# 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.

## Definition

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.

## Solving Problems

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: $2^{\frac{4}{3}}=\sqrt[a]{16}?$

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

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

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: $5^{\frac{2}{3}}=\left(\sqrt[3]{5}\right)^x?$

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

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: $\sqrt[3]{8^{\frac{2}{5}}}=\left(a^3\right)^\frac{1}{15}?$

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 $8:$ $\begin{array}{c}&(a) 4^{\frac{3}{2}} &(b) \left(\sqrt{2}\right)^6 &(c) \sqrt{4^3} &(d) 64^{\frac{1}{3}}?\end{array}$

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

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

correctsolution to this problem is as follows:$\begin{aligned} \left( (-8)^{2}\right)^{\frac{3}{2}} &= (64)^{\frac{3}{2}} \\ &= \sqrt{64^3} \\ &= 8^3 \\ &= 512 \end{aligned}$

## Note on Calculators

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.*

## See Also

**Cite as:**Rational Exponents.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/calculation-rational-exponents/