NON-INTEGER EXPONENTS
 
 

Objective:   To extend the definition of xn to non-integer exponents..

Recall 1: The square root of 5, is a number b, such that b*b = 5.
Similarly, the cube root of 5 is a number b such that b*b*b =5, and the n-th root of 5 is a number b such that

  In general the n-th root of a is denoted by

 In particular, if n = 2, then the notation is just plain sqrt(a).

Recall 2: Remember that

  This means that 1/a multiplied to itself n times gives 1/(a^n).    This number must be the n-th root of a, because that is how  the n-th root of a was defined.  So,

Recall 3: By knowing that

we also know what a^(m/n)  means,   since

Note

For this to make sense, we must have n > 0, a different from 0 and a^(1/n)  be a real number.  Notice that a^(1/n) is not a real number if n is even and a < 0.  Take the case where a = -1 and n = 2.  In that case we would be looking for a number sqrt(-1),  where  sqrt(-1)*sqrt(-1)= -1.  Remember that if you multiply any real number by itself you must get a non-negative number.

Note
 If m/n > 0, then  0^(m/n)= 0.

Examples

Example 1. Compute the values of

(a)    8^(4/3)  (b)    9^(-3/2)  (c)    (-27)^(4/3).

Solution:

(a)     8^(4/3) = [8^(1/3)]^4=2^4=16,

  (b)     9^(-3/2) =[9^(-1/2)]^3=(1/3)^3=1/27 ,

  (c)      (-27)^(4/3) =[(-27)^(1/3)]^4=(-3)^4=81

Note
 If Example 1 c) had been (-27)^(3/4), then we would have to say that it is not a real number because (-27)^(1/4) is the 4-th root of a negative number, which does not exist.

Example 2. Write the expression  as a fractional exponent.

Solution: The expression  may be rewritten as  .  Apply  Recall 3, to write it as (5*x)^(3/2).

Example 3. Write the expression  as a fractional exponent.

Solution: Work from the inside out.  Applying Recall 3 to sqrt(x^5), we may rewrite it as x^(5/2), so that the original expression becomes sqrt(x^(5/2)).
Apply Recall 3 once again, to write sqrt(x^(5/2))  as x^(5/4) . 

Example 4 Write the expression  as a fractional exponent.

Solution: Write  (x^3)/(y^6)as (x^3)*(y^( -6)),  then use Recall 3 to write the expression as x^(3/3)*y (-6/3).  This may be rewritten as

x*y^(-2).

 If you are asked to write the expression using positive exponents, then you would have to end with
x/(y^2) .

Example 5
a) 
b) 

Solution: (a) Write  sqrt(x) as x^(1/2) and sqrt(y) as y^(1/2).
Then
 sqrt(x)*sqrt(y)=x^(1/2)*y^(1/2),

                      = (x*y)^(1/2),

                      = sqrt (x*y).

  (b) Write  sqrt(x) as x^(1/2) and sqrt(y) as y^(1/2).

   Then  sqrt(x) / sqrt(y) =x^(1/2) / y^(1/2). ,

                                =(x/y)^(1/2)=sqrt (x/y)

Warning: sqrt (x)+sqrt (y) is not sqrt (x+y).

Example 6 Show that =(x+y)^(2/15) .

Solution:  From Example 5 b), we know that

  Let a = (x + y)^(2/3)  and b = (x + y)^(2/5).  Then

= sqrt((x + y)^(2/3) )/ sqrt((x + y)^(2/5) ),

                         = [(x + y)^(2/3)]^(1/2)/ [((x + y)^(2/5)]^(1/2),

                        =[(x + y)^(1/3)]/ [((x + y)^(1/5)]

                        =(x + y)^[(1/3)-(1/5)]

                        =(x+y)^(2/15)

EXERCISES

1. Evaluate 8^(2/3).
 

2. Evaluate (-8)^(1/3).
 

3  Evaluate (1/27)^(2/3)
 

4 Evaluate 
 

5 Evaluate (-8)^( -1/3).
 

6 Evaluate (1/27)^(-2/3).
 

7 Evaluate 
 

Simplify the following:

8. 
 

9. 
 

10.
 

11. 
 

12.
 
 

13. [x^(1/2)*x^(1/5)]/(x^(1/3)]
 

14 [(x*y^2)^(1/3)]*[(x^2*y)^(1/3)]
 

15 [(27*x^3)^(1/3)]

16. 

17. [x^10/y^6]^(1/4)/