Write each expression in radical form :
Problem 1 :
m3/5
Solution :
= m3/5
We can write the fractional power as product of integer and fraction. So, we get
= m3 ⋅ (1/5)
Power 1/5 can be used as 5th root.
= 5√m3
Problem 2 :
(10x)-3/4
Solution :
First let us write -3/4 as a product of integer and fraction.
(10x)-3/4 = (10x)-3 ⋅ (1/4)
Changing power 1/4 as 4th root.
= 4√(10x)-3
In order to change the negative exponent as positive exponent, we will flip the base.
= 4√1/(10x)3
= 1/4√(10x)3
Problem 3 :
(7x)3/2
Solution :
= (7x)3/2
Writing the fractional power as a product of integer and fraction, we get
= (7x)3⋅ (1/2)
Power 1/2 can be written as square root.
= √(7x)3
Here 7x is repeated 3 times inside the square root.
Accordingly the rule in square root, if we see two same terms inside the square root, we can take one out of the radical.
= √(7x) (7x) (7x)
= 7x√(7x)
Problem 4 :
(6b)-4/3
Solution :
= (6b)-4/3
Writing the fractional power as a product of integer and fraction, we get
= (6b)-4⋅ (1/3)
1/3 can be written as cube root.
= 3√(6b)-4
In order to change the negative exponent as positive exponent, we will flip the base.
= 3√1/(6b)4
Since it is cube root, we take one term out of the cube root for every three same terms.
= 3√1/(6b) ⋅ (6b) ⋅ (6b) ⋅ (6b)
= 1/(6b 3√6b)
Problem 5 :
82/3
Solution :
Method 1 :
= 82/3
= 82 ⋅ (1/3)
Writing 1/3 as cube root.
= 3√82
= 3√64
= 3√(4 ⋅ 4 ⋅ 4)
= 4
Method 2 :
When we evaluate with just numerical values in the base, we will try to write the base in exponential form.
= 82/3
Writing 8 in exponential form, we get 8 = 23
= (23)2/3
When we have power raised to another power, we will multiply both the powers.
= 23 x (2/3)
= 22
= 4
Problem 6 :
100-3/2
Solution :
Method 1 :
= 100-3/2
= 100-3(1/2)
= √(100)-3
= 1/√(100)3
= 1/100√100
= 1/100(10)
= 1/1000
Method 2 :
= 100-3/2
Writing 100 in exponential form, we get
100 = 102
= (10)2(-3/2)
= 10-3
= 1/103
= 1/1000
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