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Solving Equations with Fractional Exponents

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Solving Rational Exponent Equations
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The period (in seconds) of a pendulum with a length of L (in meters) is given by the formula P = 2\pi{(\frac{L}{9.8})}^{\frac{1}{2}} . If the period of a pendulum is 10\pi 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 2x^{\frac{3}{2}}-19=35 .

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

2(9)^{\frac{3}{2}}-19&=35 \\2 \cdot 27 -19 &= 35 \\54 - 19 &= 35

9 is a solution to this equation.

Example B

Solve 3x^{\frac{5}{2}}=96 .

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

3x^{\frac{5}{2}}&=96\\x^{\frac{5}{2}}&=32

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

\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

Check: 3(4)^{\frac{5}{2}}=3 \cdot 2^5=3 \cdot 32=96

Example C

Solve -2(x-5)^{\frac{3}{4}}+48=-202 .

Solution: Isolate (x-5)^{\frac{3}{4}} by subtracting 48 and dividing by -2.

-2(x-5)^{\frac{3}{4}}+48&=-202\\-2(x-5)^{\frac{3}{4}}&=-250\\(x-5)^{\frac{3}{4}}&=-125

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

\left[ \left(x-5 \right)^{\frac{3}{4}}\right]^{\frac{4}{3}}&=\left(-125 \right)^{\frac{4}{3}}\\x-5&=625\\x&=630

Check: -2(630-5)^{\frac{3}{4}}+48=-2 \cdot 625^{\frac{3}{4}}+48=-2 \cdot 125+48=-250+48=-202

Intro Problem Revisit We need to plug 156.8 in to the equation P = 2\pi{(\frac{L}{9.8})}^{\frac{1}{2}} for L and solve. If our answer equals 10\pi , then the given length is correct.

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

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

Guided Practice

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

1. 8(3x-1)^{\frac{2}{3}}=200

2. 6x^{\frac{3}{2}}-141=1917

Answers

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

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

Check: 8(3(42)-1)^{\frac{2}{3}}=8(126-1)^{\frac{2}{3}}=8(125)^{\frac{2}{3}}=8 \cdot 25=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.

6x^{\frac{3}{2}}-141&=1917 \\6x^{\frac{3}{2}}&=2058 \\x^{\frac{3}{2}}&=343 \\x&=343^{\frac{2}{3}}=7^2=49

Check: 6(49)^{\frac{3}{2}}-141=6 \cdot 343-141=2058-141=1917

Explore More

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

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

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

  1. 2x^{\frac{3}{2}}=54
  2. 3x^{\frac{1}{3}}+5=17
  3. (7x-3)^{\frac{2}{5}}=4
  4. (4x+5)^{\frac{1}{2}}=x-4
  5. x^{\frac{5}{2}}=16x^{\frac{1}{2}}
  6. (5x+7)^{\frac{3}{5}}=8
  7. 5x^{\frac{2}{3}}=45
  8. (7x-8)^{\frac{2}{3}}=4(x-5)^{\frac{2}{3}}
  9. 7x^{\frac{3}{7}}+9=65
  10. 4997=5x^{\frac{3}{2}}-3
  11. 2x^{\frac{3}{4}}=686
  12. x^3=(4x-3)^{\frac{3}{2}}

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