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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,bRa0,b0, and m,nN,we have the following theorems from Chapter 3;

THEOREM 1 am·an=am+n

THEOREM 2 (am)n=amn

THEOREM 3 (ab)m=am bm

THEOREM 4 aman=

{am-n,when (m)>(n)1,when (m)=(n)1an-m,when (m)<(n)

THEOREM 5 (ab)m=ambm

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

(am)1n=(a1n)m=amn

From the definition we have

(1an)n=ann=a

When m is an even number, amis 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, am 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 a1n we mean a number whose nth power is a (if a1n=b, the bn=a), with the following conditions;

1. When n is an even number and a>0a1n>0.

(16)14=2

When n is an even number and a<0a1n is not a real number.

( 4)12 is not a real number

2. When n is an odd number and a>0a1n>0.

(27)13=3

When n is an odd number and a<0a1n<0.

( 32)15= 2

DEFINITION For aR and  m,nN, whenever a1n we define amn as (a1n)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. 22·212=22+12=252

2. x·x13=x1+13=x43

3. 312·332=312+32=32=9

4. x12·x13=x12+13=x56

5. (23)23=23·23=22=4

6. (81)34=(34)34=34·34=33=27

7. (x45)10=x45·10=x8

8. (232)49=232·49=223

9. (x23)65=x23·65=x45

10. (3 x)34=334 x34

11. (xy)13=x13 y13

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

(apq brs)uv=apuqv brusv

EXAMPLE Multiply 3 x2and 2 x12 y.

Solution   3 x2 (2 x12 y)=6 x2+12=6 x52 y

EXAMPLE Multiply 2 x13 y12 and 3 x23 y52.

Solution (2 x13 y12) (3 x23 y52)=(2·3) (x13 x23) (y12 y52)

=6 x13+23 y12+52=6 xy3

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

EXAMPLE Evaluate (324)12

Solution   (324)12=(24·34)12=22·12·34·12=2·32=18

EXAMPLE Simplify (x14 y32)4.

Solution   (x14 y32)4=x14·4 y32·4=xy6

EXAMPLE Simplify (x4 y5)12

Solution   (x4 y5)12=x4·12 y5·12=x2 y52

EXAMPLE Multiply (x3 y6)23 and (x8 y4)34

Solution   (x3 y6)23 (x8 y4)34=(x2 y4) (x6 y3)=x8 y7

EXAMPLE Multiply (x76 y58)87 and (x43 y47)12

Solution   (x76 y58)87(x43 y47)12=(x43 y57) (x23 y27)=x2 y

EXAMPLE Multiply x32 (2 x12-3).

Solution   x32 (2 x12-3)=x32·2 x12-x32·3=2 x2-3 x32

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 (3 x12-2) (x12+3)

Solution   (3 x12-2)

x12+33 x-2 x12

+ 9 x12-63 x+7 x12-6

Hence (3 x12-2) (x12+3)=3 x+7 x12-6

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. 23212=23-12=252

2. x4x25=x4-25=x1815

3. 52352=152-23=1543

4. x73x4=1x4-73=1x53

5. 274234=274-34=2

6. 323312=323-12=316

7. 716756=1756-16=1723

8. x52x92=1x92-52=1x2

(5x)14=514x14

EXAMPLE Simplify x53 y25x23 y.

Solution   x53 y25x23 y=x53-23y1-25=xy35

EXAMPLE Simplify (a54 b38a38 b32)43.

Solution   (a54 b38a38 b32)43=a54·43 b38·43a38·43 b32·43=a53 b12a12 b2=a76b32

EXAMPLE Simplify (16)32(32)35

Solution   (16)32(32)35=(24)32(25)35=2623=23=8

EXAMPLE Simplify (16 x43 y76)3(8 x32 y58)4.

Solution   (16 x43 y76)3(8 x32 y58)4=(24 x43 y76)3(23 x32 y58)4=212 x4 y72212 x6 y52=yx2

EXAMPLE Simplify (9 x2 y4 z6)32(8 x6 y9 z12)23.

Solution   (9 x2 y4 z6)32(8 x6 y9 z12)23=(32 x2 y4 z6)32(23 x6 y9 z12)23=33 x3 y6 z922 x4 y6 z8=27 z4 x

EXAMPLE Simplify (x23 y512)65(x34 y12)43.

Solution   (x23 y512)65(x34 y12)43=x45 y12xy23=1x15 y16

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 a0,

am-n=1 , or a0=1

So we define

If a0 , a0=1

When a=0, we have 00, 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. 20=1

2. ( 20)0=1

3. (a2 b3)0=1

Notes 1. 2 a0=2 (1)=2

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

3. a0+b0=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 a0,

a0-n=1an-0

So we define

a0,a n=1an

According to the deﬁnition of negative exponents, a0,a n=1an

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

EXAMPLES 1. 3 2=132=19

2.  5 3=153= (1125)

3. x 4=1x4

4. x (32)=1x32

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

aman=am-n

The following are direct applications of the theorems:

1. x 2·x5=x 2+5=x3

2. x 2·x 3=x 2-3=x 5=1x5

3. (x2) 3=x2 ( 3)=x 6=1x6

4. (x 3)4=x 3 (4)=x 12=1x12

5. (x 2) 5=x 2 ( 5)=x10

6. (xy) 2=x 2 y 2=1x2 y2

7. x3x 6=x3-( 6)=x3+6=x9

8. x 4x=1x1-( 4)=1x5

9. (xy) 6=x 6y 6=y6x6

10. 0.004=41000=4103=4·10 3

Notes    1.  a n= (1an)

2. 1a n=11an=an

3. (ab) n=a nb n=bnan=(ba)n

4. (a+b) n=1(a+b)n a b

5. a n+b n=1an+1bn=bn+anan bn

EXAMPLE Express xy 2 with positive exponents.

Solution   xy 2=x·1y2=xy2

EXAMPLE Multiply x 1 y 3 and (x2 y 2) and write the answer with positive exponents.

Solution  (x 1 y 3)(x2 y 2)=(x 1 x2) (y 3 y 2)

=x 1+2 y 3-2

= xy 5

= xy5

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

Solution  (3 x 2 y)3=33 x 6 y3=33·1x6·y3=27 y3x6

EXAMPLE Simplify (2 x2 y 3) 2and write the answer with positive exponents.

Solution  (2 x2 y 3) 2=2 2 x 4 y6

=122·1x4·y6=y64 x4

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

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

=(x2 y 2 z 4) (23 x6 y 3 z9)

=23 (x2 x6) (y 2 y 3) (z 4 z9)

=8 x8 y 5 z5 =8 x8 z5y5

Let's see how our math solver simplifies this problem and other similar problems. Please click "Solve Similar" for more such examples.

EXAMPLE Simplify x 3 y2 z 2x2 y 4 z 3 and write the answer with positive exponents.

Solution  x 3 y2 z 2x2 y 4 z 3 =x 3x2·y2y 4·z 2z 3

=1x2 x3·y2 y41·z3z2=y6 zx5

EXAMPLE Simplify (x2 y 4 z3) 2(x 3 y 2 z 1)2 and write the answer with positive exponents.

Solution  (x2 y 4 z3) 2(x 3 y 2 z 1)2 =x 4 y8 z 6x 6 y 4 z 2

=x 4+61·y8+41·1z 2+6

=x2 y12z4

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

Solution  2 a 1-3 b 2a 1+2 b 2 =2a-3b21a+2b2

Multiplying the numerator and denominator of the complex fraction by ab2,
we get

2 b2-3 ab2+2 a

The last result can be accomplished by multiplying both numerator and
denominator of the original fraction by ab2.

2 a 1-3 b 2a 1+2 b 2=ab2 (2 a 1-3 b 2)ab2 (a 1+2 b 2)=2 b2-3 ab2+2 a

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=3.25·101

2. 738.6=7.386·100=7.386·102

3. 6.78=6.78·100

4. 0.976=9.6710=9.67·10 1

5. 0.064=6.4100=6.4102=6.4·10 2

6. 0.008=8.01000=8.0103=8.0·10 3

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

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