Solving Equations Involving Rational Exponents
Rational exponents are exponents that are fractions, where the numerator is a power and the denominator is a root. For example, [latex]{16}^{\frac{1}{2}}[/latex] is another way of writing [latex]\sqrt{16}[/latex]; [latex]{8}^{\frac{1}{3}}[/latex] is another way of writing [latex]\text{ }\sqrt[3]{8}[/latex]. The ability to work with rational exponents is a useful skill, as it is highly applicable in calculus. We can solve equations in which a variable is raised to a rational exponent by raising both sides of the equation to the reciprocal of the exponent. The reason we raise the equation to the reciprocal of the exponent is because we want to eliminate the exponent on the variable term, and a number multiplied by its reciprocal equals 1. For example, [latex]\frac{2}{3}\left(\frac{3}{2}\right)=1[/latex], [latex]3\left(\frac{1}{3}\right)=1[/latex], and so on.A General Note: Rational Exponents
A rational exponent indicates a power in the numerator and a root in the denominator. There are multiple ways of writing an expression, a variable, or a number with a rational exponent:[latex]{a}^{\frac{m}{n}}={\left({a}^{\frac{1}{n}}\right)}^{m}={\left({a}^{m}\right)}^{\frac{1}{n}}=\sqrt[n]{{a}^{m}}={\left(\sqrt[n]{a}\right)}^{m}[/latex]
Example 1: Evaluating a Number Raised to a Rational Exponent
Evaluate [latex]{8}^{\frac{2}{3}}[/latex].Solution
Whether we take the root first or the power first depends on the number. It is easy to find the cube root of 8, so rewrite [latex]{8}^{\frac{2}{3}}[/latex] as [latex]{\left({8}^{\frac{1}{3}}\right)}^{2}[/latex].[latex]\begin{array}{l}{\left({8}^{\frac{1}{3}}\right)}^{2}\hfill&={\left(2\right)}^{2}\hfill \\ \hfill&=4\hfill \end{array}[/latex]
Try It 1
Evaluate [latex]{64}^{-\frac{1}{3}}[/latex]. SolutionExample 2: Solve the Equation Including a Variable Raised to a Rational Exponent
Solve the equation in which a variable is raised to a rational exponent: [latex]{x}^{\frac{5}{4}}=32[/latex].Solution
The way to remove the exponent on x is by raising both sides of the equation to a power that is the reciprocal of [latex]\frac{5}{4}[/latex], which is [latex]\frac{4}{5}[/latex].[latex]\begin{array}{ll}{x}^{\frac{5}{4}}\hfill&=32\hfill & \hfill \\ {\left({x}^{\frac{5}{4}}\right)}^{\frac{4}{5}}\hfill&={\left(32\right)}^{\frac{4}{5}}\hfill & \hfill \\ x\hfill&={\left(2\right)}^{4}\hfill & \text{The fifth root of 32 is 2}.\hfill \\ \hfill&=16\hfill & \hfill \end{array}[/latex]
Try It 2
Solve the equation [latex]{x}^{\frac{3}{2}}=125[/latex]. SolutionExample 3: Solving an Equation Involving Rational Exponents and Factoring
Solve [latex]3{x}^{\frac{3}{4}}={x}^{\frac{1}{2}}[/latex].Solution
This equation involves rational exponents as well as factoring rational exponents. Let us take this one step at a time. First, put the variable terms on one side of the equal sign and set the equation equal to zero.[latex]\begin{array}{l}3{x}^{\frac{3}{4}}-\left({x}^{\frac{1}{2}}\right)\hfill&={x}^{\frac{1}{2}}-\left({x}^{\frac{1}{2}}\right)\hfill \\ 3{x}^{\frac{3}{4}}-{x}^{\frac{1}{2}}\hfill&=0\hfill \end{array}[/latex]
Now, it looks like we should factor the left side, but what do we factor out? We can always factor the term with the lowest exponent. Rewrite [latex]{x}^{\frac{1}{2}}[/latex] as [latex]{x}^{\frac{2}{4}}[/latex]. Then, factor out [latex]{x}^{\frac{2}{4}}[/latex] from both terms on the left.
[latex]\begin{array}{l}3{x}^{\frac{3}{4}}-{x}^{\frac{2}{4}}\hfill&=0\hfill \\ {x}^{\frac{2}{4}}\left(3{x}^{\frac{1}{4}}-1\right)\hfill&=0\hfill \end{array}[/latex]
Where did [latex]{x}^{\frac{1}{4}}[/latex] come from? Remember, when we multiply two numbers with the same base, we add the exponents. Therefore, if we multiply [latex]{x}^{\frac{2}{4}}[/latex] back in using the distributive property, we get the expression we had before the factoring, which is what should happen. We need an exponent such that when added to [latex]\frac{2}{4}[/latex] equals [latex]\frac{3}{4}[/latex]. Thus, the exponent on x in the parentheses is [latex]\frac{1}{4}[/latex].
Let us continue. Now we have two factors and can use the zero factor theorem.
[latex]\begin{array}{ll}{x}^{\frac{2}{4}}\left(3{x}^{\frac{1}{4}}-1\right)\hfill&=0\hfill & \hfill \\ {x}^{\frac{2}{4}}\hfill&=0\hfill & \hfill \\ x=0\hfill & \hfill \\ 3{x}^{\frac{1}{4}}-1\hfill&=0\hfill & \hfill \\ 3{x}^{\frac{1}{4}}\hfill&=1\hfill & \hfill \\ {x}^{\frac{1}{4}}\hfill&=\frac{1}{3}\hfill & \text{Divide both sides by 3}.\hfill \\ {\left({x}^{\frac{1}{4}}\right)}^{4}\hfill&={\left(\frac{1}{3}\right)}^{4}\hfill & \text{Raise both sides to the reciprocal of }\frac{1}{4}.\hfill \\ x\hfill&=\frac{1}{81}\hfill & \hfill \end{array}[/latex]
The two solutions are [latex]x=0[/latex], [latex]x=\frac{1}{81}[/latex].
Try It 3
Solve: [latex]{\left(x+5\right)}^{\frac{3}{2}}=8[/latex]. SolutionLicenses & Attributions
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- College Algebra. Provided by: OpenStax Authored by: OpenStax College Algebra. Located at: https://cnx.org/contents/[email protected]:1/Preface. License: CC BY: Attribution.
