ORCCA Radical Expressions and Rational Exponents

Section14.2Radical Expressions and Rational Exponents

Recall that in Remark 14.1.22, we learned to calculate the cube root of a number, say \(\sqrt[3]{8}\text{,}\) we can type 8^(1/3) into a calculator. This suggests that \(\sqrt[3]{8}=8^{\sfrac{1}{3}}\text{.}\) In this section, we will learn why this is true, and how to simplify expressions with rational exponents.

Many learners will find a review of exponent rules to be helpful before continuing with the current section. Section 2.7 covers an introduction to exponent rules, and there is more in Section 6.1. The basic rules are summarized in List 6.1.15. These rules are still true and we can use them throughout this section whenever they might help.

Figure14.2.1Alternative Video Lesson

Subsection14.2.1Radical Expressions and Rational Exponents

Compare the following calculations:

\begin{align*} \sqrt{9}\cdot\sqrt{9}\amp=3\cdot3\amp9^{\sfrac{1}{2}}\cdot9^{\sfrac{1}{2}}\amp=9^{\sfrac{1}{2}+\sfrac{1}{2}}\\ \amp=9\amp\amp=9^{1}\\ \amp\amp\amp=9 \end{align*}

If we rewrite the above calculations with exponents, we have:

\begin{align*} \left(\sqrt{9}\right)^2\amp=9\amp\left(9^{\sfrac{1}{2}}\right)^2\amp=9 \end{align*}

Since \(\sqrt{9}\) and \(9^{\sfrac{1}{2}}\) are both positive, and squaring either of them generates the same number, we conclude that

\begin{equation*} \sqrt{9}=9^{\sfrac{1}{2}} \end{equation*}

We can verify this result by entering 9^(1/2) into a calculator, and we get 3. In general for any non-negative real number \(a\text{,}\) we have:

\begin{equation*} \sqrt{a}=a^{\sfrac{1}{2}} \end{equation*}

Similarly, when \(a\) is non-negative we can prove:

\begin{align*} \sqrt[3]{a}\amp=a^{\sfrac{1}{3}}\amp\sqrt[4]{a}\amp=a^{\sfrac{1}{4}}\amp\sqrt[5]{a}\amp=a^{\sfrac{1}{5}}\amp\cdots \end{align*}

Let's summarize this information with a new exponent rule.

Fact14.2.2Radicals and Rational Exponents Rule

If \(n\) is any natural number, and \(a\) is any non-negative real number or function with non-negative outputs, then

\begin{equation*} a^{\sfrac{1}{n}}=\sqrt[n]{a}\text{.} \end{equation*}

Additionally, if \(n\) is an odd natural number, then even when \(a\) is negative, we still have \(a^{\sfrac{1}{n}}=\sqrt[n]{a}\text{.}\)

Warning14.2.3Exponents on Negative Bases

Some computers and calculators follow different conventions when there is an exponent on a negative base. To see an example of this, visit WolframAlpha and try entering cuberoot(-8), and then try (-8)^(1/3), and you will get different results. cuberoot(-8) will come out as \(-2\text{,}\) but (-8)^(1/3) will come out as a certain non-real complex number. Most likely, the graphing technology you are using does behave as in Fact 14.2.2, but you should confirm this.

With this relationship, we can re-write radical expressions as expressions with rational exponents.

Example14.2.4

Evaluate \(\sqrt[4]{9}\) with a calculator. Round your answer to two decimal places.

Since \(\sqrt[4]{9}=9^{\sfrac{1}{4}}\text{,}\) we press the following buttons on a calculator to get the value: 9^(1/4). So, we see that \(\sqrt[4]{9}\approx1.73\text{.}\)

For many examples that follow, we will not need a calculator. We will, however, need to recognize the roots in Table 14.2.5.

Square Roots	Cube Roots	\(4^{th}\)-Roots	\(5^{th}\)-Roots	Roots of Powers of \(2\)
\(\sqrt{1}=1\)	\(\sqrt[3]{1}=1\)	\(\sqrt[4]{1}=1\)	\(\sqrt[5]{1}=1\)
\(\sqrt{4}=2\)	\(\sqrt[3]{8}=2\)	\(\sqrt[4]{16}=2\)	\(\sqrt[5]{32}=2\)	\(\sqrt{4}=2\)
\(\sqrt{9}=3\)	\(\sqrt[3]{27}=3\)	\(\sqrt[4]{81}=3\)		\(\sqrt[3]{8}=2\)
\(\sqrt{16}=4\)	\(\sqrt[3]{64}=4\)			\(\sqrt[4]{16}=2\)
\(\sqrt{25}=5\)	\(\sqrt[3]{125}=5\)			\(\sqrt[5]{32}=2\)
\(\sqrt{36}=6\)				\(\sqrt[6]{64}=2\)
\(\sqrt{49}=7\)				\(\sqrt[7]{128}=2\)
\(\sqrt{64}=8\)				\(\sqrt[8]{256}=2\)
\(\sqrt{81}=9\)				\(\sqrt[9]{512}=2\)
\(\sqrt{100}=10\)				\(\sqrt[10]{1024}=2\)
\(\sqrt{121}=11\)
\(\sqrt{144}=12\)

Table14.2.5Small Roots of Appropriate Natural Numbers

Example14.2.6

Convert the radical expression \(\sqrt[3]{5}\) into an expression with a rational exponent and simplify it if possible.

\(\sqrt[3]{5}=5^{\sfrac{1}{3}}\text{.}\) No simplification is possible since the cube root of \(5\) is not a perfect integer appearing in Table 14.2.5.

Example14.2.7

Write the expressions in radical form using Fact 14.2.2 and simplify the results.

\(4^{\sfrac{1}{2}}\)
\((-9)^{\sfrac{1}{2}}\)
\(-16^{\sfrac{1}{4}}\)
\(64^{-\sfrac{1}{3}}\)
\((-27)^{\sfrac{1}{3}}\)
\(3^{\sfrac{1}{2}}\cdot3^{\sfrac{1}{2}}\)
\(12^0\)

Solution

\(\begin{aligned}[t] 4^{\sfrac{1}{2}}\amp=\sqrt{4}\\ \amp=2 \end{aligned}\)
\(\begin{aligned}[t] (-9)^{\sfrac{1}{2}}\amp=\sqrt{-9} \end{aligned}\)This value is non-real.
Without parentheses around \(-16\text{,}\) the negative sign in this problem should be left out of the radical.

\(\begin{aligned}[t] -16^{\sfrac{1}{4}}\amp=-\sqrt[4]{16}\\ \amp=-2 \end{aligned}\)
\(\begin{aligned}[t] 64^{-\sfrac{1}{3}}\amp=\frac{1}{64^{\sfrac{1}{3}}}\\ \amp=\frac{1}{\sqrt[3]{64}}\\ \amp=\frac{1}{4} \end{aligned}\)
\(\begin{aligned}[t] (-27)^{\sfrac{1}{3}}\amp=\sqrt[3]{-27}\\ \amp=-3 \end{aligned}\)
\(\begin{aligned}[t] 3^{\sfrac{1}{2}}\cdot3^{\sfrac{1}{2}}\amp=\sqrt{3}\cdot\sqrt{3}\\ \amp=\sqrt{3\cdot 3}\\ \amp=\sqrt{9}\\ \amp=3 \end{aligned}\)
\(\begin{aligned}[t] 12^0\amp=1 \end{aligned}\)

Fact 14.2.2 applies to variables in expressions just as much as it does to numbers.

Remark14.2.8

In general, it is easier to do algebra with rational exponents on variables than with radicals of variables. You should use Fact 14.2.2 to convert from rational exponents to radicals on variables only as a last step in simplifying.

Example14.2.9

Write the expressions as simplified as they can be using radicals.

\(2x^{-\sfrac{1}{2}}\)
\((5x)^{\sfrac{1}{3}}\)
\(\left(-27x^{12}\right)^{\sfrac{1}{3}}\)
\(\left(\frac{16x}{81y^8}\right)^{\sfrac{1}{4}}\)

Solution

Note that in this example the exponent is only applied to the \(x\text{.}\) Making this type of observation should be our first step for each of these exercises.

\begin{align*} 2x^{-\sfrac{1}{2}}\amp=\frac{2}{x^{\sfrac{1}{2}}} \amp\amp\text{by the }\knowl{./knowl/item-exponent-rules-negative-exponent-definition.html}{\text{Negative Exponent Rule}}\\ \amp=\frac{2}{\sqrt{x}} \amp\amp\text{by the } \knowl{./knowl/fact-rational-exponent-rule-i.html}{\text{Radicals and Rational Exponents Rule}} \end{align*}
In this exercise, the exponent applies to both the \(5\) and \(x\text{.}\)

\begin{align*} (5x)^{\sfrac{1}{3}}\amp=\sqrt[3]{5x} \amp\amp\text{by the } \knowl{./knowl/fact-rational-exponent-rule-i.html}{\text{Radicals and Rational Exponents Rule}} \end{align*}

We could choose to simplify our answer in a different way. Note that neither one is technically preferred over the other except that perhaps the first way is simpler.

\begin{align*} (5x)^{\sfrac{1}{3}}\amp=5^{\sfrac{1}{3}}x^{\sfrac{1}{3}} \amp\amp\text{by the } \knowl{./knowl/item-exponent-rules-product-to-a-power.html}{\text{Product to a Power Rule}}\\ \amp=\sqrt[3]{5}\sqrt[3]{x} \amp\amp\text{by the } \knowl{./knowl/fact-rational-exponent-rule-i.html}{\text{Radicals and Rational Exponents Rule}} \end{align*}
As in the previous exercise, we have a choice as to how to simplify this expression. Here we should note that we do know what the cube root of \(-27\) is, so we will take the path to splitting up the expression, using the Product to a Power Rule, before applying the root.

\begin{align*} \left(-27x^{12}\right)^{\sfrac{1}{3}}\amp=(-27)^{\sfrac{1}{3}}\cdot \left(x^{12}\right)^{\sfrac{1}{3}}\amp\amp\text{by the } \knowl{./knowl/item-exponent-rules-product-to-a-power.html}{\text{Product to a Power Rule}}\\ \amp=(-27)^{\sfrac{1}{3}}\cdot \left(x^{12\cdot\sfrac{1}{3}}\right) \amp\amp\text{by the } \knowl{./knowl/item-exponent-rules-power-to-a-power.html}{\text{Power to a Power Rule}}\\ \amp=\sqrt[3]{-27}\cdot x^4 \amp\amp\text{by the } \knowl{./knowl/fact-rational-exponent-rule-i.html}{\text{Radicals and Rational Exponents Rule}}\\ \amp=-3x^4 \end{align*}
We'll use the exponent rule for a fraction raised to a power.

\begin{align*} \left(\frac{16x}{81y^8}\right)^{\sfrac{1}{4}}\amp=\frac{\left(16x\right)^{\sfrac{1}{4}}}{\left(81y^8\right)^{\sfrac{1}{4}}} \amp\amp\text{by the } \knowl{./knowl/item-exponent-rules-quotient-to-a-power.html}{\text{Quotient to a Power Rule}}\\ \amp=\frac{16^{\sfrac{1}{4}}\cdot x^{\sfrac{1}{4}}}{81^{\sfrac{1}{4}} \cdot \left(y^8\right)^{\sfrac{1}{4}}} \amp\amp\text{by the } \knowl{./knowl/item-exponent-rules-product-to-a-power.html}{\text{Product to a Power Rule}}\\ \amp=\frac{16^{\sfrac{1}{4}}\cdot x^{\sfrac{1}{4}}}{81^{\sfrac{1}{4}}\cdot y^2}\\ \amp=\frac{\sqrt[4]{16}\cdot \sqrt[4]{x}}{\sqrt[4]{81}\cdot y^2} \amp\amp\text{by the } \knowl{./knowl/fact-rational-exponent-rule-i.html}{\text{Radicals and Rational Exponents Rule}}\\ \amp=\frac{2\sqrt[4]{x}}{3y^2} \end{align*}

Fact 14.2.2 describes what can be done when there is a fractional exponent and the numerator is a \(1\text{.}\) The numerator doesn't have to be a \(1\) though and we need guidance for that situation.

Fact14.2.10Radicals and Rational Exponents Rule

If \(m\) and \(n\) are natural numbers such that \(\frac{m}{n}\) is a reduced fraction, and \(a\) is any non-negative real number or function that takes non-negative values, then

\begin{equation*} a^{\sfrac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m\text{.} \end{equation*}

Additionally, if \(n\) is an odd natural number, then even when \(a\) is negative, we still have \(a^{\sfrac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m\text{.}\)

Remark14.2.11

By Fact 14.2.10, there are two ways to express \(a^{\sfrac{m}{n}}\) as a radical, both

\begin{align*} a^{\sfrac{m}{n}}\amp=\sqrt[n]{a^m}\amp\amp\text{and}\amp a^{\sfrac{m}{n}}\amp=\left(\sqrt[n]{a}\right)^m \end{align*}

There are different times to use each formula. In general, use \(a^{\sfrac{m}{n}}=\sqrt[n]{a^m}\) for variables and \(a^{\sfrac{m}{n}}=\left(\sqrt[n]{a}\right)^m\) for numbers.

Example14.2.12

Consider the expression \(27^{\sfrac{4}{3}}\text{.}\) Use both versions of Fact 14.2.10 to explain part of Remark 14.2.11.
Consider the expression \(x^{\sfrac{4}{3}}\text{.}\) Use both versions of Fact 14.2.10 to explain the other part of Remark 14.2.11.

Solution

The expression \(27^{\sfrac{4}{3}}\) can be evaluated in the following two ways by Fact 14.2.10.

\begin{align*} 27^{\sfrac{4}{3}}\amp=\sqrt[3]{27^4}\amp\amp\text{by the first part of the } \knowl{./knowl/fact-rational-exponent-rule-ii.html}{\text{Radicals and Rational Exponents Rule}}\\ \amp=\sqrt[3]{531441}\\ \amp=81\\ \amp\amp\text{or}\\ 27^{\sfrac{4}{3}}\amp=\left(\sqrt[3]{27}\right)^4\amp\amp\text{by the second part of the } \knowl{./knowl/fact-rational-exponent-rule-ii.html}{\text{Radicals and Rational Exponents Rule}}\\ \amp=3^4\\ \amp=81 \end{align*}

The calculations using \(a^{\sfrac{m}{n}}=\left(\sqrt[n]{a}\right)^m\) worked with smaller numbers and can be done without a calculator. This is why we made the general recommendation in Remark 14.2.11.
The expression \(x^{\sfrac{4}{3}}\) can be evaluated in the following two ways by Fact 14.2.10.

\begin{align*} x^{\sfrac{4}{3}}\amp=\sqrt[3]{x^4}\amp\amp\text{by the first part of } \knowl{./knowl/fact-rational-exponent-rule-ii.html}{\text{Radicals and Rational Exponents Rule}}\\ \amp\amp\text{or}\\ x^{\sfrac{4}{3}}\amp=\left(\sqrt[3]{x}\right)^4\amp\amp\text{by the second part of the } \knowl{./knowl/fact-rational-exponent-rule-ii.html}{\text{Radicals and Rational Exponents Rule}} \end{align*}

In this case, the simplification using \(a^{\sfrac{m}{n}}=\sqrt[n]{a^m}\) is just shorter looking and easier to write. This is why we made the general recommendation in Remark 14.2.11.

Example14.2.13

Simplify the expressions using Fact 14.2.10.

\(8^{\sfrac{2}{3}}\)
\(16^{-\sfrac{3}{2}}\)
\(-16^{\sfrac{3}{4}}\)
\(\left(-\frac{27}{64}\right)^{\sfrac{2}{3}}\)

Solution

\begin{align*} 8^{\sfrac{2}{3}}\amp=\left(\sqrt[3]{8}\right)^2\amp\amp\text{by the second part of the } \knowl{./knowl/fact-rational-exponent-rule-ii.html}{\text{Radicals and Rational Exponents Rule}}\\ \amp=2^2\\ \amp=4 \end{align*}
\begin{align*} 16^{-\sfrac{3}{2}}\amp=\frac{1}{16^{\sfrac{3}{2}}}\amp\amp\text{by the } \knowl{./knowl/item-exponent-rules-negative-exponent-definition.html}{\text{Negative Exponent Rule}}\\ \amp=\frac{1}{\left(\sqrt{16}\right)^3}\amp\amp\text{by the second part of the } \knowl{./knowl/fact-rational-exponent-rule-ii.html}{\text{Radicals and Rational Exponents Rule}}\\ \amp=\frac{1}{4^3}\\ \amp=\frac{1}{64} \end{align*}
\begin{align*} -16^{\sfrac{3}{4}}\amp=-\left(\sqrt[4]{16}\right)^3 \amp\amp\text{by the second part of the } \knowl{./knowl/fact-rational-exponent-rule-ii.html}{\text{Radicals and Rational Exponents Rule}}\\ \amp=-2^3\\ \amp=-8 \end{align*}
In this problem the negative can be associated with either the numerator or the denominator, but not both. We choose the numerator.

\begin{align*} \left(-\frac{27}{64}\right)^{\sfrac{2}{3}}\amp=\left(\sqrt[3]{-\frac{27}{64}}\right)^2 \amp\amp\text{by the second part of the } \knowl{./knowl/fact-rational-exponent-rule-ii.html}{\text{Radicals and Rational Exponents Rule}}\\ \amp=\left(\frac{\sqrt[3]{-27}}{\sqrt[3]{64}}\right)^2\\ \amp=\left(\frac{-3}{4}\right)^2\\ \amp=\frac{(-3)^2}{(4)^2}\\ \amp=\frac{9}{16} \end{align*}

While we are looking at the algebra of \(x^{\sfrac{m}{n}}\text{,}\) we should briefly examine a graph to see what this type of function can look like.

Fractional powers can make some fairly interesting graphs. We invite you to play with these graphs on your favorite graphing program.

Figure14.2.14A Graph of \(y=x^{\sfrac{2}{3}}\)

Subsection14.2.2More Expressions with Rational Exponents

To recap, here is a “complete” list of exponent rules.

Product Rule: \(a^{n} \cdot a^{m} = a^{n+m}\)
Power to a Power Rule: \((a^{n})^{m} = a^{n\cdot m}\)
Product to a Power Rule: \((ab)^{n} = a^{n} \cdot b^{n}\)
Quotient Rule: \(\dfrac{a^{n}}{a^{m}} = a^{n-m}\text{,}\) as long as \(a \neq 0\)
Quotient to a Power Rule: \(\left( \dfrac{a}{b} \right)^{n} = \dfrac{a^{n}}{b^{n}}\text{,}\) as long as \(b \neq 0\)
Zero Exponent Rule: \(a^{0} = 1\) for \(a\neq0\)
Negative Exponent Rule: \(a^{-n} = \frac{1}{a^n}\)
Negative Exponent Reciprocal Rule: \(\frac{1}{a^{-n}} = a^n\)
Negative Exponent on Fraction Rule: \(\left(\frac{x}{y}\right)^{-n}=\left(\frac{y}{x}\right)^{n}\)
Radical and Rational Exponent Rule: \(x^{\sfrac{1}{n}}=\sqrt[n]{x}\)
Radical and Rational Exponent Rule: \(x^{\sfrac{m}{n}}=\left(\sqrt[n]{x}\right)^m\text{,}\) usually for numbers
Radical and Rational Exponent Rule: \(x^{\sfrac{m}{n}}=\sqrt[n]{x^m}\text{,}\) usually for variables

List14.2.15Summary of Exponent Rules

Example14.2.16

Convert the following radical expressions into expressions with rational exponents, and simplify them if possible.

\(\dfrac{1}{\sqrt{x}}\)
\(\dfrac{1}{\sqrt[3]{25}}\)

Solution

\begin{align*} \frac{1}{\sqrt{x}}\amp=\frac{1}{x^{\sfrac{1}{2}}}\amp\amp\text{by the } \knowl{./knowl/fact-rational-exponent-rule-i.html}{\text{Radicals and Rational Exponents Rule}}\\ \amp=x^{-\sfrac{1}{2}}\amp\amp\text{by the } \knowl{./knowl/item-exponent-rules-negative-exponent-definition.html}{\text{Negative Exponent Rule}} \end{align*}
\begin{align*} \frac{1}{\sqrt[3]{25}}\amp=\frac{1}{25^{\sfrac{1}{3}}}\amp\amp\text{by the } \knowl{./knowl/fact-rational-exponent-rule-i.html}{\text{Radicals and Rational Exponents Rule}}\\ \amp=\frac{1}{\left(5^2\right)^{\sfrac{1}{3}}}\\ \amp=\frac{1}{5^{2\cdot\sfrac{1}{3}}}\amp\amp\text{by the } \knowl{./knowl/item-exponent-rules-power-to-a-power.html}{\text{Power to a Power Rule}}\\ \amp=\frac{1}{5^{\sfrac{2}{3}}}\\ \amp=5^{-\sfrac{2}{3}}\amp\amp\text{by the } \knowl{./knowl/item-exponent-rules-negative-exponent-definition.html}{\text{Negative Exponent Rule}} \end{align*}

Learners of these simplifications often find it challenging, so we now include a plethora of examples of varying difficulty.

Example14.2.17

Use exponent properties in List 14.2.15 to simplify the expressions, and write all final versions using radicals.

\(2w^{\sfrac{7}{8}}\)
\(\frac{1}{2}y^{-\sfrac{1}{2}}\)
\(\left(27b\right)^{\sfrac{2}{3}}\)
\(\left(-8p^6\right)^{\sfrac{5}{3}}\)
\(\sqrt{x^3}\cdot\sqrt[4]{x}\)
\(h^{\sfrac{1}{3}}+h^{\sfrac{1}{3}}+h^{\sfrac{1}{3}}\)
\(\frac{\sqrt{z}}{\sqrt[3]{z}}\)
\(\sqrt{\sqrt[4]{q}}\)
\(3\left(c^{\sfrac{1}{2}}+d^{\sfrac{1}{2}}\right)^2\)
\(3\left(4k^{\sfrac{2}{3}}\right)^{-\sfrac{1}{2}}\)

Solution

\begin{align*} 2w^{\sfrac{7}{8}}\amp=2\sqrt[8]{w^7}\amp\amp\text{by the } \knowl{./knowl/fact-rational-exponent-rule-ii.html}{\text{Radicals and Rational Exponents Rule}} \end{align*}
\begin{align*} \frac{1}{2}y^{-\sfrac{1}{2}}\amp=\frac{1}{2}\frac{1}{y^{\sfrac{1}{2}}}\amp\amp\text{by the } \knowl{./knowl/item-exponent-rules-negative-exponent-definition.html}{\text{Negative Exponent Rule}}\\ \amp=\frac{1}{2}\frac{1}{\sqrt{y}}\amp\amp\text{by the } \knowl{./knowl/fact-rational-exponent-rule-ii.html}{\text{Radicals and Rational Exponents Rule}}\\ \amp=\frac{1}{2\sqrt{y}} \end{align*}
\begin{align*} \left(27b\right)^{\sfrac{2}{3}}\amp=\left(27\right)^{\sfrac{2}{3}}\cdot\left(b\right)^{\sfrac{2}{3}}\amp\amp\text{by the } \knowl{./knowl/item-exponent-rules-product-to-a-power.html}{\text{Product to a Power Rule}}\\ \amp=\left(\sqrt[3]{27}\right)^2\cdot\sqrt[3]{b^2}\amp\amp\text{by the } \knowl{./knowl/fact-rational-exponent-rule-ii.html}{\text{Radicals and Rational Exponents Rule}}\\ \amp=3^2\cdot\sqrt[3]{b^2}\\ \amp=9\sqrt[3]{b^2} \end{align*}
\begin{align*} \left(-8p^6\right)^{\sfrac{5}{3}}\amp=\left(-8\right)^{\sfrac{5}{3}}\cdot\left(p^6\right)^{\sfrac{5}{3}}\amp\amp\text{by the } \knowl{./knowl/item-exponent-rules-product-to-a-power.html}{\text{Product to a Power Rule}}\\ \amp=\left(-8\right)^{\sfrac{5}{3}}\cdot p^{6\cdot\sfrac{5}{3}}\amp\amp\text{by the } \knowl{./knowl/item-exponent-rules-power-to-a-power.html}{\text{Power to a Power Rule}}\\ \amp=\left(\sqrt[3]{-8}\right)^5\cdot p^{10}\amp\amp\text{by the } \knowl{./knowl/fact-rational-exponent-rule-ii.html}{\text{Radicals and Rational Exponents Rule}}\\ \amp=(-2)^5\cdot p^{10}\\ \amp=-32p^{10} \end{align*}
\begin{align*} \sqrt{x^3}\cdot\sqrt[4]{x}\amp=x^{\sfrac{3}{2}}\cdot x^{\sfrac{1}{4}}\amp\amp\text{by the } \knowl{./knowl/fact-rational-exponent-rule-ii.html}{\text{Radicals and Rational Exponents Rule}}\\ \amp=x^{\sfrac{3}{2}+\sfrac{1}{4}}\amp\amp\text{by the } \knowl{./knowl/item-exponent-rules-product.html}{\text{Product Rule}}\\ \amp=x^{\sfrac{6}{4}+\sfrac{1}{4}}\\ \amp=x^{\sfrac{7}{4}}\\ \amp=\sqrt[4]{x^7}\amp\amp\text{by the } \knowl{./knowl/fact-rational-exponent-rule-ii.html}{\text{Radicals and Rational Exponents Rule}} \end{align*}
\begin{align*} h^{\sfrac{1}{3}}+h^{\sfrac{1}{3}}+h^{\sfrac{1}{3}}\amp=3h^{\sfrac{1}{3}}\\ \amp=3\sqrt[3]{h}\amp\amp\text{by the } \knowl{./knowl/fact-rational-exponent-rule-i.html}{\text{Radicals and Rational Exponents Rule}} \end{align*}
\begin{align*} \frac{\sqrt{z}}{\sqrt[3]{z}}\amp=\frac{z^{\sfrac{1}{2}}}{z^{\sfrac{1}{3}}}\amp\amp\text{by the } \knowl{./knowl/fact-rational-exponent-rule-i.html}{\text{Radicals and Rational Exponents Rule}}\\ \amp=z^{\sfrac{1}{2}-\sfrac{1}{3}}\amp\amp\text{by the } \knowl{./knowl/item-exponent-rules-quotient.html}{\text{Quotient Rule}}\\ \amp=z^{\sfrac{3}{6}-\sfrac{2}{6}}\\ \amp=z^{\sfrac{1}{6}}\\ \amp=\sqrt[6]{z}\amp\amp\text{by the } \knowl{./knowl/fact-rational-exponent-rule-i.html}{\text{Radicals and Rational Exponents Rule}} \end{align*}
\begin{align*} \sqrt{\sqrt[4]{q}}\amp=\sqrt{q^{\sfrac{1}{4}}}\amp\amp\text{by the } \knowl{./knowl/fact-rational-exponent-rule-i.html}{\text{Radicals and Rational Exponents Rule}}\\ \amp=\left(q^{\sfrac{1}{4}}\right)^{\sfrac{1}{2}}\amp\amp\text{by the } \knowl{./knowl/fact-rational-exponent-rule-i.html}{\text{Radicals and Rational Exponents Rule}}\\ \amp=q^{\sfrac{1}{4}\cdot\sfrac{1}{2}}\amp\amp\text{by the } \knowl{./knowl/item-exponent-rules-power-to-a-power.html}{\text{Power to a Power Rule}}\\ \amp=q^{\sfrac{1}{8}}\\ \amp=\sqrt[8]{q}\amp\amp\text{by the } \knowl{./knowl/fact-rational-exponent-rule-i.html}{\text{Radicals and Rational Exponents Rule}} \end{align*}
\begin{alignat*}{2} 3\left(c^{\sfrac{1}{2}}+d^{\sfrac{1}{2}}\right)^2\amp=3\left(c^{\sfrac{1}{2}}+d^{\sfrac{1}{2}}\right)\left(c^{\sfrac{1}{2}}+d^{\sfrac{1}{2}}\right)\\ \amp=3\left(\left(c^{\sfrac{1}{2}}\right)^2+2c^{\sfrac{1}{2}}\cdot d^{\sfrac{1}{2}}+\left(d^{\sfrac{1}{2}}\right)^2\right)\\ \amp=3\left(c^{\sfrac{1}{2}\cdot 2}+2c^{\sfrac{1}{2}}\cdot d^{\sfrac{1}{2}}+d^{\sfrac{1}{2}\cdot 2}\right)\\ \amp=3\left(c+2c^{\sfrac{1}{2}}\cdot d^{\sfrac{1}{2}}+d\right)\\ \amp=3\left(c+2(cd)^{\sfrac{1}{2}}+d\right)\amp\amp\text{by the } \knowl{./knowl/item-exponent-rules-product-to-a-power.html}{\text{Product to a Power Rule}}\\ \amp=3\left(c+2\sqrt{cd}+d\right)\amp\amp\text{by the } \knowl{./knowl/fact-rational-exponent-rule-i.html}{\text{Radicals and Rational Exponents Rule}}\\ \amp=3c+6\sqrt{cd}+3d \end{alignat*}
\begin{align*} 3\left(4k^{\sfrac{2}{3}}\right)^{-\sfrac{1}{2}}\amp=\frac{3}{\left(4k^{\sfrac{2}{3}}\right)^{\sfrac{1}{2}}}\amp\amp\text{by the } \knowl{./knowl/item-exponent-rules-negative-exponent-definition.html}{\text{Negative Exponent Rule}}\\ \amp=\frac{3}{4^{\sfrac{1}{2}}\left(k^{\sfrac{2}{3}}\right)^{\sfrac{1}{2}}}\amp\amp\text{by the } \knowl{./knowl/item-exponent-rules-product-to-a-power.html}{\text{Product to a Power Rule}}\\ \amp=\frac{3}{4^{\sfrac{1}{2}}k^{\sfrac{2}{3}\cdot\sfrac{1}{2}}}\amp\amp\text{by the } \knowl{./knowl/item-exponent-rules-power-to-a-power.html}{\text{Power to a Power Rule}}\\ \amp=\frac{3}{4^{\sfrac{1}{2}}k^{\sfrac{1}{3}}}\\ \amp=\frac{3}{\sqrt{4}\sqrt[3]{k}}\amp\amp\text{by the } \knowl{./knowl/fact-rational-exponent-rule-i.html}{\text{Radicals and Rational Exponents Rule}}\\ \amp=\frac{3}{2\sqrt[3]{k}} \end{align*}

We will end a with a short application on rational exponents. Kepler's Laws of Orbital Motion ¹en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion describe how planets orbit stars and how satellites orbit planets. In particular, his third law has a rational exponent, which we will now explore.

Example14.2.18Kepler and the Satellite

Kepler's third law of motion says that for objects with a roughly circular orbit that the time (in hours) that it takes to make one full revolution around the planet, \(T\text{,}\) is proportional to three-halves power of the distance (in kilometers) from the center of the planet to the satellite, \(r\text{.}\) For the Earth, it looks like this:

\begin{equation*} T=\frac{2\pi}{\sqrt{G\cdot M_E}}r^{\sfrac{3}{2}} \end{equation*}

In this case, both \(G\) and \(M_E\) are constants. \(G\) stands for the universal gravitational constant ²en.wikipedia.org/wiki/Gravitational_constant where \(G\) is about \(8.65\times 10^{-13}\) ^km³⁄_kg h² and \(M_E\) stands for the mass of the Earth ³en.wikipedia.org/wiki/Earth_mass where \(M_E\) is about \(5.972\times 10^{24}\) kg. Inputting these values into this formula yields a simplified version that looks like this:

\begin{equation*} T\approx2.76\times 10^{-6}r^{\sfrac{3}{2}} \end{equation*}

Most satellites orbit in what is called low Earth orbit ⁴en.wikipedia.org/wiki/Low_Earth_orbit, including the international space station which orbits at about 340 km above from Earth's surface. The Earth's average radius is about 6380 km. Find the period of the international space station.

Solution

The formula has already been identified, but the input takes just a little thought. The formula uses \(r\) as the distance from the center of the Earth to the satellite, so to find \(r\) we need to combine the radius of the Earth and the distance to the satellite above the surface of the Earth.

\begin{align*} r\amp=340+6380\\ \amp=6720 \end{align*}

Now we can input this value into the formula and evaluate.

\begin{align*} T\amp\approx2.76\cdot 10^{-6}r^{\sfrac{3}{2}}\\ \amp\approx2.76\cdot 10^{-6}(\substitute{6720})^{\sfrac{3}{2}}\\ \amp\approx2.76\cdot 10^{-6}\left(\sqrt{6720}\right)^3\\ \amp\approx1.52 \end{align*}

The formula tells us that it takes a little more than an hour and a half for the ISS to orbit the Earth! That works out to 15 or 16 sunrises per day.