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Positive fractional exponents & zero and negative exponents

The purpose of this chapter is to extend the scope of the rules of exponents, discussed in Chapter 3 , and to study some of their applications in algebra.

When a,bRa!=0,b!=0, and m,nN,we have the following theorems from Chapter 3;

THEOREM 1 a^m*a^n=a^(m+n)

THEOREM 2 (a^m)^n=a^(mn)

THEOREM 3 (ab)^m=a^(m)b^m

THEOREM 4 a^m/a^n=

        a^(m-n), when (m)>(n) 1, when (m)=(n) 1/a^(n-m), when (m)<(n)

THEOREM 5 (a/b)^m=a^m/b^m

9.1 Positive Fractional Exponents

For Theorem 2 for exponents to hold for positive fractional exponents, we must have the following definition:

DEFINITION When aR and m,nN, we define

 (a^m)^(1/n)=(a^(1/n))^m=a^(m/n)

  From the definition we have

  (1/a^n)^n=a^(n/n)=a

   When m is an even number, a^mis a positive number if ais either a positive number or a negative number; for example,

         (+2)^4=16 and (-2)^4=16

When m is an odd number, a^m is a positive number when a is a positive number, and a negative number when a is a negative number; for example,

         (+3)^3=27 and (-3)^3=-27

DEFINITION By a^(1/n) we mean a number whose nth power is a (if a^(1/n)=b, the b^n=a), with the following conditions;

1. When n is an even number and a>0a^(1/n)>0.

    (16)^(1/4)=2

     When n is an even number and a<0a^(1/n) is not a real number.

    (-4)^(1/2) is not a real number

2. When n is an odd number and a>0a^(1/n)>0.

     (27)^(1/3)=3

     When n is an odd number and a<0a^(1/n)<0.

  (-32)^(1/5)=-2

DEFINITION For aR and  m,nN, whenever a^(1/n) we define a^(m/n) as (a^(1/n))^m

 According to the definitions above, it can be shown that Theorems 1-3 are valid when a>0, b>0. and m,n are positive fractional exponents.

Note Theorems 1-3 are true for positive fractional exponents when a and b are positive numbers. Hence the literal numbers may not be assigned negative specific values.

The following are direct applications of the theorems

1. 2^2*2^(1/2)=2^(2+1/2)=2^(5/2)

2. x*x^(1/3)=x^(1+1/3)=x^(4/3)

3. 3^(1/2)*3^(3/2)=3^(1/2+3/2=3^2=9

4. x^(1/2)*x^(1/3)=x^(1/2+1/3)=x^(5/6)

5. (2^3)^(2/3)=2^(3*2/3)=2^2=4

6. (81)^(3/4)=(3^4)^(3/4)=3^(4*3/4)=3^3=27

7. (x^(4/5))^10=x^(4/5*10)=x^8

8. (2^(3/2))^(4/9)=2^(3/2*4/9)=2^(2/3)

9. (x^(2/3))^(6/5)=x^(2/3*6/5)=x^(4/5)

10. (3x)^(3/4)=3^(3/4)x^(3/4)

11. (xy)^(1/3)=x^(1/3)y^(1/3)

Note When a,bRa>0b>0, and p,q,r,s,u,vN, we have

    ((a^(p/q)b^(r/s))^(u/v)=a^((pu)/(qv))b^((ru)/(sv)

EXAMPLE Multiply 3x^2and 2x^(1/2)y.

Solution   3x^2(2x^(1/2)y)=6x^(2+1/2)=6x^(5/2)y

EXAMPLE Multiply 2x^(1/3)y^(1/2) and 3x^(2/3)y^(5/2).

Solution (2x^(1/3)y^(1/2))(3x^(2/3)y^(5/2)=(2*3)(x^(1/3)x^(2/3))(y^(1/2)y^(5/2))

  =6x^(1/3+2/3)y^(1/2+5/2)=6xy^3

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EXAMPLE Evaluate (324)^(1/2)

Solution   (324)^(1/2)=(2^4*3^4)^(1/2)=2^(2*1/2)*3^(4*1/2)=2*3^2=18

EXAMPLE Simplify (x^(1/4)y^(3/2))^4.

Solution   (x^(1/4)y^(3/2))^4=x^(1/4*4)y^(3/2*4)=xy^6

EXAMPLE Simplify (x^4y^5)^(1/2)

Solution   (x^4y^5)^(1/2)=x^(4*1/2)y^(5*1/2)=x^2y^(5/2)

EXAMPLE Multiply (x^3y^6)^(2/3) and (x^8y^4)^(3/4)

Solution   (x^3y^6)^(2/3)(x^8y^4)^(3/4)=(x^2y^4)(x^6y^3)=x^8y^7

EXAMPLE Multiply (x^(7/6)y^(5/8))^(8/7) and (x^(4/3)y^(4/7))^(1/2

Solution   (x^(7/6)y^(5/8))^(8/7)(x^(4/3)y^(4/7))^(1/2=(x^(4/3)y^(5/7))(x^(2/3)y^(2/7))=x^2y

EXAMPLE Multiply x^(3/2)(2x^(1/2)-3).

Solution   x^(3/2)(2x^(1/2)-3)=x^(3/2)*2x^(1/2)-x^(3/2)*3=2x^2-3x^(3/2)

Let’s see how our step by step solver solve this and similar problems. Click on "Solve Similar" button to see more examples.

EXAMPLE Multiply (3x^(1/2)-2)(x^(1/2)+3)

Solution   (3x^(1/2)-2)

   (x^(1/2)+3)/(3x-2x^(1/2))

   (+9x^(1/2)-6)/(3x+7x^(1/2)-6)

   Hence (3x^(1/2)-2)(x^(1/2)+3)=3x+7x^(1/2)-6

According to the definition, Theorems 4 and 5 are valid when a>0b>0, and m,n are positive fractional exponents.

The following are direct applications of the theorems:

1. (2^3)/(2^(1/2))=2^(3-1/2)=2^(5/2)

2. (x^4)/(x^(2/5))=x^(4-2/5)=x^(18/15)

3. (5^(2/3))/(5^2)=1/(5^(2-2/3))=1/(5^(4/3))

4. (x^(7/3))/(x^4)=1/(x^(4-7/3))=1/(x^(5/3))

5. (2^(7/4))/(2^(3/4))=2^(7/4-3/4)=2

6. (3^(2/3))/(3^(1/2))=3^(2/3-1/2)=3^(1/6)

7. (7^(1/6))/(7^(5/6))=1/(7^(5/6-1/6)=1/(7^(2/3))

8. (x^(5/2))/(x^(9/2))=1/(x^(9/2-5/2)=1/x^2

(5/x)^(1/4)=(5^(1/4))/(x^(1/4))

EXAMPLE Simplify (x^(5/3)y^(2/5))/(x^(2/3)y).

Solution   (x^(5/3)y^(2/5))/(x^(2/3)y)=(x^(5/3-2/3))/(y^(1-2/5))=x/(y^(3/5))

EXAMPLE Simplify ((a^(5/4)b^(3/8))/(a^(3/8)b^(3/2)))^(4/3).

Solution   ((a^(5/4)b^(3/8))/(a^(3/8)b^(3/2)))^(4/3)=(a^(5/4*4/3)b^(3/8*4/3))/(a^(3/8*4/3)b^(3/2*4/3))=(a^(5/3)b^(1/2))/(a^(1/2)b^2)=(a^(7/6))/(b^(3/2))

EXAMPLE Simplify ((16)^(3/2))/((32)^(3/5))

Solution   ((16)^(3/2))/((32)^(3/5))=(2^4)^(3/2)/((2^5)^(3/5))=(2^6)/(2^3)=2^3=8

EXAMPLE Simplify ((16x^(4/3)y^(7/6))^3)/((8x^(3/2)y^(5/8))^4).

Solution   ((16x^(4/3)y^(7/6))^3)/((8x^(3/2)y^(5/8))^4)=((2^4x^(4/3)y^(7/6))^3)/((2^3x^(3/2)y^(5/8))^4=(2^12x^4y^(7/2))/(2^12x^6y^(5/2))=y/x^2

EXAMPLE Simplify ((9x^2y^4z^6)^(3/2))/((8x^6y^9z^12)^(2/3)).

Solution   ((9x^2y^4z^6)^(3/2))/((8x^6y^9z^12)^(2/3))=((3^2x^2y^4z^6)^(3/2))/((2^3x^6y^9z^12)^(2/3))=(3^3x^3y^6z^9)/(2^2x^4y^6z^8)=(27z)/(4x)

EXAMPLE Simplify (x^(2/3)y^(5/12))^(6/5))/((x^(3/4)y^(1/2))^(4/3)).

Solution   (x^(2/3)y^(5/12))^(6/5))/((x^(3/4)y^(1/2))^(4/3))=(x^(4/5)y^(1/2))/(xy^(2/3)=1/(x^(1/5)y^(1/6))

9.2 Zero and Negative Exponents

For the first and second parts of Theorem 4 for exponents (page 320) to be consistent, we must have, for n=m and a!=0,

    a^(m-n)=1 , or a^0=1

So we define

If a!=0 , a^0=1

When a=0, we have 0^0, which is indeterminate.

According lo this definition. it can be shown that the previous theorems for exponents are valid when a zero exponent occurs.

EXAMPLES 1. 2^0=1

2. (-20)^0=1

3. (a^2b^3)^0=1

Notes 1. 2a^0=2(1)=2

2. if a!=-b,(a+b)^0=1

3. a^0+b^0=1+1=2

Again, for the first and third parts of Theorem 4 for exponents lu be consistent, we must have, when m=0 and a!=0,

 a^(0-n)=1/(a^(n-0))

So we define

 a!=0, a^-n=1/a^n

According to the definition of negative exponents, a!=0, a^-n=1/a^n

, it can be shown that the theorems for exponents are still valid.

EXAMPLES 1. 3^-2=1/3^2=1/9

2. -5^-3=1/5^3=-(1/125

3. x^-4=1/x^4

4. x^-(3/2)=1/(x^(3/2))

Remarks Theorems 1-5 on page 320 are true when a>0b>0, and m,n are rational numbers.

  Now we can write Theorem 4 as

  (a^m)/(a^n)=a^(m-n)

The following are direct applications of the theorems:

1. x^-2*x^5=x^(-2+5)=x^3

2. x^-2*x^-3=x^(-2-3)=x^-5=1/x^5

3. (x^2)^-3=x^(2(-3))=x^-6=1/x^6

4. (x^-3)^4=x^(-3(4))=x^-12=1/x^12

5. (x^-2)^-5=x^(-2(-5))=x^10

6. (xy)^-2=x^-2y^-2=1/(x^2y^2)

7. (x^3)/(x^-6)=x^(3-(-6))=x^(3+6)=x^9

8. (x^-4)/(x)=1/(x^(1-(-4))=1/x^5

9. (x/y)^-6=x^-6/y^-6=y^6/x^6

10. 0.004=4/1000=4/10^3=4×10^-3

Notes    1. -a^-n=-(1/a^n)

2. 1/(a^-n)=1/(1/a^n)=a^n

3. (a/b)^-n=(a^-n)/(b^-n)=(b^n)/(a^n)=(b/a)^n

4. (a+b)^-n=1/(a+b)^n a!=-b

5. a^-n+b^-n=1/a^n+1/b^n=(b^n+a^n)/(a^(n)b^n)

EXAMPLE Express xy^-2 with positive exponents.

Solution   xy^-2=x*1/y^2=x/y^2

EXAMPLE Multiply x^-1y^-3 and (x^2y^-2) and write the answer with positive exponents.

Solution  (x^-1y^-3)(x^2y^-2)=(x^-1x^2)(y^-3y^-2)

=x^(-1+2)y^(-3-2)

   = xy^-5

   = x/y^5

EXAMPLE Simplify (3x^-2y)^3 and write the answer with positive exponents.

Solution  (3x^-2y)^3=3^3x^-6y^3=3^3*1/x^6*y^3=(27y^3)/(x^6)

EXAMPLE Simplify (2x^2y^-3)^-2and write the answer with positive exponents.

 Solution  (2x^2y^-3)^-2=2^-2x^-4y^6

         =1/2^2*1/x^4*y^6=y^6/(4x^4)

EXAMPLE Simplify(xy^-1z^-2)^2(2^-1x^-2yz^-3)^-3 and write the answer with positive exponents.

Solution  (xy^-1z^-2)^2(2^-1x^-2yz^-3)^-3

 =(x^2y^-2z^-4)(2^3x^6y^-3z^9)

=2^3(x^2x^6)(y^-2y^-3)(z^-4z^9)

=8x^8y^-5z^5 =(8x^8z^5)/(y^5)

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EXAMPLE Simplify (x^-3y^2z^-2)/(x^2y^-4z^-3) and write the answer with positive exponents.

Solution  (x^-3y^2z^-2)/(x^2y^-4z^-3) =x^-3/x^2*y^2/y^-4*z^-2/z^-3

  =1/(x^2x^3)*(y^2y^4)/1*z^3/z^2=(y^6z)/x^5

  

EXAMPLE Simplify (x^2y^-4z^3)^-2/(x^-3y^-2z^-1)^2 and write the answer with positive exponents.

Solution  (x^2y^-4z^3)^-2/(x^-3y^-2z^-1)^2 =(x^-4y^8z^-6)/(x^-6y^-4z^-2)

  =(x^(-4+6))/1*(y^(8+4))/1*1/(z^(-2+6))

  =(x^2y^12)/z^4

EXAMPLE Simplify (2a^-1-3b^-2)/(a^-1+2b^-2)and write the answer with positive exponents.

 Solution  (2a^-1-3b^-2)/(a^-1+2b^-2) =(2/a-3/b^2)/(1/a+2/b^2)

Multiplying the numerator and denominator of the complex fraction by ab^2,
we get

  (2b^2-3a)/(b^2+2a)

The last result can be accomplished by multiplying both numerator and
denominator of the original fraction by ab^2.

  (2a^-1-3b^-2)/(a^-1+2b^-2)=(ab^2(2a^-1-3b^-2))/(ab^2(a^-1+2b^-2)=(2b^2-3a)/(b^2+2a)

Any positive number in decimal notation can be written as the product of a number between 1 and 10 and a power of 10. For example:

1. 32.5=3.25*10^1

2. 738.6=7.386*100=7.386*10^2

3. 6.78=6.78*10^0

4. 0.976=(9.67)/10=9.67*10^-1

5. 0.064=6.4/100=6.4/10^2=6.4*10^-2

6. 0.008=8.0/1000=8.0/10^3=8.0*10^-3

The decimal point is always placed after the leftmost digit. This notation is called the scientific notation for a number.