- By definition, \(9^\frac{1}{2}\) is the positive real number \(y\) such that \(y^2=9^1\text{.}\) Because \(3^2 = 9^1\text{,}\) we conclude that \(9^\frac{1}{2}=3\text{,}\) just as in Section 2.4.
- Similarly, \(8^\frac{2}{3}\) is the positive real number \(y\) such that \(y^3=8^2\text{.}\) Because \(4^3 = 64 = 8^2\text{,}\) we conclude that \(8^\frac{2}{3}=4\text{.}\)
Section 3.2 Rational Exponents
For a positive real number \(x\) and positive integers \(m\) and \(n\text{,}\) we define \(x^{\frac{m}{n}}\) to be the positive real number \(y\) such that \(y^n = x^m\text{,}\) if such a number \(y\) exists. Notice that this coincides with the definition of reciprocal exponents \(\frac{1}{n}\) when \(m = 1\text{.}\) We define \(0^\frac{m}{n}\) to be \(0\text{.}\)
Example 3.2.1. Working out Rational Exponents.
It follows from this definition above, and those in Section 2.4 that for positive \(x\)
\begin{equation*}
x^\frac{m}{n} = \left(x^m\right)^\frac{1}{n}\text{.}
\end{equation*}
It is more involved to demonstrate carefully, but it is also true that
\begin{equation*}
x^\frac{m}{n} = \left(x^\frac{1}{n}\right)^m.
\end{equation*}
Example 3.2.2. Rational Exponents Read Both Ways.
Consider the number \(9^\frac{3}{2}\text{,}\) which is the positive real number \(y\) such that \(y^2=9^3\text{,}\) or equivalently \(y^2=729\text{.}\) This is precisely the definition of \(729^\frac{1}{2}\text{,}\) and so we have shown that \(9^\frac{3}{2}=\left(9^3\right)^\frac{1}{2}\text{.}\)
If we want to work with \(9^\frac{3}{2}\text{,}\) we might find it useful to think of it as \(\left(9^3\right)^\frac{1}{2}=729^\frac{1}{2}=\sqrt{729} = 27\text{.}\) On the other hand, we might prefer to think of it as \(\left(9^\frac{1}{2}\right)^3=\left(\sqrt{9}\right)^3 = 3^3=27\text{.}\) Which one you prefer probably depends on whether you recognized \(\sqrt{729}\text{.}\)
Notice that \(x^\frac{m}{n}=\left(x^\frac{1}{n}\right)^m\) is consistent with the powers of powers rule. In fact, all of the familiar rules of exponents apply to expressions involving rational exponents, provided that the base is positive. When exponential expressions with the same base are multiplied, the exponents are added, as in:
\begin{equation*}
2^\frac{2}{3} 2^\frac{4}{5} = 2^{\frac{2}{3} + \frac{4}{5}} = 2^\frac{22}{15}
\end{equation*}
When an exponential expression is raised to a power, the exponents are multiplied:
\begin{equation*}
\left(3^\frac{3}{4}\right)^\frac{2}{5} = 3^\frac{6}{20}\text{.}
\end{equation*}
When a rational exponent is not in lowest terms, as in the above example, we can reduce it:
\begin{equation*}
3^\frac{6}{20}=3^\frac{3}{10}\text{.}
\end{equation*}
Finally, negative rational exponents behave as expected as long as the base is positive. Following the negative exponents rule:
\begin{equation*}
\begin{aligned}
x^{-\frac{m}{n}} &= \left(x^{-1}\right)^\frac{m}{n} \\
&= \left(\frac{1}{x}\right)^\frac{m}{n} \\
&= \left(\left(\frac{1}{x}\right)^m\right)^\frac{1}{n} \\
&= \left(\frac{1}{x^m}\right)^\frac{1}{n}
&= \frac{1}{x^\frac{m}{n}}
\end{aligned}
\end{equation*}
when \(x^{\frac{m}{n}}\) is defined and not \(0\text{.}\)
Example 3.2.3. The problem with negative bases.
Why have we been so careful to exclude negative bases? The short answer is that no definition gives us everything we want from them. If you are intereseted in the long answer, here it is - but you do not need to go this far and should not let this digression cause you confusion!
Consider as an example \(\left(-3\right)^\frac{6}{2}\text{.}\) If we want everything we have said so far to be true, then this number should be the same as
\begin{equation*}
\left(\left(-3\right)^6\right)^\frac{1}{2} = \left(729\right)^\frac{1}{2}=27\text{.}
\end{equation*}
On the other hand, however, it should also be the same as
\begin{equation*}
\left(\left(-3\right)^\frac{1}{2}\right)^6,
\end{equation*}
which is undefined because \((-3)^\frac{1}{2}\) is not a real number.
Maybe we could change our definition to attempt to address this problem? Perhaps we should decide that \(x^\frac{m}{n}\) is defined to be \(\left(x^\frac{1}{n}\right)^m\text{.}\) Sadly, this alternative introduces a new problem: \(\left(\left(-3\right)^\frac{1}{2}\right)^6\) is undefined, but \(\left(-3\right)^3=-27\text{;}\) we have lost the ability to reduce exponents to lowest terms.
Perhaps we should decide that \(x^\frac{m}{n}\) is defined to be \(\left(x^m\right)^\frac{1}{n}\text{?}\) Unfortunately, we have the same problem: \(\left(\left(-3\right)^6\right)^\frac{1}{2}=\left(729\right)^\frac{1}{2}=27\text{,}\) but \(\left(-3\right)^3=-27\text{.}\)
We are now equipped to work with any sort of rational exponent as long as the base is not negative.
Example 3.2.4. Working with Rational Exponents.
- Look at \(8^\frac{2}{3} = (8^2)^\frac{1}{3}=(64)^\frac{1}{3}=4\text{.}\) Because \(8^\frac{1}{3}\) is defined, we can also compute \((8^\frac{1}{3})^2 = 2^2 = 4\text{.}\)
- Look at \(625^\frac{3}{4} = (625^3)^\frac{1}{4}=(244140625)^\frac{1}{4}=125\text{.}\) Because \(625^\frac{1}{4}\) is defined, we can also compute \((625^\frac{1}{4})^3 = 5^3 = 125\text{,}\) which is arguably an easier calculation.
- Look at \(25^{-\frac{3}{2}} = \left(\frac{1}{25}\right)^\frac{3}{2} = \left(\left(\frac{1}{25}\right)^\frac{1}{2}\right)^3 = \left(\frac{1}{5}\right)^3 = \frac{1}{125} \text{.}\)
- Look at \(\left(\frac{144}{81}\right)^\frac{4}{8} = \left(\frac{144}{81}\right)^\frac{1}{2}=\frac{144^\frac{1}{2}}{25^\frac{1}{2}}=\frac{12}{5}.\)
Example 3.2.5. Finding a Base Given a Rational Exponent.
Find \(x\) if \(x^\frac{3}{5} = -7\text{.}\)
Solution.
Since \(x^\frac{3}{5} = -7\text{,}\) we have that \(x^3 = \left(x^\frac{3}{5}\right)^5 = (-7)^5\text{,}\) so that
\begin{equation*}
x = \left(x^3\right)^\frac{1}{3} = \left((-7)^5\right)^\frac{1}{3} = (-7)^\frac{5}{3}
\end{equation*}
Exercises Practice Problems
In questions 1 to 5, find a simpler form of each expression.
1.
\(-16^\frac{3}{2}\)Answer.
\(-64\)
2.
\((\frac{8}{343})^{-\frac{2}{3}}\)Answer.
\(\frac{49}{4}\)
3.
\((27)^\frac{5}{3}\)Answer.
\(243\)
4.
\((6)^{\frac{6}{2}}\)Answer.
\(216\)
5.
\((\frac{32}{243})^\frac{2}{5}\)Answer.
\(\frac{4}{9}\)
6.
Find \(x\) if \(x^{\frac{7}{4}} = 9\text{.}\)
Answer.
\(9^\frac{4}{7} = 3^\frac{8}{7}\)
