6.3 指数的权力属性
Section outline
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There are 1,000 bacteria present in a culture. When the culture is treated with an antibiotic, the bacteria count is halved every 4 hours. How many bacteria remained 24 hours later?
::文化中存在1000种细菌。当文化被用抗生素处理时,细菌数每4小时减半。24小时后还剩下多少细菌?Power Properties of Exponents
::指数的功率属性The last set of properties to explore are the power properties. Let’s investigate what happens when a power is raised to another power.
::最后一组要探索的属性是功率属性。让我们来调查当一个权力被提升到另一个权力时会发生什么。Power of a Power Property
::权力财产权Step 1: Rewrite ( 2 3 ) 5 as 2 3 five times.
::第1步:重写(23)5为23次5次。( 2 3 ) 5 = 2 3 ⋅ 2 3 ⋅ 2 3 ⋅ 2 3 ⋅ 2 3
Step 2: Expand each 2 3 . How many 2’s are there?
::第二步:扩大每个23个,有多少2个?( 2 3 ) 5 = 2 ⋅ 2 ⋅ 2 ⏟ 2 3 ⋅ 2 ⋅ 2 ⋅ 2 ⏟ 2 3 ⋅ 2 ⋅ 2 ⋅ 2 ⏟ 2 3 ⋅ 2 ⋅ 2 ⋅ 2 ⏟ 2 3 ⋅ 2 ⋅ 2 ⋅ 2 ⏟ 2 3 = 2 15
Step 3: What is the product of the powers?
::第3步:权力的产物是什么?3 ⋅ 5 = 15
Step 4: Fill in the blank. ( a m ) n = a − ⋅ −
::第4步:填空。 (am)n=a( a m ) n = a m n
:am)n=amn(上午)
The other two exponent properties are a form of the .
::其他两种指数属性是...的一种形式。Power of a Product Property : ( a b ) m = a m b m
::产品产权的功率ab)m=ambm
Power of a Quotient Property : ( a b ) m = a m b m
::引号属性的功率ab)m=ambm
Simplify ( 3 4 ) 2
::简化( 3434) 2( 3 4 ) 2 = 3 4 ⋅ 2 = 3 8 = 6561
Simplify ( x 2 y ) 5
::简化 (x2y) 5( x 2 y ) 5 = x 2 ⋅ 5 y 5 = x 10 y 5
:x2y) 5=x2=x2=5y5=x10y5
Simplify ( 3 a − 6 2 2 a 2 ) 4 (do not leave any negative exponents)
::简化 (3a- 622a2) 4 (不要留下任何负指数)This problem uses the Negative Exponent Property . Distribute the 4 t h power first and then move the negative power of a from the numerator to the denominator.
::这个问题使用负指数属性。 先分配第4次功率, 然后将分子的负功率从分子移动到分母 。( 3 a − 6 2 2 a 2 ) 4 = 3 4 a − 6 ⋅ 4 2 2 ⋅ 4 a 2 ⋅ 4 = 81 a − 24 2 8 a 8 = 81 256 a 8 + 24 = 81 256 a 32
:3a-622a2)4=34a-64224a24=81a-2428a8=81256a8+24=81256a32
Simplify 4 x − 3 y 4 z 6 12 x 2 y ÷ ( 5 x y − 1 15 x 3 z − 2 ) 2 (do not leave any negative exponents)
::简化 4x- 3y4z4z612x2y( 5xy- 115x3z-2) 2 (不要留下任何负指数)This problem is definitely as complicated as these types of problems get. Here, all the properties of exponents will be used. Remember that dividing by a fraction is the same as multiplying by its reciprocal .
::这个问题肯定和这类问题一样复杂。 在这里, 所有的指数属性都将被使用。 记住, 分数除以一个分数与对等乘法相同 。4 x − 3 y 4 z 6 12 x 2 y ÷ ( 5 x y − 1 15 x 3 z − 2 ) 2 = 4 x − 3 y 4 z 6 12 x 2 y ⋅ 225 x 6 z − 4 25 x 2 y − 2 = y 3 z 6 3 x 5 ⋅ 9 x 4 y 2 z 4 = 3 x 4 y 5 z 6 x 5 z 4 = 3 y 5 z 2 x
::4-3y4z612x2y(5xy-115x3z-2)2=4x-3y4z612x2y225x6z-425x2y=425x2y2=y3z63x5}9x4y2z4=3x4y5z6x5z4=3y5z2xExamples
::实例Example 1
::例1Earlier, you were asked to find the number of bacteria that remained 24 hours later.
::早些时候,有人要求你找到24小时后留下的细菌数量。To find the number of bacteria remaining, we use the exponential expression 1000 ( 1 2 ) n where n is the number of four-hour periods.
::为了找到剩余细菌的数量, 我们使用指数表达式 1000( 12)n, 其中 n 是四个小时的间隔数 。There are 6 four-hour periods in 24 hours, so we set n equal to 6 and solve.
::24小时有6个4小时的时段 所以我们设定了等于6的时段 然后解决1000 ( 1 2 ) 6
Applying the Power of a Quotient Property, we get:
::运用引号属性的力量,我们得到:1000 ( 1 6 2 6 ) = 1000 ⋅ 1 2 6 = 1000 64 = 15.625
Therefore , there are 15.625 bacteria remaining after 24 hours.
::因此,24小时后还剩下15.625种细菌。Example 2
::例2Simplify without negative exponents: ( 5 a 3 b 4 ) 7 .
:5a3b4)7。
Distribute the 7 to every power within the parenthesis.
::将7号分配到括号内的所有权力( 5 a 3 b 4 ) 7 = 5 7 a 21 b 28 = 78 , 125 a 21 b 28
:5a3b4)7=57a21b28=78,125a21b28
Example 3
::例3Simplify without negative exponents: ( 2 x 5 ) − 3 ( 3 x 9 ) 2 .
::简化无负指数2x5)-3(3x9)2。
Distribute the -3 and 2 to their respective parenthesis and then use the properties of negative exponents, quotient and product properties to simplify.
::将 - 3 和 2 分配给各自的括号,然后使用负指数、商数和产品属性的特性进行简化。( 2 x 5 ) − 3 ( 3 x 9 ) 2 = 2 − 3 x − 15 3 2 x 18 = 9 x 3 8
:2x5)-3(3x9)2=2-3x-1532x18=9x38)
Example 4
::例4Simplify without negative exponents: ( 5 x 2 y − 1 ) 3 10 y 6 ⋅ ( 16 x 8 y 5 4 x 7 ) − 1 .
::简化时没有负引号 : (5x2y- 1) 310y6}(16x8y54x7) - 1 。Distribute the exponents that are outside the parenthesis and use the other properties of exponents to simplify. Anytime a fraction is raised to the -1 power, it is equal to the reciprocal of that fraction to the first power.
::分配括号外的推数, 并使用推数的其他属性来简化。 每当一个分数被提升到 - 1 功率时, 它等于该分数与第一个功率的对等值 。( 5 x 2 y − 1 ) 3 10 y 6 ⋅ ( 16 x 8 y 5 4 x 7 ) − 1 = 5 3 x − 6 y − 3 10 y 6 ⋅ 4 x 7 16 x 8 y 5 = 500 x y − 3 160 x 8 y 11 = 25 8 x 7 y 14
:5x2y-1)310y6(16x8y54x7)-1=53x-6y-310y64x716x8y5=500xy-3160x8y11=258x7y14)
Review
::回顾Simplify the following expressions without negative exponents.
::简化以下表达式,无负引号。- ( 2 5 ) 3
-
(
3
x
)
4
:3x)4
- ( 4 5 ) 2
-
(
6
x
3
)
3
:6x3)3
-
(
2
a
3
b
5
)
7
:2a3b5)7
-
(
4
x
8
)
−
2
:4x8)-2
-
(
1
7
2
h
9
)
−
1
:172h9)-1
-
(
2
x
4
y
2
5
x
−
3
y
5
)
3
:2x4y25x-3y5)3
-
(
9
m
5
n
−
7
27
m
6
n
5
)
−
4
:9m5n-7727m6n5)-4
-
(
4
x
)
2
(
5
y
)
−
3
(
2
x
3
y
5
)
2
:4x)2(5y)-3(2x3y5)2
-
(
5
r
6
)
4
(
1
3
r
−
2
)
5
:5r6)4(13r-2)5
-
(
4
t
−
1
s
)
3
(
2
−
1
t
s
−
2
)
−
3
:4吨-1s)3(2吨-1吨-2)-3
-
6
a
2
b
4
18
a
−
3
b
4
⋅
(
8
b
12
40
a
−
8
b
5
)
2
::6a2b418a-3b4(8b1240a-8b5)2 -
2
(
x
4
y
4
)
0
2
4
x
3
y
5
z
÷
8
z
10
32
x
−
2
y
5
::2(x4y4)024x3y5z8z1032x-2y5 -
5
g
6
15
g
0
h
−
1
⋅
(
h
9
g
15
j
7
)
−
3
::5g615g0h-1(h9g15j7)-3 -
Challenge
a
7
b
10
4
a
−
5
b
−
2
⋅
[
(
6
a
b
12
)
2
12
a
9
b
−
3
]
2
÷
(
3
a
5
b
−
4
)
3
::a7b104a-5b-2[(6ab12)21212a9b-33]2(3a5b-4)3 -
Rewrite
4
3
as a power of 2.
::重写43是2的功率 -
Rewrite
9
2
as a power of 3.
::将92重写为3的功率 -
Solve the equation for
x
.
3
2
⋅
3
x
=
3
8
::解析 x. 323x=38 的方程 -
Solve the equation for
x
.
(
2
x
)
4
=
4
8
::解析 x. (2x)4=48 的方程
Review (Answers)
::回顾(答复)Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键 。