Example: 2x-1=y,2y+3=x
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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
When a,b∈R, a!=0,b!=0, and m,n∈N
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 a∈R and m,n∈N, 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^m
(+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
1. When n is an even number and a>0, a^(1/n)>0.
(16)^(1/4)=2
When n is an even number and a<0, a^(1/n) is not a real number.
(-4)^(1/2) is not a real number
2. When n is an odd number and a>0, a^(1/n)>0.
(27)^(1/3)=3
When n is an odd number and a<0, a^(1/n)<0.
(-32)^(1/5)=-2
DEFINITION For a∈R and m,n∈N, 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,b∈R, a>0, b>0, and p,q,r,s,u,v∈N, 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)
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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>0, b>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
9 (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
When a=0, we have 0^0, which is indeterminate.
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
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>0, b>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)
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
(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.