Equations with Fractional Exponents - Purdue University

16-week Lesson 15 (8-week Lesson 11)

Equations with Fractional Exponents

We have seen already when covering Lesson 3 that fractional exponents are simply an alternate way of expressing radicals.

6 = (6)

So a square root is equivalent to a power of 12, which is the reciprocal of the index 2. The same is true for any radical; to express a radical as an

exponent, we simply need to take the reciprocal of the index of the radical.

5

=

1

(5)3

4

=

1

(4)4

3

=

1

(3)5

2

=

1

(2)6

This led us to our notation for expressing radicals using fractional exponents, as well as converting fractional exponents back to radicals, which we will be focusing on in this lesson.

Converting

an

exponent

1

( )

to

a

radical

( )

- to write a fractional exponent as a radical, write the denominator of

the exponent as the index of the radical and the base of the expression

as the radicand

3

o the expression 5 can be written as a radical in two ways, both

of which are equivalent

3

5

=

53

3

5

= (5)3

o regardless of which way you choose to write the radical

expression, the index is the same (in this case 5)

- it is imperative that the radical exists, otherwise the expression is

meaningless; in other words, if the radical has an even root, the

radicand must be non-negative

1

16-week Lesson 15 (8-week Lesson 11)

Equations with Fractional Exponents

Steps for Solving Equations with Fractional Exponents: 1. isolate the variable that has a fractional exponent 2. convert from a fractional exponent to a radical 3. solve for the variable by using roots and/or exponents (principle of powers)

Example 1: Solve the following equations for and enter exact answers

only (no decimal approximations). If there is more than one solution,

separate your answers with commas. If there are no real solutions, enter

NO SOLUTION.

3

a. 5 = 27

5

b. 24 + 10 = 74

5

24 = 64

5

4 = 32

(4)5 = 32

To eliminate the exponent of 5 that's outside the parentheses, take the fifth root of both sides of the equation:

5(4)5 = 532

Keep in mind that taking the odd root of both sides of an equation does not produce both a positive and negative root; that only happens with an even root:

4 = 2

Now to eliminate the fourth root, take both sides of the equation to the power of 4:

(4)4 = (2)4

=

2

16-week Lesson 15 (8-week Lesson 11)

Equations with Fractional Exponents

Example 2: Solve the following equations for and enter exact answers

only (no decimal approximations). If there is more than one solution,

separate your answers with commas. If there are no real solutions, enter

NO SOLUTION. a. 2 = 16

2

b. 3 = 16

= ?16

= ?

When taking the square root of both sides of an equation, there is a positive root and a negative root. This is because solving an equation such as 2 = 16 by extracting roots must produce the same answer as if we had solved by factoring.

2 = 16 2 - 16 = 0 ( - 4)( + 4) = 0 - 4 = 0 ; + 4 = 0 = ; = -

3

16-week Lesson 15 (8-week Lesson 11)

Equations with Fractional Exponents

Example 3: Solve the following equations for and enter exact answers

only (no decimal approximations). If there is more than one solution,

separate your answers with commas. If there are no real solutions, enter

NO SOLUTION. a. 2 = -25

2

b. 3 = -25

Using real numbers, it is not possible to have something squared equal to a negative value. So in this case, there is no real number that you can raise to the power of 2, and produce - 25. Therefore 2 = -25 has no real solutions.

If you did not realize that 2 = -25 did not have any real solutions, you could still attempt to take the square root of both sides of the equation in order to solve for .

2 = -25

= ?-25

Since it is not possible to take the square root of a negative value using real numbers, now you should see there is no real solution for this equation.

4

16-week Lesson 15 (8-week Lesson 11)

Equations with Fractional Exponents

Example 4: Solve the following equations for and enter exact answers

only (no decimal approximations). If there is more than one solution,

separate your answers with commas. If there are no real solutions, enter

NO SOLUTION. a. = -64

3

b. 2 = -64

Using real numbers, it is not possible to have the square root of something equal to a negative value. So in this case, there is no real number that you can take the square root of to produce - 64. Therefore

= -64 has no real solutions.

If you did not realize that

= -64 has no real solutions, you could still attempt to solve the problem by squaring both sides of the equation.

= -64

= 4096

However the answer of 4096 does not make the original equation true, because the square root of 4096 is 64, not - 64.

5

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