Rational Exponents

Rational exponents (also called fractional exponents) are expressions with exponents that are rational numbers (as opposed to integers ).

While all the standard rules of exponents apply, it is helpful to think about rational exponents carefully.

A rational exponent is written as $x^{ \frac{a}{b}},$ with base $x$ and exponent $\frac{a}{b}$. This expression is equivalent to the $b^{\text{th}}$-root of $x$ raised to the $a^{\text{th}}$-power, or $\sqrt[b]{x^a}$. For example, $8^{\frac{2}{3}}$could be rewritten as $\sqrt[3]{8^2}.$

Note that the exponent is applied first, before the radical, and also that if the base is negative, taking roots is no longer simple, and requires Complex Number Exponentiation.

When solving questions like this, it is often easier to try to first represent the number inside the radical as an exponent. For example, $8$ can be rewritten as $2^3$. So,

$$\begin{aligned} 8^{\frac{2}{3}} &= \sqrt[3]{8^2} \\ & = \left(\sqrt[3]{8}\right)^2 \\ & = \left(\sqrt[3]{2^3}\right)^2 \\ &= 2^2 \\ &= 4 \end{aligned}$$

What is $a$ in the following equation: $$2^{\frac{4}{3}}=\sqrt[a]{16}?$$

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Since $2^{\frac{4}{3}}=\sqrt[3]{2^4}=\sqrt[3]{16},$ equating this with $\sqrt[a]{16}$ gives $a=3.$ $_\square$

Evaluate $\sqrt{125^{\frac{2}{3}}}.$

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We have $\sqrt{125^{\frac{2}{3}}}=\sqrt[2]{125^{\frac{2}{3}}}=125^{\frac{1}{2}\times\frac{2}{3}}=\left(5^3\right)^{\frac{1}{2}\times\frac{2}{3}}=\left(5^3\right)^\frac{1}{3}=5^{3\times\frac{1}{3}}=5.$ $_\square$

What is $x$ in the following equation: $$5^{\frac{2}{3}}=\left(\sqrt[3]{5}\right)^x?$$

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Since $5^{\frac{2}{3}}=\left(5^{\frac{1}{3}}\right)^2=\left(\sqrt[3]{5^1}\right)^2=\left(\sqrt[3]{5}\right)^2,$ equating this with $\left(\sqrt[3]{5}\right)^x$ gives $x=2.$ $_\square$

Evaluate $27^{-\frac{2}{3}}.$

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We have $27^{-\frac{2}{3}}=\left(3^3\right)^{-\frac{2}{3}}=3^{3\times \left(-\frac{2}{3}\right)}=3^{-2}=\left(3^{2}\right)^{-1}=\frac{1}{3^2}=\frac{1}{9}.$ $_\square$

What is $a$ in the following equation: $$\sqrt[3]{8^{\frac{2}{5}}}=\left(a^3\right)^\frac{1}{15}?$$

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Since $\sqrt[3]{8^{\frac{2}{5}}}=8^{\frac{1}{3}\times\frac{2}{5}}=8^{\frac{2}{15}}=\left(8^2\right)^{\frac{1}{15}}=\left(64\right)^{\frac{1}{15}}=\left(4^3\right)^{\frac{1}{15}}.$ Equating this with $\left(a^3\right)^\frac{1}{15}$ gives $a=4.$ $_\square$

Which of the following is NOT equal to $8:$ $$\begin{array}{c}&(a) 4^{\frac{3}{2}} &(b) \left(\sqrt{2}\right)^6 &(c) \sqrt{4^3} &(d) 64^{\frac{1}{3}}?\end{array}$$

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Since $x^{ \frac{y}{n}}=\sqrt[n]{x^y},$ thus $4^{\frac{3}{2}}=\sqrt[2]{4^3}=\sqrt{4^3}=\sqrt{64}=8.$ This implies that neither of $(a)$ and $(c)$ is the answer. Now, since $x^{ \frac{y}{n}}=\left(x^{ \frac{1}{n}}\right)^y=\left(\sqrt[n]{x}\right)^y=\sqrt[n]{x^y},$ thus $\left(\sqrt{2}\right)^6=\sqrt{2^6}=\sqrt{64}=8,$ implying that $(b)$ is not the answer, either. Finally, since $64^{\frac{1}{3}}=\left(2^6\right)^{\frac{1}{3}}=2^{6\times {\frac{1}{3}}}=2^2=4 \ne 8,$ thus the answer is $(d).$ $_\square$

Simplify $\left( (-8)^{2}\right)^{\frac{3}{2}}$.

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We might be tempted to simplify the exponents using the power rule, so that we would say $\left( (-8)^{2}\right)^{\frac{3}{2}} = (-8)^{2 \times \frac{3}{2}} = (-8)^3 = -512.$ However, this is incorrect!

We have to remember that order of operations requires us to perform operations inside parentheses before all other operations. Thus, the correct solution to this problem is as follows:

$$\begin{aligned} \left( (-8)^{2}\right)^{\frac{3}{2}} &= (64)^{\frac{3}{2}} \\ &= \sqrt{64^3} \\ &= 8^3 \\ &= 512 \end{aligned}$$

When entering an expression like the following into a calculator,

(-27)^(2/3)

the results may or may not be correct, depending on how your calculator functions. The correct answer is 9, but some calculators may return an error or complex number as a result of shortcuts in the calculator's approach to rational exponents. Be careful when using your calculator on these sorts of problems.

• Rules of Exponents
• Rational Numbers