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# 7.9: Solving Rational Exponent Equations

Created by: CK-12

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

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. $x^3=(4x-3)^{\frac{3}{2}}$

Mar 12, 2013

Oct 28, 2014

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