Section 2.4 Reciprocal Exponents
For a positive integer \(n\text{,}\) a real number \(y\) is an \(n\)-th root of \(x\) if \(y^n=x\text{.}\)
If \(n\) is odd, every real number \(x\) has exactly one \(n\)-th root.
If \(n\) is even, a real number \(x\) has no \(n\)-th root when \(x\lt 0\text{,}\) exactly one \(n\)-th root when \(x=0\text{,}\) and exactly two \(n\)-th roots when \(x>0\text{.}\) If \(x\) has two \(n\)-th roots, one is positive and the other is its negative.
The symbol \(\sqrt[n]{x}\) is defined to be the \(n\)-th root of \(x\) when \(n\) is odd, or the non-negative \(n\)-th root of \(x\text{,}\) if one exists, when \(n\) is even.
Notice that if \(n\) is odd, \(\sqrt[n]{x}\) exists and has the same sign as \(x\text{.}\)
- \(\displaystyle \sqrt[3]{64} = 4\)
- \(\displaystyle \sqrt[3]{-64} = -4\)
- \(\displaystyle \sqrt[2]{64} = 8\)
- \(\displaystyle -\sqrt[2]{64} = (-1) \sqrt[2]{64} =-8\)
- \(\displaystyle \sqrt[2]{-64} \text{ is not defined. }\)
The expression \(x^\frac{1}{n}\) is just another way of writing \(\sqrt[n]{x}\text{.}\) Notice that if \(x^\frac{1}{n}\) is defined, then \(\left(x^\frac{1}{n}\right)^n=x\text{.}\)
If \(n\) is even and \(x\) is non-negative, then \((x^n)^\frac{1}{n} = x\text{,}\) because \(x^n\) is non-negative and \((x^n)^\frac{1}{n}\) is the non-negative \(n\)-th root of \(x^n\text{.}\) If \(n\) is even and \(x\) is negative, then \(x^n\) is positive, so that \((x^n)^\frac{1}{n}=\mid x \mid\text{.}\)
For example \(((-3)^2)^\frac{1}{2}= 9^\frac{1}{2}=3\text{.}\) Notice that, in this case, the expression \(\left((-3)^\frac{1}{2}\right)^2\) does not make sense because \((-3)^\frac{1}{2}\) is not defined.
When \(n\) is odd, \(x^n\) has the same sign as \(x\text{,}\) so that \((x^n)^\frac{1}{n}\) also has the same sign as \(x\text{.}\) Therefore \((x^n)^\frac{1}{n}=x\text{.}\) For example \(((-3)^3)^{\frac{1}{3}} = (-27)^{\frac{1}{3}}=-3\) and, since \((-3)^{\frac{1}{3}}\) is defined, \(\left((-3)^\frac{1}{3}\right)^3=-3\text{.}\)
Example 2.4.2. Finding an \(n\)-th root.
Find \(x\) if \(x^\frac{1}{4} = \frac{3}{2}\text{.}\)
Solution.
Since \(x^\frac{1}{4} = \frac{3}{2}\text{,}\) we know that
\begin{equation*}
\begin{aligned}
x &= \left(x^\frac{1}{4}\right)^4 \\
&= \left(\frac{3}{2}\right)^4 \\
&= \frac{3^4}{2^4} \\
&= \frac{81}{16}
\end{aligned}
\end{equation*}
Example 2.4.3. Finding another \(n\)-th root.
Find \(x\) if \(x^\frac{1}{3} = \left(\frac{5}{4}\right)^\frac{1}{7}\text{.}\)
Solution.
Since \(x^\frac{1}{3} = \left(\frac{5}{4}\right)^\frac{1}{7}\text{,}\) we know that
\begin{equation*}
\begin{aligned}
x &= \left(x^\frac{1}{3}\right)^{3} \\
&= \left(\left(\frac{5}{4}\right)^\frac{1}{7}\right)^{3} \\
&= \left(\sqrt[7]{\frac{5}{4}}\right)^{3} \\
&= \sqrt[7]{\frac{125}{64}}
\end{aligned}
\end{equation*}
Exercises Practice Problems
In questions 1 - 5, use the rules of exponents to find an expression equivalent to the given expression.
1.
\(36^\frac{1}{2}\)
Solution.
\(6\)
2.
\((-125)^\frac{1}{3}\)
Solution.
\(-5\)
3.
\((\frac{9}{4})^\frac{1}{2}\)
Solution.
\(\frac{3}{2}\)
4.
\(x^\frac{1}{4}x^{-\frac{1}{5}}\)
Solution.
\(x^\frac{1}{20}\)
5.
\((m^\frac{1}{2}n^{-\frac{1}{3}})^{-\frac{1}{6}}\)
Solution.
\(\frac{ n^{1/18}} { m^{1/12}}\)
6.
Find \(x\) if \(x^\frac{1}{5} = \left(\frac{2}{3}\right)^\frac{1}{2}\text{.}\)
Solution.
\(\sqrt{\frac{32}{243}} = \frac{4}{9}\sqrt{\frac{2}{3}}\)
