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7.9: Solving Rational Exponent Equations

Difficulty Level: At Grade Created by: CK-12

The period (in seconds) of a pendulum with a length of L (in meters) is given by the formula P=2π(L9.8)12 . If the period of a pendulum is 10π is the length of the pendulum 156.8?

Guidance

This concept is very similar to the previous two. When solving a rational exponent equation, isolate the variable. Then, to eliminate the exponent, you will need to raise everything to the reciprocal power.

Example A

Determine if x = 9 is a solution to 2x3219=35 .

Solution: Substitute in x and see if the equation holds.

2(9)3219227195419=35=35=35

9 is a solution to this equation.

Example B

Solve 3x52=96 .

Solution: First, divide both sides by 3 to isolate x .

3x52x52=96=32

x is raised to the five-halves power. To cancel out this exponent, we need to raise everything to the two-fifths power.

(x52)25xx=3225=3225=3252=22=4

Check: 3(4)52=325=332=96

Example C

Solve 2(x5)34+48=202 .

Solution: Isolate (x5)34 by subtracting 48 and dividing by -2.

2(x5)34+482(x5)34(x5)34=202=250=125

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

[(x5)34]43x5x=(125)43=625=630

Check: 2(6305)34+48=262534+48=2125+48=250+48=202

Intro Problem Revisit We need to plug 156.8 in to the equation P=2π(L9.8)12 for L and solve. If our answer equals 10π , then the given length is correct.

P=2π(L9.8)122π(156.89.8)122π(16)122π(4)=8π

8π does not equal 10π , so the length cannot be 156.8.

Guided Practice

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

1. 8(3x1)23=200

2. 6x32141=1917

Answers

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

8(3x1)23[(3x1)23]323x13xx=200=(25)32=125=126=42

Check: 8(3(42)1)23=8(1261)23=8(125)23=825=200

2. Here, only the 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.

6x321416x32x32x=1917=2058=343=34323=72=49

Check: 6(49)32141=6343141=2058141=1917

Explore More

Determine if the following values of x are solutions to the equation 3x35=24

  1. x=32
  2. x=32
  3. x=8

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

  1. 2x32=54
  2. 3x13+5=17
  3. (7x3)25=4
  4. (4x+5)12=x4
  5. x52=16x12
  6. (5x+7)35=8
  7. 5x23=45
  8. (7x8)23=4(x5)23
  9. 7x37+9=65
  10. 4997=5x323
  11. 2x34=686
  12. x3=(4x3)32

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Difficulty Level:

At Grade

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Date Created:

Mar 12, 2013

Last Modified:

Jun 04, 2015

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