# EVALUATING EXPRESSIONS WITH NEGATIVE EXPONENTS

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 a⋅ 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 a⋅ a= am+n, we get

= 4 ⋅ v2/3 - 1

= 4v-1/3

By using the negative exponent a-x = 1/ax

= (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

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/⋅ y1/3 ⋅ 1/x4/3  1/y-7/4

By combining the like terms, we get

x1/⋅ 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

Example 6 :

u-5/4v (u3/2)-3/2

Solution :

= u-5/4v (u3/2)-3/2

= u-5/4 ⋅ v (u3/2)-3/2

By using the power rule (am)= amn, we get

=  u-5/4 ⋅ v u(3/2 ⋅ (-3/2))

u-5/4 ⋅ v⋅ u-9/4

By combining the like terms, we get

u-5/4 ⋅ u-9/4 ⋅ v2

By using the product rule a⋅ a= 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

Example 7 :

uv ⋅ u ⋅ (v3/2)3

Solution :

= uv ⋅ u ⋅ (v3/2)3

= u ⋅ ⋅ u ⋅ (v3/2)3

By using the power rule (am)= amn, we get

⋅ ⋅ u ⋅ v9/2

By combining the like terms, we get

= u ⋅ u ⋅ v ⋅ v9/2

By using the product rule a⋅ a= am+n, we get

= u1+1 ⋅ v1+9/2

= u⋅ v11/2

So, the answer is u⋅ 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

⋅ 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

⋅ x-2y5/3 ⋅ x5/4 ⋅ y5/3  1/xy1/2

By using the negative exponent 1/ax a-x, we get

⋅ x-2y5/3 ⋅ x5/4 ⋅ y5/3  x-1 ⋅ y-1/2

By combining the like terms, we get

⋅ x-2 ⋅ x5/4  x-1 ⋅ y5/3 ⋅  y5/3 ⋅ y-1/2

By using the product rule a⋅ a= 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 a⋅ a= am+n, we get

= 2 ⋅ x-3/4-1 ⋅ y10/3-1/2

⋅ x-7/4 ⋅ y17/6

By using the negative exponent a-x = 1/ax

= 2  y17/6 ⋅ 1/x7/4

= (2y17/6)/x7/4

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 a⋅ a= 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

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

y-5/4 ⋅ y 1/2  y1/3

By using the product rule a⋅ a= am+n, we get

⋅ y-5/4+1  1/2  y1/3

= 3 ⋅ y-1/4  y1/3  1/2

= 3  y-1/4+1/3  1/2

⋅ y1/12  1/2

= (3y1/12)/2

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