a^nb^n\end{eqnarray*}\]

\({\Large \displaystyle\frac{a^n}{a^m}=a^{n-m}}\) show me why


This follows immediately from the previous rules:

\[\begin{eqnarray*}\displaystyle\frac{a^n}{a^m} & = & a^n\displaystyle\frac{1}{a^m} \\ & & \\ & = & a^na^{-m} \\ & & \\ & = & a^{n-m}\end{eqnarray*}\]

\({\Large \left(\displaystyle\frac{a}{b}\right)^n=\displaystyle\frac{a^n}{b^n}}\) show me why


By the previous rules:

\[\begin{eqnarray*}\left(\displaystyle\frac{a}{b}\right)^n & = & \left(a \times b^{-1}\right)^n \\ & &\\ & = & a^n \times \left(b^{-1}\right)^n \\ & &\\ & = & a^n \times b^{-n} \\ & & \\ & = & \displaystyle\frac{a^n}{b^n}\end{eqnarray*}\]

\({\Large a^{\frac{p}{q}}=\sqrt[q]{a^p}=\left(\sqrt[q]{a}\right)^p}\) show me why


We observe that since we may write 1 as \(\frac{n}{n}\) for any non-zero \(n\), we have that

\[a = a^1 = a^{\frac{n}{n}} = \left(a^{\frac{1}{n}}\right)^n\]

and consequently \(a^{\frac{1}{n}}\) must represent the \(n\)th root of \(a\). Since also

\[a^{\frac{p}{q}} = \left(a^{\frac{1}{q}}\right)^p = \left(a^p\right)^{\frac{1}{q}}\]

we have that \(a^{\frac{p}{q}}\) is the \(q\)th root of \(a^p\), or equivalently the \(p\)th power of \(a^{\frac{1}{q}}\).

Since any exponential expression can be written as an equivalent logarithmic expression, these rules have their counterparts in the laws of logarithms, which any good student of mathematics should also seek to master.