# CONVERTING BETWEEN RADICAL FORM AND EXPONENTIAL FORM

Write each expression in radical form.

Problem 1 :

71/2

Solution:

= 71/2

= √7

Problem 2 :

44/3

Solution:

= 44/3

Writing 4 in exponential form, we get 4 = 22

= (22)4/3

When we have power raised to another power, we will multiply both the powers.

= 22 × (4/3)

= 28/3

= ∛(2 × 2 × 2 × 2 × 2 × 2 × 2 × 2)

= 4∛4

Problem 3 :

25/3

Solution:

= 25/3

We can write the fractional power as product of integer and fraction. So, we get

= 25 (1/3)

1/3 can be written as cube root.

= ∛ (2)5

= ∛ (2 × 2 × 2 × 2 × 2)

= 2 ∛4

Problem 4 :

74/3

Solution:

= 74/3

We can write the fractional power as product of integer and fraction. So, we get

= 74 (1/3)

1/3 can be written as cube root.

= ∛ (7)4

= ∛ (7 × 7 × 7 × 7)

= 7 ∛7

Problem 5 :

63/2

Solution:

= 63/2

We can write the fractional power as product of integer and fraction. So, we get

= (6)3 (1/2)

1/2 can be written as square root.

= √(6)3

= √(6 × 6 × 6)

= 6√6

Problem 6 :

21/6

Solution:

= 21/6

Power 1/6 can be written as 6th root.

6√2

Write each expression in exponential form.

Problem 7 :

(√10)3

Solution:

Square root can be written as power 1/2.

= (√103)1/2

= (10)3/2

Problem 8 :

6√2

Solution:

6√2

6th root can be written as power 1/6.

= (2)1/6

Problem 9 :

(∜2)5

Solution:

= (∜2)5

4th root can be written as power 1/4.

= (2(1/4))5

= (2)5/4

Problem 10 :

(∜5)5

Solution:

= (∜5)5

4th root can be written as power 1/4.

= (5(1/4))5

= (5)5/4

Problem 11 :

∛2

Solution:

= ∛2

cube root can be written as power 1/3.

= (2)1/3

Problem 12 :

6√10

Solution:

6√10

6th root can be written as power 1/6.

= (10)1/6

Write each expression in radical form.

Problem 13 :

(5x)-5/4

Solution:

Let us write -5/4 as a product of integer and fraction.

(5x)-5/4 = (5x)-5 · (1/4)

Changing power 1/4 as 4th root.

= ∜(5x)-5

Inorder to change the negative exponent as positive exponent, we will flip the base.

= ∜1/(5x)5

= 1/∜(5x)5

Problem 14 :

(5x)-1/2

Solution:

Let us write -1/2 as a product of integer and fraction.

(5x)-1/2 = (5x)-1 · (1/2)

Changing power 1/2 as square root.

= √(5x)-1

In order to change the negative exponent as positive exponent, we will flip the base.

= 1/√(5x)

Problem 15 :

(10n)3/2

Solution:

= (10n)3/2

Writing the fractional power as a product of integer and fraction, we get

= (10n)3 (1/2)

Power 1/2 can be written as square root.

= √(10n)3

Here 10n is repeated 3 times inside the square root.

= √(10n) (10n) (10n)

= 10n√(10n)

Problem 16 :

a6/5

Solution:

Writing the fractional power as a product of integer and fraction, we get

= a6/5

= a6 x (1/5)

1/5 can be written as 5th root.

= 5√a6

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