10.7: Solving with rational exponents
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In this section, we take solving with radicals one step further and apply radicals and rational exponents to solve equations with exponents. Since radicals have some restrictions on the radicand, we will also have some restrictions here when applying a rational exponents in the solving process.
The Odd Root Property
Let's start with the case that we take an odd root of an equation.
If \(x^n=p\), where \(n\) is odd, then \(\sqrt{x}=\sqrt[n]{p}\). Note, the radicand can be any real number, i.e., \(p\) is any number in \((-\infty, \infty)\).
Solve: \(x^5=32\)
Solution
We can easily apply the odd root property to solve for \(x\).
\[\begin{array}{rl}x^5=32&\text{Apply odd root property} \\ \sqrt[5]{x^5}=\sqrt[5]{32}&\text{Simplify} \\ x=2&\text{Solution}\end{array}\nonumber\]
Solve: \(4r^3-2=106\)
Solution
We can easily apply the odd root property to solve for \(r\).
\[\begin{array}{rl}4r^3-2=106&\text{Isolate the variable term} \\ 4r^3=108&\text{Isolate }r^3 \\ r^3=27&\text{Apply odd root property} \\ \sqrt[3]{r^3}=\sqrt[3]{27}&\text{Simplify} \\ r=3&\text{Solution}\end{array}\nonumber\]
The Even Root Property
With even roots, we have the restriction on the radicand where the radicand is required to be non-negative here. We discussed this in the previous section, e.g., \(\sqrt{-4}\) is not a real number. We continue this restriction when taking even roots of an equation.
If \(x^n = p\), where \(n\) is even, then \(x =\sqrt[n]{p}\) or\( x = −\sqrt[n]{p}\) or we can write \(x = \pm\sqrt[n]{p}\). Note, the radicand can be any real non-negative number, i.e., \(p\geq 0\).
Solve: \(x^4=16\)
Solution
We can easily apply the even root property to solve for \(x\).
\[\begin{array}{rl}x^4=16&\text{Apply even root property} \\ \sqrt[4]{x^4}=\sqrt[4]{16}&\text{Simplify} \\ |x|=\pm 2 \\ x=\pm 2&\text{Solution}\end{array}\nonumber\]
Notice, it wasn’t given that \(x ≥ 0\). Hence, we cannot assume it is, so we put absolute value around \(x\). Once we verify the solution(s), then we can remove the absolute value around \(x\).
In 1545, French mathematician Gerolamo Cardano published his book The Great Art, or the Rules of Algebra, which included the solution to an equation with a fourth power, but it was considered absurd by many to take a quantity to the fourth power because there are only three dimensions!
Solve: \((2x + 4)^2 = 36\) Find and verify all solutions that satisfy the equation.
Solution
We can easily apply the even root property to solve for \(x\).
\[\begin{array}{rl}(2x+4)^2=36&\text{Apply even root property} \\ \sqrt{(2x+4)^2}=\pm\sqrt{36} &\text{Simplify }\sqrt{36} \\ 2x+4=\pm 6&\text{Rewrite into two equations} \\ 2x+4=6\quad\text{or}\quad 2x+4=-6&\text{Isolate the variable term in each equation} \\ 2x=2\quad\text{or}\quad 2x=-10&\text{Solve each equation} \\ x=1\quad\text{or}\quad x=-5&\text{Solutions}\end{array}\nonumber\]
We can always verify the solutions by substituting back in \(1\), \(−5\) into the original equation:
\[\begin{array}{rl} (2x+4)^2=36&\text{Plug-n-chug }x=1 \\ (2(\color{blue}{1}\color{black}{)}+4)^2\stackrel{?}{=}36&\text{Simplify each side} \\ (2+4)^2\stackrel{?}{=}36 \\ 6^2\stackrel{?}{=}36 \\ 36=36&\checkmark\text{True}\end{array}\nonumber\]
Let's try the next solution \(x=-5\):
\[\begin{array}{rl}(2x+4)^2=36&\text{Plug-n-chug }x=-5 \\ (2(\color{blue}{-5}\color{black}{)}+4)^2\stackrel{?}{=}36&\text{Simplify each side} \\ (-10+4)^2\stackrel{?}{=}36 \\ (-6)^2\stackrel{?}{=}36 \\ 36=36&\checkmark\text{True}\end{array}\nonumber\]
Thus, \(1\), (-5) are, in fact, solutions to the original equation.
Solve: \((6x − 9)^2 = 45\) Find and verify all solutions that satisfy the equation.
Solution
\[\begin{array}{rl}(6x-9)^2=45&\text{Apply even root property} \\ \sqrt{(6x-9)^2}=\pm \sqrt{45}&\text{Simplify }\sqrt{45} \\ 6x-9=\pm 3\sqrt{5}&\text{Isolate the variable term} \\ 6x=9\pm 3\sqrt{5}&\text{Divide both sides by }6 \\ x=\dfrac{9\pm 3\sqrt{5}}{6}&\text{Factor a GCF from numerator} \\ x=\dfrac{\color{blue}{\cancel{3}}\color{black}{9}3\pm\sqrt{5})}{\color{blue}{\cancelto{2}{6}}} &\color{black}{\text{Simplify}} \\ x=\dfrac{3\pm\sqrt{5}}{2}&\text{Solution}\end{array}\nonumber\]
Notice, we didn’t split the equation into two different equations and solve. Since \(\sqrt{45}\) is an irrational number, we can leave the \(±\) and solve as usual. We leave verifying the solutions to the student.
Solve: \(256w^8+40=41\)
Solution
We have to isolate the variable term first, then we can apply the even root property.
\[\begin{array}{rl}256w^8+40=41&\text{Isolate the variable term.} \\ 256w^8=1&\text{Divide each side by }256\\ w^8=\dfrac{1}{256}&\text{Apply even root property} \\ \sqrt[8]{w^8}=\pm\sqrt[8]{\dfrac{1}{256}}&\text{Simplify the radicals} \\ |w|=\pm\dfrac{1}{2} \\ w=\pm\dfrac{1}{2}&\text{Solution}\end{array}\nonumber\]
Notice, it wasn't given that \(w\geq 0\). Hence, we cannot assume it is and we put absolute value around \(w\). Once we verify the solution(s), then we can remove the absolute value around \(w\).
Solving Equations with Rational Exponents
When exponents are fractions, we convert the rational exponent into a radical expression to solve. Recall, \(a^{\dfrac{m}{n}}=(\sqrt[n]{a})^m\). Then we clear the exponent by applying either the even or odd root property and solve as usual.
Given an equation with rational exponents, we can follow the following steps to solve.
Step 1. Rewrite any rational exponents as radicals.
Step 2. Apply the odd or even root property. Recall, even roots require the radicand to be positive unless otherwise noted.
Step 3. Raise each side to the power of the root.
Step 4. Solve. Verify the solutions, especially when there is an even root.
Solve: \((4x+1)^{\dfrac{2}{5}}=9\). Assume all variables are positive.
Solution
We follow the steps in order to solve the equation with a rational exponent.
Step 1. Rewrite any rational exponents as radicals. \[\begin{aligned} (4x+1)^{\dfrac{2}{5}}&=9 \\ (\sqrt[5]{4x+1})^2&=9\end{aligned}\]
Step 2. Apply the odd or even root property. Recall, even roots require the radicand to be positive unless otherwise noted.
Since we are taking the square root, which is even, then we apply the even root property: \[\begin{aligned}(\sqrt[5]{4x+1})^2&=9 \\ \sqrt[5]{4x+1}&=\pm\sqrt{9} \\ \sqrt[5]{4x+1}=\pm 3\end{aligned}\]
Step 3. Raise each side to the power of the root.
Since the root is \(5\), then we can raise each side to the fifth power: \[\begin{aligned}\sqrt[5]{4x+1}&=\pm 3 \\ (\sqrt[5]{4x+1})^5&=(\pm 3)^5 \\ 4x+1&=\pm 243\end{aligned}\]
Step 4. Solve. Verify the solutions, especially when there is an even root. \[\begin{array}{rll}4x+1=243&\text{or}&4x+1=-243 \\ 4x=242&\text{or}&4x=-244 \\ x=\dfrac{242}{4}&\text{or}&x=-61 \\ x=\dfrac{121}{2}&\text{or}&x=-61\end{array}\nonumber\] We can verify all solutions. Let's start by verifying that \(x=-61\) is a solution. \[\begin{aligned}(\sqrt[5]{4x+1})^2&=9 \\ (\sqrt[5]{4(\color{blue}{-61}\color{black}{)}+61})^2&\stackrel{?}{=}9 \\ (\sqrt[5]{-244+1})^2&\stackrel{?}{=}9 \\ (\sqrt[5]{-243})^2&\stackrel{?}{=}9 \\ (-3)^2&\stackrel{?}{=}9 \\ 9&=9\:\checkmark\text{ True}\end{aligned}\] We leave the verification of the second solution to the student.
Thus, the solutions to the equation are \(\dfrac{121}{2},-61\).
Solve: \((3x-2)^{\dfrac{3}{4}}=64\)
Solution
We follow the steps in order to solve the equation with a rational exponent.
Step 1. Rewrite any rational exponents as radicals. \[\begin{aligned}(3x-2)^{\dfrac{3}{4}}&=64 \\ (\sqrt[4]{3x-2})^3&=64 \end{aligned}\]
Step 2. Apply the odd or even root property. Recall, even roots require the radicand to be positive unless otherwise noted.
Since we are taking the cube root, which is odd, then we apply the odd root property: \[\begin{aligned}(\sqrt[4]{3x-2})^3&=64 \\ \sqrt[4]{3x-2}&=\sqrt[3]{64} \\ \sqrt[4]{3x-2}&=4\end{aligned}\]
Step 3. Raise each side to the power of the root.
Since the root is \(4\), then we can raise each side to the fourth power: \[\begin{aligned}\sqrt[4]{3x-2}&=4 \\ (\sqrt[4]{3x-2})^4&=4^4 \\ 3x-2&=256\end{aligned}\]
Step 4. Solve. Verify the solutions, especially when there is an even root. \[\begin{aligned}3x-2&=256 \\ 3x&=258 \\ x&=86\end{aligned}\] Since there is an even root in the original equation, we should verify the solution. \[\begin{aligned}(\sqrt[4]{3x-2})^3&=64 \\ (\sqrt[4]{3(\color{blue}{86}\color{black}{)}-2})^3&\stackrel{?}{=}64 \\ (\sqrt[4]{258-2})^3&\stackrel{?}{=}64 \\ (\sqrt[4]{256})^3&\stackrel{?}{=}64 \\ (4)^3&\stackrel{?}{=}64 \\ 64&=64\:\checkmark\text{ True}\end{aligned}\]
Thus, the solution is \(86\).
When solving equations with rational exponents, it is very helpful to convert the equations into their radical form so we can see which property we need to use and to identify whether we need to verify the solutions due to an even root in the original equation.
Solving with Rational Exponents Homework
Solve.
\(x^2=75\)
\(x^2+5=13\)
\(3x^2+1=73\)
\((x+2)^5=-243\)
\((2x+5)^3-6=21\)
\((x-1)^{\dfrac{2}{3}}=4\)
\((2-x)^{\dfrac{3}{2}}=27\)
\((2x-3)^{\dfrac{2}{3}}=4\)
\((x+\dfrac{1}{2})^{-\dfrac{2}{3}}=4\)
\((x-1)^{-\dfrac{5}{2}}=32\)
\((3x-2)^{\dfrac{4}{5}}=16\)
\((4x+2)^{\dfrac{3}{5}}=-8\)
\(x^3=-8\)
\(4x^3-2=106\)
\((x-4)^2=49\)
\((5x+1)^4=16\)
\((2x+1)^2+3=21\)
\((x-1)^{\dfrac{3}{2}}=8\)
\((2x+3)^{\dfrac{4}{3}}=16\)
\((x+3)^{-\dfrac{1}{3}}=4\)
\((x-1)^{-\dfrac{5}{3}}=32\)
\((x+3)^{\dfrac{3}{2}}=-8\)
\((2x+3)^{\dfrac{3}{2}}=27\)
\((3-2x)^{\dfrac{4}{3}}=-81\)