<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Dismiss
Skip Navigation

7.9: Solving Rational Exponent Equations

Difficulty Level: At Grade Created by: CK-12
Atoms Practice
Estimated10 minsto complete
%
Progress
Practice Solving Equations with Fractional Exponents
Practice
Progress
Estimated10 minsto complete
%
Practice Now
Turn In

The period (in seconds) of a pendulum with a length of L (in meters) is given by the formula \begin{align*}P = 2\pi{(\frac{L}{9.8})}^{\frac{1}{2}}\end{align*}P=2π(L9.8)12. If the period of a pendulum is \begin{align*}10\pi\end{align*}10π is the length of the pendulum 156.8?

Solving Rational Exponent Equations

When solving a rational exponent equation, you first isolate the variable. Then, to eliminate the exponent, you will need to raise everything to the reciprocal power.

Let's determine if x = 9 is a solution to \begin{align*}2x^{\frac{3}{2}}-19=35\end{align*}2x3219=35.

Substitute in x and see if the equation holds.

\begin{align*}2(9)^{\frac{3}{2}}-19&=35 \\ 2 \cdot 27 -19 &= 35 \\ 54 - 19 &= 35 \end{align*}2(9)3219227195419=35=35=35

9 is a solution to this equation.

Now, let's solve the following equations for x.

  1. \begin{align*}3x^{\frac{5}{2}}=96\end{align*}3x52=96

First, divide both sides by 3 to isolate \begin{align*}x\end{align*}x.

\begin{align*}3x^{\frac{5}{2}}&=96\\ x^{\frac{5}{2}}&=32 \end{align*}3x52x52=96=32

\begin{align*}x\end{align*}x is raised to the five-halves power. To cancel out this exponent, we need to raise everything to the two-fifths power.

\begin{align*}\left(x^{\frac{5}{2}}\right)^{\frac{2}{5}}&=32^{\frac{2}{5}}\\ x&=32^{\frac{2}{5}}\\ x&=\sqrt[5]{32}^2=2^2=4\end{align*}(x52)25xx=3225=3225=3252=22=4

Check: \begin{align*}3(4)^{\frac{5}{2}}=3 \cdot 2^5=3 \cdot 32=96\end{align*}3(4)52=325=332=96

  1. \begin{align*}-2(x-5)^{\frac{3}{4}}+48=-202\end{align*}2(x5)34+48=202

Isolate \begin{align*}(x-5)^{\frac{3}{4}}\end{align*}(x5)34 by subtracting 48 and dividing by -2.

\begin{align*}-2(x-5)^{\frac{3}{4}}+48&=-202\\ -2(x-5)^{\frac{3}{4}}&=-250\\ (x-5)^{\frac{3}{4}}&=-125\end{align*}2(x5)34+482(x5)34(x5)34=202=250=125

To undo the three-fourths power, raise everything to the four-thirds power.

\begin{align*}\left[ \left(x-5 \right)^{\frac{3}{4}}\right]^{\frac{4}{3}}&=\left(-125 \right)^{\frac{4}{3}}\\ x-5&=625\\ x&=630\end{align*}[(x5)34]43x5x=(125)43=625=630

Check: \begin{align*}-2(630-5)^{\frac{3}{4}}+48=-2 \cdot 625^{\frac{3}{4}}+48=-2 \cdot 125+48=-250+48=-202\end{align*}2(6305)34+48=262534+48=2125+48=250+48=202

Examples

Example 1

Earlier, you were asked to verify the length of the pendulum. 

We need to plug 156.8 in to the equation \begin{align*}P = 2\pi{(\frac{L}{9.8})}^{\frac{1}{2}}\end{align*}P=2π(L9.8)12 for L and solve. If our answer equals \begin{align*}10\pi\end{align*}10π, then the given length is correct.

\begin{align*}P = 2\pi{(\frac{L}{9.8})}^{\frac{1}{2}}\\ 2\pi{(\frac{156.8}{9.8})}^{\frac{1}{2}}\\ 2\pi (16)^{\frac{1}{2}}\\ 2\pi (4) = 8 \pi\end{align*}P=2π(L9.8)122π(156.89.8)122π(16)122π(4)=8π

\begin{align*}8\pi\end{align*}8π does not equal \begin{align*}10\pi\end{align*}10π, so the length cannot be 156.8.

Solve the following rational exponent equations and check for extraneous solutions.

Example 2

\begin{align*}8(3x-1)^{\frac{2}{3}}=200\end{align*}8(3x1)23=200

Divide both sides by 8 and raise everything to the three-halves power.

\begin{align*}8(3x-1)^{\frac{2}{3}}&=200\\ \left[ \left(3x-1 \right)^{\frac{2}{3}}\right]^{\frac{3}{2}}&=(25)^{\frac{3}{2}}\\ 3x-1&=125\\ 3x&=126\\ x&=42\end{align*}8(3x1)23[(3x1)23]323x13xx=200=(25)32=125=126=42

Check: \begin{align*}8(3(42)-1)^{\frac{2}{3}}=8(126-1)^{\frac{2}{3}}=8(125)^{\frac{2}{3}}=8 \cdot 25=200\end{align*}8(3(42)1)23=8(1261)23=8(125)23=825=200

Example 3

\begin{align*}6x^{\frac{3}{2}}-141=1917\end{align*}6x32141=19172. 

Here, only the \begin{align*}x\end{align*}x is raised to the three-halves power. Subtract 141 from both sides and divide by 6. Then, eliminate the exponent by raising both sides to the two-thirds power.

\begin{align*}6x^{\frac{3}{2}}-141&=1917 \\ 6x^{\frac{3}{2}}&=2058 \\ x^{\frac{3}{2}}&=343 \\ x&=343^{\frac{2}{3}}=7^2=49\end{align*}6x321416x32x32x=1917=2058=343=34323=72=49

Check: \begin{align*}6(49)^{\frac{3}{2}}-141=6 \cdot 343-141=2058-141=1917\end{align*}6(49)32141=6343141=2058141=1917

Review

Determine if the following values of x are solutions to the equation \begin{align*}3x^{\frac{3}{5}}=-24\end{align*}3x35=24

  1. \begin{align*}x=32\end{align*}x=32
  2. \begin{align*}x=-32\end{align*}x=32
  3. \begin{align*}x=8\end{align*}x=8

Solve the following equations. Round any decimal answers to 2 decimal places.

  1. \begin{align*}2x^{\frac{3}{2}}=54\end{align*}2x32=54
  2. \begin{align*}3x^{\frac{1}{3}}+5=17\end{align*}3x13+5=17
  3. \begin{align*}(7x-3)^{\frac{2}{5}}=4\end{align*}(7x3)25=4
  4. \begin{align*}(4x+5)^{\frac{1}{2}}=x-4\end{align*}(4x+5)12=x4
  5. \begin{align*}x^{\frac{5}{2}}=16x^{\frac{1}{2}}\end{align*}x52=16x12
  6. \begin{align*}(5x+7)^{\frac{3}{5}}=8\end{align*}(5x+7)35=8
  7. \begin{align*}5x^{\frac{2}{3}}=45\end{align*}5x23=45
  8. \begin{align*}(7x-8)^{\frac{2}{3}}=4(x-5)^{\frac{2}{3}}\end{align*}(7x8)23=4(x5)23
  9. \begin{align*}7x^{\frac{3}{7}}+9=65\end{align*}
  10. \begin{align*}4997=5x^{\frac{3}{2}}-3\end{align*}
  11. \begin{align*}2x^{\frac{3}{4}}=686\end{align*}
  12. \begin{align*}x^3=(4x-3)^{\frac{3}{2}}\end{align*}

Answers for Review Problems

To see the Review answers, open this PDF file and look for section 7.9. 

Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes
Show More

Image Attributions

Show Hide Details
Description
Difficulty Level:
At Grade
Grades:
Date Created:
Mar 12, 2013
Last Modified:
Jun 07, 2016
Save or share your relevant files like activites, homework and worksheet.
To add resources, you must be the owner of the Modality. Click Customize to make your own copy.
Please wait...
Please wait...
Image Detail
Sizes: Medium | Original
 
MAT.ALG.933.1.L.1
Here