Write each expression in radical form.
Problem 1 :
7^{1/2}
Solution:
= 7^{1/2}
= √7
Problem 2 :
4^{4/3}
Solution:
= 4^{4/3}
Writing 4 in exponential form, we get 4 = 2^{2}
= (2^{2})^{4/3}
When we have power raised to another power, we will multiply both the powers.
= 2^{2} ^{×} ^{(4/3)}
= 2^{8/3}
= ∛(2 × 2 × 2 × 2 × 2 × 2 × 2 × 2)
= 4∛4
Problem 3 :
2^{5/3}
Solution:
= 2^{5/3}
We can write the fractional power as product of integer and fraction. So, we get
= 2^{5 (1/3)}
1/3 can be written as cube root.
= ∛ (2)^{5}
= ∛ (2 × 2 × 2 × 2 × 2)
= 2 ∛4
Problem 4 :
7^{4/3}
Solution:
= 7^{4/3}
We can write the fractional power as product of integer and fraction. So, we get
= 7^{4} ^{(1/3)}
1/3 can be written as cube root.
= ∛ (7)^{4}
= ∛ (7 × 7 × 7 × 7)
= 7 ∛7
Problem 5 :
6^{3/2}
Solution:
= 6^{3/2}
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 :
2^{1/6}
Solution:
= 2^{1/6}
Power 1/6 can be written as 6^{th} root.
= ^{6}√2
Write each expression in exponential form.
Problem 7 :
(√10)^{3}
Solution:
Square root can be written as power 1/2.
= (√10^{3})^{1/2}
= (10)^{3/2}
Problem 8 :
^{6}√2
Solution:
= ^{6}√2
6^{th} root can be written as power 1/6.
= (2)^{1/6}
Problem 9 :
(∜2)^{5}
Solution:
= (∜2)^{5}
4^{th} root can be written as power 1/4.
= (2^{(1/4)})^{5}
= (2)^{5/4}
Problem 10 :
(∜5)^{5}
Solution:
= (∜5)^{5}
4^{th} 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
6^{th} 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 4^{th} 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 :
a^{6/5}
Solution:
Writing the fractional power as a product of integer and fraction, we get
= a^{6/5}
= a^{6 x (1/5)}
1/5 can be written as 5th root.
= ^{5}√a^{6}
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM