SIMPLIFYING EXPRESSIONS WITH FRACTIONAL EXPONENTS

The following properties of exponents can be used to simplify expressions with fractional exponents. 

xm ⋅ xn  =  xm+n

xm ÷ xn  =  xm-n

(xm)n  =  xmn

(xy)m  =  xm ⋅ ym

(x / y)m  =  xm / ym

x-m  =  1 / xm

xm/n  =  y -----> x  =  yn/m

(x / y)-m  =  (y / x)m

√x  =  x1/2

n√x  =  x1/n

n√xn  =  x

Example 1 : 

Simplify :

3432/3

Solution : 

=  3432/3

=  3431/3 ⋅ 2

Power of a Power Property.

=  (3√343)2

=  (3√73)2

=  72

=  49

Example 2 : 

Simplify :

641/6 + 1213/2

Solution : 

=  641/6 + 1213/2 

=  641/6 + 1211/2 ⋅ 3

Power of a Power Property.

=  641/6 + (1211/2)3

Write the fractional exponents as radicals.

664 + (121)3

=  6√26 + (√112)3

=  2 + (11)3

=  2 + 1331

=  1333

Example 3 : 

Simplify :

√(x6y6)

Solution : 

=  √(x6y6)

Write the square root as exponent. 

=  (x6y6)1/2

Power of a Product Property.

=   (x6)1/2 ⋅ (y6)1/2

Power of a Power Property.

=   (x⋅ 1/2⋅ (y⋅ 1/2)

Simplify exponents. 

=   x3 ⋅ y3

=   x3y3

Example 4 : 

Simplify :

3√(a12b6)

Solution : 

=  3√(a12b6)

Write the cube root as exponent. 

=  (a12b6)1/3

Power of a Product Property.

=  (a12)1/3 ⋅ (b6)1/3

Power of a Power Property.

 (a12 ⋅ 1/3⋅ (b⋅ 1/3)

Simplify exponents. 

=  a4 ⋅ b2

=  a4b2

Example 5 : 

Simplify :

(b1/2)2 ⋅ √b2

Solution : 

=  (b1/2)2 ⋅ √b2

Power of a Power Property.

 (b1/2 ⋅ 2⋅ b

Simplify exponents. 

 b ⋅ b

Product of Powers Property.

=  b1 + 1

=  b2

Example 6 : 

Simplify :

[(x1/3y1/6)12⋅ [4√y4]

Solution : 

=  [(x1/3y1/6)12⋅ [4√y4]

=  (x1/3y1/6)12 ⋅ y

Power of a Product Property.

=   (x1/3)12 ⋅ (y1/6)12 ⋅ y

Power of a Power Property.

=   (x1/3 ⋅ 12⋅ (y1/6 ⋅ 12⋅ y

=   x4 ⋅ y2 ⋅ y

Product of powers property.

=   x4 ⋅ y2 + 1

=   x4y3

Example 7 : 

Simplify :

(ab1/2)2 / 5√a5

Solution : 

=  (ab1/2)2 / 5√a5

Power of a Product Property.

=  (a)2 ⋅ (b1/2)2 / a

Power of a Power Property.

=  a2 ⋅ b1/2 ⋅ 2 / a

=  a2 ⋅ b / a

Division of powers property.

=  a2 - 1 ⋅ b

=  ab

Example 8 : 

The approximate number of Calories C that an animal needs each day is given by C = 72m2/3, where m is the mass of animal in kilograms. Find the number of Calories that a 27 kg dog needs each day.  

Solution : 

C  =  72m2/3

Substitute 27 for m. 

=  72(27)2/3

=  72(33)2/3

Power of a Power Property.

=  72(3⋅ 2/3)

=  72(32)

=  72(9)

=  648

The dog needs 648 Calories per day to maintain health. 

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. SAT Math Resources (Videos, Concepts, Worksheets and More)

    Dec 23, 24 03:47 AM

    SAT Math Resources (Videos, Concepts, Worksheets and More)

    Read More

  2. Digital SAT Math Problems and Solutions (Part - 91)

    Dec 23, 24 03:40 AM

    Digital SAT Math Problems and Solutions (Part - 91)

    Read More

  3. Digital SAT Math Problems and Solutions (Part - 90)

    Dec 21, 24 02:19 AM

    Digital SAT Math Problems and Solutions (Part - 90)

    Read More