To evaluate an expression with exponents, we should to be aware of the rules of exponents.
The Required Rules of Exponents are below.
Example 1 :
yx1/3 ⋅ xy3/2
Solution :
= yx1/3 ⋅ xy3/2
By writing each term separately, we get
= y ⋅ x1/3 ⋅ x ⋅ y3/2
By combining the like terms, we get
= x1/3 ⋅ x ⋅ y3/2 ⋅ y
By using the product rule am ⋅ an = am+n, we get
= x1/3+1 ⋅ y3/2+1
= x4/3 ⋅ y5/2
So, the answer is x4/3 ⋅ y5/2
Example 2 :
4v2/3 ⋅ v-1
Solution :
= 4v2/3 ⋅ v-1
By writing each term separately, we get
= 4 ⋅ v2/3 ⋅ v-1
By using the product rule am ⋅ an = am+n, we get
= 4 ⋅ v2/3 - 1
= 4v-1/3
By using the negative exponent a-x = 1/ax
= (4)/v1/3
So, the answer is (4)/v1/3.
Example 3 :
(a1/2 b1/2)-1
Solution :
= (a1/2 b1/2)-1
By using the negative exponent a-x = 1/ax
= 1/(a1/2 b1/2)
So, the answer is 1/(a1/2 b1/2).
Example 4 :
(x0y1/3)3/2 x0
Solution :
= (x0y1/3)3/2 x0
By using the zero exponent a0 = 1, we get
= (1y1/3)3/2 1
= (y1/3)3/2
By using the power rule (am)n = amn, we get
= y(1/3 ⋅ 3/2)
= y3/6
= y1/2
So, the answer is y1/2.
Example 5 :
(2x1/2y1/3)/(2x4/3y-7/4)
Solution :
= (2x1/2y1/3)/(2x4/3y-7/4)
By writing each term separately, we get
= x1/2 ⋅ y1/3 ⋅ 1/x4/3 ⋅ 1/y-7/4
By combining the like terms, we get
= x1/2 ⋅ 1/x4/3 ⋅ y1/3 ⋅ 1/y-7/4
By using the quotient rule am/an = am-n, we get
= x1/2-4/3 ⋅ y1/3+7/4
= x-5/6⋅ y25/12
By using the negative exponent a-x = 1/ax
= y25/12 ⋅ 1/x5/6
= y25/12/x5/6
So, the answer is y25/12/x5/6.
Example 6 :
u-5/4v2 ⋅ (u3/2)-3/2
Solution :
= u-5/4v2 ⋅ (u3/2)-3/2
= u-5/4 ⋅ v2 ⋅ (u3/2)-3/2
By using the power rule (am)n = amn, we get
= u-5/4 ⋅ v2 ⋅ u(3/2 ⋅ (-3/2))
= u-5/4 ⋅ v2 ⋅ u-9/4
By combining the like terms, we get
= u-5/4 ⋅ u-9/4 ⋅ v2
By using the product rule am ⋅ an = am+n, we get
= u-5/4-9/4 ⋅ v2
= u-14/4 ⋅ v2
= u-7/2 ⋅ v2
By using the negative exponent a-x = 1/ax
= 1/u7/2 ⋅ v2
= v2/u7/2
So, the answer is v2/u7/2.
Example 7 :
uv ⋅ u ⋅ (v3/2)3
Solution :
= uv ⋅ u ⋅ (v3/2)3
= u ⋅ v ⋅ u ⋅ (v3/2)3
By using the power rule (am)n = amn, we get
= u ⋅ v ⋅ u ⋅ v9/2
By combining the like terms, we get
= u ⋅ u ⋅ v ⋅ v9/2
By using the product rule am ⋅ an = am+n, we get
= u1+1 ⋅ v1+9/2
= u2 ⋅ v11/2
So, the answer is u2 ⋅ v11/2.
Example 8 :
(2x-2y5/3)/x-5/4y-5/3 ⋅ xy1/2
Solution :
= (2x-2y5/3)/x-5/4y-5/3 ⋅ xy1/2
By writing each term separately, we get
= 2 ⋅ x-2y5/3 ⋅ 1/x-5/4 ⋅ 1/y-5/3 ⋅ 1/xy1/2
By using the negative exponent 1/a-x = ax, we get
= 2 ⋅ x-2y5/3 ⋅ x5/4 ⋅ y5/3 ⋅ 1/xy1/2
By using the negative exponent 1/ax = a-x, we get
= 2 ⋅ x-2y5/3 ⋅ x5/4 ⋅ y5/3 ⋅ x-1 ⋅ y-1/2
By combining the like terms, we get
= 2 ⋅ x-2 ⋅ x5/4 ⋅ x-1 ⋅ y5/3 ⋅ y5/3 ⋅ y-1/2
By using the product rule am ⋅ an = am+n, we get
= 2 ⋅ x-2+5/4 ⋅ x-1 ⋅ y5/3+5/3 ⋅ y-1/2
= 2 ⋅ x-3/4 ⋅ x-1 ⋅ y10/3 ⋅ y-1/2
By using the product rule am ⋅ an = am+n, we get
= 2 ⋅ x-3/4-1 ⋅ y10/3-1/2
= 2 ⋅ x-7/4 ⋅ y17/6
By using the negative exponent a-x = 1/ax
= 2 ⋅ y17/6 ⋅ 1/x7/4
= (2y17/6)/x7/4
So, the answer is (2y17/6)/x7/4.
Example 9 :
(a3/4b-1 ⋅ b7/4)/3b-1
Solution :
= (a3/4b-1 ⋅ b7/4)/3b-1
By writing each term separately, we get
= a3/4 ⋅ b-1 ⋅ b7/4 ⋅ 1/3 ⋅ b-1
By using the negative exponent 1/a-x = ax, we get
= a3/4 ⋅ b-1 ⋅ b7/4 ⋅ 1/3 ⋅ b
By using the product rule am ⋅ an = am+n, we get
= a3/4 ⋅ b-1+7/4 ⋅ 1/3 ⋅ b
= a3/4 ⋅ b3/4 ⋅ b ⋅ 1/3
= a3/4 ⋅ b3/4+1 ⋅ 1/3
= a3/4 ⋅ b7/4 ⋅ 1/3
= (a3/4b7/4)/3
So, the answer is (a3/4b7/4)/3.
Example 10 :
(3y-5/4)/y-1 ⋅ 2y-1/3
Solution :
= (3y-5/4)/y-1 ⋅ 2y-1/3
By writing each term separately, we get
= 3 ⋅ y-5/4 ⋅ 1/y-1 ⋅ 1/2 ⋅ 1/y-1/3
By using the negative exponent 1/a-x = ax, we get
= 3 ⋅ y-5/4 ⋅ y1 ⋅ 1/2 ⋅ y1/3
By using the product rule am ⋅ an = am+n, we get
= 3 ⋅ y-5/4+1 ⋅ 1/2 ⋅ y1/3
= 3 ⋅ y-1/4 ⋅ y1/3 ⋅ 1/2
= 3 ⋅ y-1/4+1/3 ⋅ 1/2
= 3 ⋅ y1/12 ⋅ 1/2
= (3y1/12)/2
So, the answer is (3y1/12)/2.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
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