2.3 指数表达式-interactive

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• 选择节指数表达式

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  指数表达式

  Demonstrate the ability to multiply and divide exponents.

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  Understand how to raise a power to a power.
  ::懂得如何将权力提升到权力。

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  Recognize rational exponents in the context of numerical and algebraic expressions.
  ::在数字和代数表达式方面,承认合理的推论。

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  The Purpose of this Lesson 
  ::本课程的目的

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  In this lesson, you will first explore exponential expressions that feature integer exponents, including zero and negative numbers. Then you'll explore rational exponents more generally, including . The ability to work with all rational numbers as exponents will improve your and add skills to your problem-solving toolbox.
  ::在此课中, 您将首先探索具有整数指数的指数表达式, 包括零和负数。 然后您将更广义地探索理性的指数表达式, 包括 。 以所有合理数字作为指数来工作的能力将提高您的能力, 并将技能添加到您解决问题的工具箱中 。

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  Activity 1: Integer Exponents
  ::活动1:整数指数

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  Example 1-1
  ::例1-1

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  Nora is an ichthyologist performing experiments on a fish in a simulation. The fish starts with a mass of 1 kilogram, and Nora feeds it blue pills and red pills. When she feeds the fish a blue pill, it doubles in size. When she feeds it a red pill, it halves in size. If Nora first feeds the fish 7 blue pills, then 5 red pills, what's the mass of the fish?
  ::诺拉是一个在模拟中实验鱼类的理疗师。 鱼从1公斤的重量开始, 诺拉喂养蓝药丸和红药丸。 当她喂养鱼时, 它的体积翻了一番。 当她喂养一个红药丸时, 它的体积减半。 如果诺拉首先喂养鱼, 7个蓝药丸, 5个红药丸, 鱼的重量是多少?

  

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  Solution:   Since Nora feeds the fish 7 blue pills, the fish's 1-kilogram mass doubles 7 times:
  ::溶液:自从诺拉喂养鱼七颗蓝药丸以来, 鱼的1公斤质量翻了七倍:

  $$\begin{array}{r}1\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\end{array}$$

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  That's a big fish! But Nora then feeds the fish 5 red pills, so this big fish halves in size five times:
  ::那是条大鱼!

  $$\begin{array}{r}1\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2÷2÷2÷2÷2÷2\end{array}$$

   

  $$\begin{array}{r}1\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2÷2÷2÷2÷2÷2\end{array}$$

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  Canceling 2's, 5 times, gives:
  ::取消 2 秒, 5 次, 给:

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  $$\begin{array}{r}\begin{array}{c}1\cdot 2\cdot 2\\ 4\text{kilograms}\end{array}\end{array}$$

  
  ::1224公斤

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  $$\begin{array}{r}\begin{array}{c}1\cdot 2\cdot 2\\ 4\text{kilograms}\end{array}\end{array}$$

   
  ::1224公斤

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  Your previous work with may have led you to a simpler way of solving this problem. The enlargement of the fish can be written more simply in exponential form :
  ::你以前的工作或许使你找到了 解决这个问题的更简单的方法。 鱼的膨胀可以用指数形式写得更简单:

  $$\begin{array}{r}1(2{)}^7\end{array}$$

   

  $$\begin{array}{r}1(2{)}^7\end{array}$$

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  The subsequent reduction of the big fish can also be written more simply:
  ::大鱼随后的减少也可以写得更简单些:

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  $$\begin{array}{r}\begin{array}{ll}\text{Expression}& \text{Explanation}\\ 1\cdot 2^7÷2^5& \text{Exponential form.}\\ 1\cdot \frac{2^7}{2^5}& \text{Fractions represent division.}\\ 1(2{)}^2& \text{Canceling 5 2's on the top with 5 on the bottom.}\end{array}\end{array}$$

  
  ::Explantation12725 Explantation12725 Explantial form.12725Fractions 代表除法1(2)2 上方5 2 和底部5。

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  $$\begin{array}{r}\begin{array}{ll}\text{Expression}& \text{Explanation}\\ 1\cdot 2^7÷2^5& \text{Exponential form.}\\ 1\cdot \frac{2^7}{2^5}& \text{Fractions represent division.}\\ 1(2{)}^2& \text{Canceling 5 2's on the top with 5 on the bottom.}\end{array}\end{array}$$

  
  ::Explantation12725 Explantation12725 Explantial form.12725Fractions 代表除法1(2)2 上方5 2 和底部5。

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    The canceling process above can also be accomplished by subtracting exponents:
  ::上述取消过程也可以通过减去表率来实现:

  $$\begin{array}{r}\begin{array}{l}\frac{2^7}{2^5}\\ 2^{(7-5)}\\ 2^2\end{array}\end{array}$$

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  Division and Multiplication of Exponents
  ::指数的分化和乘数

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  When dividing powers with the same base, subtract exponents.
  ::当以相同基数进行分权时,应减去引言人。

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  When multiplying powers with the same base, add exponents.
  ::当以相同基数乘以功率时,加上引号。

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  Interactive
  ::交互式互动

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  Use the interactive to explore the rule for dividing powers with the same base.
  ::利用互动方式探讨以同一基础划分权力的规则。

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  Example 1-2
  ::例1-2

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  Nora performs another experiment with a fish whose mass starts at 1 kilogram. First, she feeds this fish 3 blue pills, then 5 red pills. What's the new mass of the fish?
  ::诺拉对一公斤质量开始的鱼进行另一次实验。 首先,她喂鱼吃三片蓝药丸,然后吃五片红药丸。鱼的新质量是多少?

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  Solution:
  ::解决方案 :

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  $$\begin{array}{r}\begin{array}{ll}\text{Expression}& \text{Explanation}\\ 1\cdot \frac{2^3}{2^5}& \text{Exponential form.}\\ 1\cdot 2^{(3-5)}& \text{Rule of exponents.}\\ 1\cdot 2^{-2}& \text{Simplifying.}\end{array}\end{array}$$

  
  ::Explaination12325 Explication form.1§2(3-5) 引言规则12-2简化。

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  That's an interesting result! What does  $\begin{array}{r}{2}^{-2}\end{array}$  mean? The fish doubled 3 times, then halved 5 times:
  ::这结果很有趣!2 -2是什么意思?鱼翻了3倍,然后又翻了5倍:

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  $$\begin{array}{r}\begin{array}{ll}\text{Expression}& \text{Explanation}\\ 1\cdot 2\cdot 2\cdot 2\cdot \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}& \text{Factors shown.}\\ 1\cdot \frac{1}{2}\cdot \frac{1}{2}& \text{Canceled 2's and halves.}\\ 1\cdot \frac{1}{2^2}& \text{Multiplying fractions.}\end{array}\end{array}$$

  
  ::显示11212 12分的2分之一半的1122分。

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  $$\begin{array}{r}\begin{array}{ll}\text{Expression}& \text{Explanation}\\ 1\cdot 2\cdot 2\cdot 2\cdot \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}& \text{Factors shown.}\\ 1\cdot \frac{1}{2}\cdot \frac{1}{2}& \text{Canceled 2's and halves.}\\ 1\cdot \frac{1}{2^2}& \text{Multiplying fractions.}\end{array}\end{array}$$

  
  ::显示11212 12分的2分之一半的1122分。

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  Multiplying the fish's mass by  $\begin{array}{r}{2}^{-2}\end{array}$  is the same as dividing by  $\begin{array}{r}{2}^2\end{array}$ , which is four. The fish has a final mass of  a quarter kilogram. 
  
  ::将鱼质量乘以 2-2 等于 22 等于 22 等于 4 等于 22 等于 4。 鱼的最终质量为 四分之一 公斤。

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    Negative Exponents
  ::负指数

  A negative exponent in the denominator of a fraction is equal to a positive exponent in the numerator, and vice versa. As an expression , this looks like:  $\begin{array}{r}{b}^{-n}=\frac{1}{b^n}\end{array}$  

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  Interactive
  ::交互式互动

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  Activity 2: Exponents of Zero and One
  ::活动2:零和一的指数

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  Example 2-1
  ::例2-1

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  Nora uses another 1-kilogram fish for a third experiment. This time she promises herself that after the experiment, the fish's mass will be 2 kilograms. What are three examples of experiments that will return this result?
  ::诺拉在第三次实验中使用了另一只1公斤的鱼。这次她保证实验结束后,鱼的重量将为2公斤。三个实验中有哪些可以返回这一结果的例子?

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  Solution:   Nora has many options, but the simplest ones involve feeding the fish one more blue pill than red pills:
  ::解决:诺拉有很多选择, 但最简单的选择是喂鱼吃一种比红药丸更多的蓝药丸:

  $$\begin{array}{r}\begin{array}{r}1\cdot 2^3\cdot 2^{-2}=1\cdot 2^1\\ 1\cdot 2^5\cdot 2^{-4}=1\cdot 2^1\\ 1\cdot 2^{-7}\cdot 2^8=1\cdot 2^1\\ 1\cdot 2^{10}\cdot 2^{-11}\cdot 2^2=1\cdot 2^1\end{array}\end{array}$$

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  From the rules of exponents, the fish ends up with a mass of   $\begin{array}{r}1\cdot 2^1\end{array}$ . The exponent of 1 means 1 doubling. So the mass of the fish is 2 kilograms.
  ::根据鱼群规则,鱼群最后质量为1 21。1的指数意味着1倍。所以鱼群质量为2公斤。

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  Example 2-2
  ::例2-2

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  Nora thinks that if she feeds a 1-kilogram fish 4 blue pills and 4 red pills, the resulting mass will be 1 kilogram. Is she right?
  ::诺拉认为如果她喂食一公斤的鱼 4个蓝药丸和4个红药丸 其质量将是1公斤

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  Solution: 
  ::解决方案 :

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  What does  $\begin{array}{r}1\cdot 2^0\end{array}$  mean? The exponent of 0 means 0 doublings. So the fish has a mass of 1 kilogram.
  ::120是什么意思?0的指数等于0倍。所以鱼质量为1公斤。

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    Exponents of 1 and 0
  ::1和0的指数值

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  Any number to the power of one is itself:  $\begin{array}{r}{b}^1=b\end{array}$
  ::一个功率的任意数字是本身: b1=b

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  Any number (except zero) to the power of zero is 1:  $\begin{array}{r}{b}^0=1\end{array}$
  ::零功率为 1 : b0=1 的任何数字(零除外)为 1 : b0=1

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  Work it Out
  ::工作出来

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  Simplify each of the following. Leave your answer in the form of a base to a power.
  $\begin{array}{r}\begin{array}{ll}\text{a.}& 2^3\cdot 2^4\\ \text{b.}& 3^6÷3^7\\ \text{c.}& 4^{-5}\cdot 4^6÷4^2\\ \text{d.}& 1^7\\ \text{e.}& {500}^0\\ \text{f.}& \frac{8^3}{8^5}\\ \text{g.}& \frac{8^5}{8^3}\\ \text{h.}& \frac{8^3}{8^{-5}}\\ \text{i.}& \frac{8^{-3}}{2^{-5}}\\ \text{j.}& \frac{x^3{x}^4}{x^5}\\ \text{k.}& \frac{x^3{2}^{-4}}{x^{-7}}\\ \text{l.}& \frac{x^3{x}^{-6}}{x^{-7}}\\ \text{m.}& \frac{1}{x^{-6}}\\ \text{n.}& \frac{1}{x^{-n}}\\ \text{o.}& \frac{x^m}{x^n}\\ \text{p.}& \frac{x^{-m}}{x^{-n}}\end{array}\end{array}$
  ::将以下各样的简化。 请将您以基数形式回答的回答留到一个电源上。 a.2324b.3637c.4-54642d.17e.5000f.8385g.8583h.838-5i.8-32-5j.x3x4x5k.x32-4x-7l.x3x-6x-6x-7m1x-6m6n.1x-no.xmxnp.x-mx-n

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  Tips for Simplifying
  ::用于简化的提示

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  Given a fraction of powers with negative exponents, you have the option to move powers with the same base across the division line to ease your simplification. For example:  $\begin{array}{r}\frac{x^{-3}}{x^{-4}}=\frac{x^4}{x^3}\end{array}$ .
  ::如果有部分权力带有负引力,您可以选择将相同基数的权力移动到分线另一侧,以方便您的简化。例如: x-3x-4=x4x3。

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  In general, solutions should be written with positive exponents instead of negative exponents. For example:  $\begin{array}{r}\frac{x^5}{x^7}=\frac{1}{x^2}.\end{array}$ It is not incorrect to write  $\begin{array}{r}{x}^{-2}\end{array}$ , but final simplifications should usually feature only positive exponents.
  ::一般来说,解决方案应该用正面的提示而不是负的提示来书写。 例如: x5x7=1x2. 写 x-2 并不错误, 但最后的简化通常只以正面提示为主 。

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  Activity 3: Rational Bases and Rational Exponents
  ::活动3:有理基础和有理指数

  

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  Example 3-1
  ::例3-1

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  Praveen is a scientist studying amoebas. Some amoebas are so big that they are visible to the naked eye! Praveen has amoebas of different sizes, and he defines their lengths in terms of the average amoeba. The average amoeba has a length of 1 amoeba unit (au). Praveen has discovered a special growth chemical that increases the length of an amoeba by raising it to the power of 4! Complete the table to determine the lengths of the amoebas after applying the chemical. What do  $\begin{array}{r}x\end{array}$   and  $\begin{array}{r}y\end{array}$   represent? Write the equation expressing length after the chemical application as a function of original length. For which value(s) does the growth chemical have no impact? For which value(s) does the growth chemical actually decrease the length of the amoeba? Why?
  

  $$\begin{array}{r}\begin{array}{cc}x& y\\ \frac{1}{10}& \\ \frac{1}{5}& \\ \frac{1}{2}& \\ \frac{3}{4}& \\ 1& \\ 2& \\ 3& \\ 4& \\ 5& \end{array}\end{array}$$

  
  ::Praveen是研究阿莫埃巴的科学家。 有些 Amoebas 是如此巨大, 以至于肉眼可以看到它们! Praveen 有不同大小的阿莫埃巴 。 他用平均阿莫埃巴来定义其长度 。 平均阿莫埃巴 长度为 1 amieba 单位 (au) 。 Praveen 发现了一种特殊的生长化学物质, 该化学物质通过将它提升到 4 的功率而增加其长度 。 完成表格以确定应用该化学之后该阿莫埃巴的长度 。 什么是 x 和 y 代表 ? 写在化学应用之后表示长度的方程式是原始长度的函数 。 生长化学的值对哪些值没有影响? 该化学的生长值是否真的降低了阿莫埃巴的长度? 为什么? xy110151234 123

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  Solution:
  ::解决方案 :

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  $$\begin{array}{r}\begin{array}{cc}x& y\\ \frac{1}{10}& \frac{1}{10,000}\\ \frac{1}{5}& \frac{1}{625}\\ \frac{1}{2}& \frac{1}{16}\\ \frac{3}{4}& \frac{81}{256}\\ 1& 1\\ 2& 16\\ 3& 81\\ 4& 256\\ 5& 625\end{array}\end{array}$$

  
  ::xy110110,00015162512163481256112163814455625

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  The variable  $\begin{array}{r}x\end{array}$  represents the original length, and  $\begin{array}{r}y\end{array}$  represents the length after chemical application. The equation is  $\begin{array}{r}y=x^4\end{array}$ . For $\begin{array}{r}x\end{array}$ -values less than 1 in the table, the function returns  even smaller  $\begin{array}{r}y\end{array}$ -values. Squaring a fraction means you are taking a fraction of a fraction, for example:
  ::变量 x 代表原始长度, Y 代表化学应用后的长度。 方程式是 y=x4。 表格中的福克斯值小于 1, 函数返回的值甚至小于 Y 值。 缩放一个分数表示您正在采取分数的一小部分, 例如 :

   

  $$\begin{array}{r}{\left(\frac{1}{2}\right)}^4=\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{16}\end{array}$$

   

  $$\begin{array}{r}{\left(\frac{1}{2}\right)}^4=\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{16}\end{array}$$

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  Half of a half of a half of a half is pretty small!
  ::半个半半个半是相当小的!

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  Note that the numerator does not change, since 1 to the fourth power is still 1. 
  
  ::请注意,分子不会改变,因为1是第4位数还是1。

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  Example 3-2
  ::例3-2

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  Some of Praveen's amoebae are too large. Praveen wants to reduce them back to their original size. What can Praveen do to reverse the mathematical operation of "raising to the fourth power?"
  ::Praveen想把它们缩回原来的体积。 Praveen能做些什么来扭转“向第四强加注”的数学操作呢?

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  Solution:  To understand his options, it is helpful to simplify the following expression:
  ::解决办法:为了理解他的选择,简化以下表述方式是有益的:

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  $$\begin{array}{r}\begin{array}{ll}\text{Expression}& \text{Explanation}\\ {\left(7^5\right)}^3& \text{Given expression.}\\ 7^5\cdot 7^5\cdot 7^5& \text{Raising to the 3rd power is the same as multiplying by itself 3 times.}\\ 7^{15}& \text{For powers (of same base) multiplied, add exponents.}\end{array}\end{array}$$

  
  ::表达式Explantation( 753) 给定表达式 757575 升至第3次功率 等于自动乘以3乘以715 乘以( 同一基数) 功率乘以( 相同基数) 乘以( 相同基数) , 加加引号 。

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  $$\begin{array}{r}\begin{array}{ll}\text{Expression}& \text{Explanation}\\ {\left(7^5\right)}^3& \text{Given expression.}\\ 7^5\cdot 7^5\cdot 7^5& \text{Raising to the 3rd power is the same as multiplying by itself 3 times.}\\ 7^{15}& \text{For powers (of same base) multiplied, add exponents.}\end{array}\end{array}$$

  
  ::表达式Explantation( 753) 给定表达式 757575 升至第3次功率 等于自动乘以3乘以715 乘以( 同一基数) 功率乘以( 相同基数) 乘以( 相同基数) , 加加引号 。

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  Instead of writing out  $\begin{array}{r}{7}^5\end{array}$  three times, you can use a shortcut:
  ::您可以使用捷径, 而不是写出75次, 而不是写出75次 :

  $$\begin{array}{r}\begin{array}{c}{\left(7^5\right)}^3\\ 7^{(5\cdot 3)}\\ 7^{15}\end{array}\end{array}$$

  $$\begin{array}{r}\begin{array}{c}{\left(7^5\right)}^3\\ 7^{(5\cdot 3)}\\ 7^{15}\end{array}\end{array}$$

  
  

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  Praveen's last amoeba has a length of 625 au's, or  $\begin{array}{r}{5}^4\end{array}$ . To what power can he raise  $\begin{array}{r}{5}^4\end{array}$  in order to get 5?
  ::Praveen最后的阿莫埃巴的长度是625 au's,或者54。 为了5个,他能提高54分的多少权力?

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  PLIX Interactive
  ::PLIX 交互式互动

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  Power to a Power
  ::权力对一权力国的权力

  To raise a power to a power, multiply the exponents:  $\begin{array}{r}{\left(b^m\right)}^n=b^{mn}\end{array}$

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  Interactive
  ::交互式互动

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  Example 3-3
  ::例3-3

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  Given the expression  $\begin{array}{r}{5}^2\end{array}$ , state two different operations that can be performed to return a value of 5.
  ::鉴于第52段的表达式,请说明两种不同的操作,可以返回值为5的数值。

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  Solution:  The option we've been using so far is to raise  $\begin{array}{r}{5}^2\end{array}$  to a power:
  ::解决方案:我们目前采用的选项是将52升至权力:

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  $$\begin{array}{r}\begin{array}{ll}\text{Expression}& \text{Explanation}\\ 5^2& \text{Given expression.}\\ {\left(5^2\right)}^{\frac{1}{2}}& \text{Raising to the power of one-half to cancel with the 2.}\\ 5^1& \text{Simplifying.}\\ 5& \text{Simplifying.}\end{array}\end{array}$$

   
  ::表达式Explanation52Felenten 表达式。 (52) 12 使用 2. 51 简化5 简化后, 将自动取消 。

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   A second option is below:
  ::第二个备选方案如下:

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  $$\begin{array}{r}\begin{array}{ll}\text{Expression}& \text{Explanation}\\ 5^2& \text{Given expression.}\\ 25& \text{Squaring 5 gives 25.}\\ \sqrt{25}& \text{Square rooting.}\\ 5& \text{The square root of 25 is 5.}\end{array}\end{array}$$

  
  ::25 Squaring 5 提供了25.25Square根根。 25的平方根是5。

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  $$\begin{array}{r}\begin{array}{ll}\text{Expression}& \text{Explanation}\\ 5^2& \text{Given expression.}\\ 25& \text{Squaring 5 gives 25.}\\ \sqrt{25}& \text{Square rooting.}\\ 5& \text{The square root of 25 is 5.}\end{array}\end{array}$$

  
  ::25 Squaring 5 提供了25.25Square根根。 25的平方根是5。

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  This shows that square rooting is equivalent to raising to the one-half power. In order to return Praveen's amoeba to its original size.
  ::这表明平方根相当于将半功率提升到半功率。 为了让Praveen的阿莫埃巴恢复到原来的大小 。

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  $$\begin{array}{r}\begin{array}{cc}\text{Using Rational Exponents}& \text{Using Rooting}\\ 625& 625\\ 5^4& 5^4\\ {\left(5^4\right)}^{\frac{1}{4}}& \sqrt[4]{5^4}\\ 5& 5\end{array}\end{array}$$

  
  ::使用有理指数来根植6256255454(54)1454455

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    Rational Exponents
  ::理性指数

  If the numerator of a rational exponent is thought of as the power, then the denominator is the root   

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  Work it Out
  ::工作出来

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  1. State the power to which each of the following expression must be raised in order to return the base:
     ::说明为了返回基地,必须使下列每一种表达方式得到何种权力:

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  $\begin{array}{r}\begin{array}{ll}\text{a.}& 2^3\\ \text{b.}& 3^5\\ \text{c.}& x^{-4}\\ \text{d.}& b^{\frac{1}{3}}\\ \text{e.}& {17}^{16}\\ \text{f.}& 5^{\frac{2}{3}}\\ \text{g.}& x^{0.2}\end{array}\end{array}$
  ::a.23b.35c.x-4d.b.13e.1716f.523g.x0.2

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  2. Simplify each of the following expressions
     ::简化下列表达式中的每个表达式

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  $\begin{array}{r}\begin{array}{ll}\text{a.}& {\left(6^7\right)}^{\frac{1}{7}}\\ \text{b.}& {\left(2^4\right)}^{\frac{3}{4}}\\ \text{c.}& {\left(3^{\frac{2}{3}}\right)}^3\\ \text{d.}& {\left(2^4{2}^3\right)}^{\frac{1}{7}}\\ \text{e.}& {\left(9^2{9}^3\right)}^{\frac{1}{10}}\\ \text{f.}& {\left({25}^{-2}{25}^6\right)}^{\frac{1}{8}}\end{array}\end{array}$
  :6717b.(24)34c.(323)3d.(2423)17e.(9293)110f.(25-2256)18

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    Summary
  ::摘要

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  • To multiply the same base to different powers, add the powers: $\begin{array}{r}{b}^m{b}^n=b^{(m+n)}\end{array}$  
    ::要将同一基数乘以不同的功率,加入功率: bmbn=b(m+n)
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  • To divide the same base with different powers, subtract the powers: $\begin{array}{r}\frac{b^m}{b^n}=b^{(m-n)}\end{array}$  
    
    ::要用不同的权力划分同一基数,请减去以下权力:bmbn=b(m-n)
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  • Any number to the power of 1 is itself: $\begin{array}{r}{b}^1=b\end{array}$  
    
    ::1 功率的任意数字本身为: b1=b
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  • Any number (except zero) to the power of zero is one: $\begin{array}{r}{b}^0=1\end{array}$  
    
    ::零功率为 1 的数值(零除外)为 1: b0=1
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  • A negative power in the numerator equals a positive power in the denominator: $\begin{array}{r}{b}^{-n}=\frac{1}{b^n}\end{array}$  
    
    ::分子中的负能量等于分母中的正能量: b-n=1bn
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  • When you raise a power to a power, multiply the exponents: $\begin{array}{r}{\left(b^m\right)}^n=b^{mn}\end{array}$  
    
    ::当你们将一个权柄升到一个权柄的时候,要乘以引力bm)n=bmn