11.9 简化和扩大对数

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  简化和扩展对数

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  You go to a concert and want to know how loud it is in decibels. The decibel level of a sound is found by 1st assigning an intensity,  $\begin{array}{r}{I}_0,\end{array}$  to a very soft sound, or the threshold. The decibel level can then be measured with the formula $\begin{array}{r}L=10\log (\frac{I}{I_0}),\end{array}$  where $\begin{array}{r}I\end{array}$  is the intensity of the sound, and $\begin{array}{r}L\end{array}$  is the level in decibels. Since $\begin{array}{r}{I}_0\end{array}$  is often a fixed quantity, it may make our calculation easier to expand the logarithm ¹ .
  ::您去听音乐会, 并想知道音频在分贝中的音量有多响。 音频的分贝级别通过第1次指定一个强度, I0 或非常软的音频, 或阈值来找到。 然后可以使用公式 L=10log( II0) 来测量分贝级别, 我就是音频的强度, L 是分贝中的音频。 由于I0 通常是一个固定数量, 它可以让我们更容易计算扩大对数 1 。

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  We discuss how to expand and simplify logarithms using the logarithmic properties in this section. 
  ::我们讨论如何利用本节中的对数属性扩大和简化对数。

  

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  Expanding Logarithms
  ::正在展开对数

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  It is sometimes helpful to  expand logarithms—that is, write them as a sum or difference of logarithms with  the power rule applied. This can make some calculations easier. 
  ::有时扩大对数是有用的,也就是说,把它们写成对数的总和或对数与所应用的对数规则的差数。 这样可以使一些计算容易一些。

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  While this is not always the case, if you try to apply the rules in the order quotient rule of logarithms, product rule of logarithms, and power rule of logarithms, you will be able to expand many logarithms. Let's consider some examples.
  ::虽然情况并非总是这样,但如果尝试应用对数的顺序商数规则、对数的产物规则以及对数的权力规则,那么可以扩展许多对数。让我们举几个例子。

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  Example 1
  ::例1

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  Expand $\begin{array}{r}{\log }_{6}17x^5.\end{array}$
  ::展开对数 6\\\\\\17x5 。

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  Solution: To expand this log, we need to use the product rule  and the power rule since there is no quotient here.
  ::解决方案:要扩展此日志, 我们需要使用产品规则和权力规则, 因为这里没有商数 。

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  $$\begin{array}{rlrl}{\log }_{6}17x^5& ={\log }_{6}17+{\log }_6{x}^5& & \text{product rule}\\ & ={\log }_{6}17+5{\log }_{6}x& & \text{power rule}\end{array}$$

  
  ::对数 6\\\\ 17x5=log6\\\ 17+log6\\\ x5 product rolog6\\\\ 17+5log6\ x power rule

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  We have written this logarithm as a sum with the power rule applied where possible.
  ::我们把这个对数写成一个总和 尽可能应用的权力规则

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  Example 2
  ::例2

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  Expand $\begin{array}{r}\ln {\left(\frac{2x}{y^3}\right)}^4.\end{array}$
  ::展开( 2xy3) 4 。

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  Solution: We will need to use all three properties to expand this example. Because the expression within the natural log is in parentheses, start with moving the $\begin{array}{r}{4}^{th}\end{array}$ power to the front of the log. Then we can proceed by applying the rules in the order quotient, product, and then power. 
  ::解决方案 : 我们需要使用所有三个属性来扩展此示例。 因为自然日志中的表达方式在括号中, 开始将第4个功率移到日志前端 。 然后我们可以按照顺序、 产品、 和电源来应用规则 。

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  $$\begin{array}{rlrl}\ln {\left(\frac{2x}{y^3}\right)}^4& =4\ln \frac{2x}{y^3}& & \text{power rule}\\ & =4(\ln 2x-\ln y^3)& & \text{quotient rule}\\ & =4(\ln 2+\ln x-3\ln y)& & \text{product rule and power rule}\\ & =4\ln 2+4\ln x-12\ln y& & \text{distributive property}\end{array}$$

  
  ::\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\分配属性属性属性属性属性属性属性属性。

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  Without a calculator, you can evaluate $\begin{array}{r}4\ln 2\approx 2.77,\end{array}$  making the final answer $\begin{array}{r}2.77+4\ln x-12\ln y.\end{array}$
  ::没有计算器, 您可以对 4ln222. 77 作出最后答案 2. 77+4lnx- 12lny 。

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  by CK-12 demonstrates how to expand a logarithm. 
  ::CK-12 演示如何扩展对数 。

   

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  Example 3
  ::例3

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  Expand   $\begin{array}{r}{\log }_8\frac{16x}{y^2}.\end{array}$
  ::展开对数816xy2。

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  Solution:  Here, we can proceed with the rules in the order quotient, product, and then power.
  ::解决方案:在这里,我们可以按照规则,按商序、产品和权力顺序进行。

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  $$\begin{array}{rlrl}{\log }_8\frac{16x}{y^2}& ={\log }_{8}16x-{\log }_8{y}^2& & \text{quotient rule}\\ & ={\log }_{8}16+{\log }_{8}x-{\log }_8{y}^2& & \text{product rule}\\ & =\frac{4}{3}+{\log }_{8}x-2{\log }_{8}y& & \text{power rule}\end{array}$$

  
  ::log8\\\ 16xy2 = log8\\ 16x- log8\\\ y2 引用规则= log8\ 16+log8\ x- log8\\\ y2 产品规则= 43+log8\ x-2log8\ ypower 规则

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  To determine $\begin{array}{r}{\log }_{8}16\end{array}$ , use the definition and the fact that 8 and 16 are both powers of 2: $\begin{array}{r}{8}^n=16\to 2^{3n}=2^4\to 3n=4\to n=\frac{4}{3}\text{so}{\log }_{8}16=\frac{4}{3}.\end{array}$
  ::为确定log816, 使用该定义, 8和16是2: 8n=1623n=243n=443, 所以log816=43的两种权力。

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  Example 4
  ::例4

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  Expand   $\begin{array}{r}{\log }_{16}(\frac{x^{2}y}{32z^5}).\end{array}$
  
  ::展开对数 16 (x2y32z5) 。

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  Solution:   Let's start with using the quotient rule .
  ::解决方案:让我们从使用商数规则开始。

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  $$\begin{array}{r}{\log }_{16}(\frac{x^{2}y}{32z^5})={\log }_{16}{x}^{2}y-{\log }_{16}32z^5\end{array}$$

  
  ::log16(x2y32z5) =log16x2y-log1632z5

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  Now apply the product rule, followed by the power rule .
  ::现在应用产品规则,然后是权力规则。

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  $$\begin{array}{rl}& ={\log }_{16}{x}^2+{\log }_{16}y-\left({\log }_{16}32+{\log }_{16}{z}^5\right)\\ & =2{\log }_{16}x+{\log }_{16}y-\frac{5}{4}-5{\log }_{16}z\end{array}$$

  
  ::=log16_x2+log16_y_(log16_32+log16_z5)=2log16_x+log16_y_54_5log16_z

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  $\begin{array}{r}{\log }_{16}32\to {16}^n=32\to 2^{4n}=2^5\to n=\frac{5}{4}\text{so}{\log }_{16}32=\frac{5}{4}.\end{array}$  Also, notice that we put parentheses around the 2nd log once it was expanded, to ensure that the log with  $\begin{array}{r}{z}^5\end{array}$ would also be subtracted (because it was in the denominator of the original expression).
  ::log163216n=3224n=25n=54 so log1632=54。此外,请注意,一旦扩展了第二个日志,我们将在它周围加括号,以确保也减去Z5的日志(因为它是原表达式的分母)。

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  Example 5
  ::例5

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  You go a concert and want to know how loud it is in decibels. The decibel level of a sound is found by 1st assigning an intensity $\begin{array}{r}{I}_0\end{array}$  to a very soft sound, or the threshold, usually $\begin{array}{r}{10}^{\text{-}12}\end{array}$  watts per square meter. The decibel level can then be measured with the formula $\begin{array}{r}L=10\log (\frac{I}{I_0}),\end{array}$ where $\begin{array}{r}I\end{array}$  is the intensity of the sound, and $\begin{array}{r}L\end{array}$  is the level in decibels. Expand the logarithm and find the decibel level of a rock concert with an intensity of 1 watt per square meter, and normal conversation with an intensity of $\begin{array}{r}{10}^{\text{-}5}\end{array}$  watts per square meter ¹ . 
  ::您去听音乐会, 并想知道音频在分贝中的响度。 音频的分贝级别通过将强度 I0 指定为非常软的音频或阈值( 通常是每平方米10-12瓦) 来发现。 然后, 分贝级别可以用公式 L= 10log( II0) 来测量, 其中我就是音频的强度, L 是分贝中的强度。 扩展对数, 并找到每平方米1瓦的岩石音乐的分贝级别, 正常对话的强度为每平方米10-5瓦1 。

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  Solution:  First we expand the logarithm. 
  ::解决方案:首先,我们扩大对数。

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  $$\begin{array}{rl}L& =10\log (\frac{I}{I_0})\\ & =10(\log I-\log I_0)\\ & =10\log I-10\log I_0\end{array}$$

  
  ::L=10log(II0)=10(logI-logI0)=10logI-10logI0

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  Now we can set up the formula for multiple calculations by inputting $\begin{array}{r}{I}_0\end{array}$ .
  ::现在我们可以通过输入 I0 来设置多重计算公式。

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  $$\begin{array}{r}L=10\log I-10\log I_0=10\log I-10\log ({10}^{\text{-}12})=10\log I+120\end{array}$$

  
  ::L=10log_I-10log_I0=10log_I-10log_(10-12)=10log_I+120

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  We can use this to find the decibel levels.
  ::我们可以用这个来找到分贝水平。

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  $$\begin{array}{rl}& \text{Rock Concert}L=10\log 1+120=10(0)+120=120\\ \\ & \text{Normal Conversation}L=10\log ({10}^{\text{-}5})+120=\text{-}50+120=70\end{array}$$

  
  ::摇滚音乐乐L=10log=10log=1+120=10(0)+120=120 热电解L=10log=10(10-5)+120=-50+120=70

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  The difference between these is quite significant, 50 decibels.   
  ::两者的差别相当大,50分贝。

   

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  Simplifying Logarithms
  ::简化对数

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  As we will see in a following section, it is helpful to be able to simplify logarithms—that is, combine a sum or difference of logarithms into a single logarithm. This will help us solve logarithmic equations. 
  ::如下文一节所示,简化对数是有益的,即将对数的总和或差数合并成单一对数。这将有助于我们解析对数方程。

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  While it is not always the case, applying the rules in the order power rule, product rule, and then quotient rule will be helpful in many cases. 
  ::虽然情况并非总是如此,但在许多情况下,适用规则的顺序权力规则、产品规则以及随后的商数规则将会有所帮助。

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  Example 6  
  ::例6

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  Simplify  $\begin{array}{r}{\log }_{7}8+{\log }_7{x}^2+{\log }_{7}3y\end{array}$ .
  ::简化对数 7\ _8+log7\\\\\\\\\\\\\\\\\\\\2+log7\\\\\\\\\\\\\\\3y 。

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  Solution:   Since there are no factors in front of the logarithms, we use the product rule next. There are no logs that are subtracted, so there is no need to use the quotient rule.
  ::解决方案 : 由于对数前没有因素, 我们接下来使用产品规则 。 没有日志可以减去, 因此不需要使用商数规则 。

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  $$\begin{array}{rlrl}{\log }_{7}8+{\log }_7{x}^2+{\log }_{7}3y& ={\log }_{7}8x^{2}3y& & \text{product rule}\\ & ={\log }_{7}24x^{2}y\end{array}$$

  
  ::对数 7\\\ 8+log7\\ x2+log7\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  Example 7
  ::例7

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  Simplify  $\begin{array}{r}\log y-\log 20+\log 8x\end{array}$ .
  ::简化对数“y-log”20+log8x。

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  Solution:   There are no factors in front of the logarithms, so we use the product rule and the quotient rule. 
  ::解决方案:在对数面前没有因素, 所以我们使用产品规则和商数规则。

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  $$\begin{array}{rl}\log y-\log 20+\log 8x& =\log (\frac{y}{20}\cdot 8x)\\ & =\log \frac{2xy}{5}\end{array}$$

  
  ::log- log_ log_ 20+log_ 8x=log_ (y20_ 8x) =log_ 2xy5

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  by CK-12 demonstrates how to simplify a logarithm.
  ::CK-12 显示如何简化对数 。

   

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  Example 8
  ::例8

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  Simplify  $\begin{array}{r}\log 9-4\log 5-4\log x+2\log 7+2\log y.\end{array}$
  ::简化对数9 - 4log5 - 4logx+2log7+2logy。

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  Solution:   Now, there are factors in front of the logarithms . We use the power rule 1st. 
  ::解答: 现在, 在对数前面有一些因素。 我们使用权力规则 1 。

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  $$\begin{array}{rl}& \log 9-4\log 5-4\log x+2\log 7+2\log y\\ & =\log 9-\log 5^4-\log x^4+\log 7^2+\log y^2\end{array}$$

  
  ::log*9 - 4log*5 - 4log*x+2log*7+2log=log*9 - log*9*54 - log*4+log* @72+log*y2

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  Next, start changing things to division and multiplication within one log.
  ::下一步,开始把事情改变成一个日志内部的分割和乘法。

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  $$\begin{array}{r}\log (\frac{9\cdot 7^2{y}^2}{5^4{x}^4})\end{array}$$

  
  ::log ( 972y254x4)

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  Lastly, multiply the terms in the fraction .
  ::最后,在分数中乘以术语。

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  $$\begin{array}{r}\log (\frac{441y^2}{625x^4})\end{array}$$

  
  :441y2625x4)

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  Example 9
  ::例9

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  Condense into one log: $\begin{array}{r}\ln 5-7\ln x^4+2\ln y.\end{array}$
  ::整合成一个日志 : 5 - 7ln x4+2ln y 。

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  Solution:   To condense this expression into one log, you will need to use all three rules .
  ::解答: 要将这个表达式压缩成一个日志, 您需要使用所有三个规则 。

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  $$\begin{array}{rl}\ln 5-7\ln x^4+2\ln y& =\ln 5-\ln x^{28}+\ln y^2\\ & =\ln (\frac{5y^2}{x^{28}})\end{array}$$

  
  ::5 - 7ln_x4+2ln_ y=ln_ 5 - ln_ x28+ln_ y2=ln_ (5y2x28)

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     WARNING
  ::警告

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  • If the problem above was $\begin{array}{r}\ln 5-(7\ln x^4+2\ln y),\end{array}$  then the answer would have been $\begin{array}{r}\ln (\frac{5}{x^{28}{y}^2}).\end{array}$  But, because there are no parentheses, the $\begin{array}{r}{y}^2\end{array}$ is in the numerator.
    ::如果上面的问题是 In5-( 7lnx4+2lny), 那么答案应该是in( 5x28y2), 但是, 因为没有括号, y2 在数字中 。
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  • You cannot cancel expressions that you are taking the log of and expressions outside of the logarithm. The following is incorrect:
    ::您无法取消在对数之外使用的表达式日志和表达式。 以下不正确 :

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  $$\begin{array}{r}\frac{\log \cancel{x}}{\cancel{x}}=\log .\end{array}$$

  
  ::log_xx=log.

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  Summary
  ::摘要

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  • To expand logarithms, write them as a sum or difference of logarithms where the power rule is applied if necessary. Often, using the rules in the order quotient rule, product rule, and then power rule will be helpful.
    ::要扩展对数, 请在需要应用权力规则的地方将其写成对数的总和或对数的差数。 通常, 使用顺序商数规则的规则、 产品规则以及权力规则将会有所帮助 。
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  • To simplify logarithms, write them as a single logarithm. Often, using the rules in the order power rule, product rule, and quotient rule will be helpful.  
    ::为了简化对数,请把它们写成单对数。 通常,使用顺序权力规则的规则、产品规则和商数规则会有所帮助。

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  Review
  ::回顾

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  Expand the following logarithmic expressions:
  ::展开以下对数表达式:

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  1.  $\begin{array}{r}\log {\left(\frac{3x}{y}\right)}^2\end{array}$
  ::1. 对数(3xy)2

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  2.  $\begin{array}{r}{\log }_8\frac{x^3{y}^2}{z^4}\end{array}$
  ::2. 对数8x3y2z4

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  3.  $\begin{array}{r}{\log }_5{\left(\frac{25x^4}{y}\right)}^2\end{array}$
  ::3. log5(25x4y)2

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  4.  $\begin{array}{r}\ln {\left(\frac{6x}{y^3}\right)}^{\text{-}2}\end{array}$
  ::4. *(6xy3)-2

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  5.  $\begin{array}{r}\ln {\left(\frac{e^5{x}^{\text{-}2}}{y^3}\right)}^6\end{array}$
  ::5. *(e5x-2y3)6

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  Simplify the following logarithmic expressions.
  ::简化以下对数表达式。

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  6.  $\begin{array}{r}2{\log }_{6}x+5{\log }_{6}y\end{array}$
  ::6. 2log6x+5log6y

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  7.  $\begin{array}{r}{\log }_{3}6+{\log }_{3}y-{\log }_{3}4\end{array}$
  ::7. 对数3=6+log3=y-log3=4

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  8.  $\begin{array}{r}\log 12-\log x+\log y^2\end{array}$
  ::8. 对数12 - logx+logy2

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  9.  $\begin{array}{r}{\log }_6{x}^2-{\log }_{6}x-{\log }_{6}y\end{array}$
  ::9. log6x2 - log6_x- log6_ y

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  10.  $\begin{array}{r}3(\log x-\log y)\end{array}$
  ::10. (3(logx-logy))

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  11.   $\begin{array}{r}\frac{1}{2}\log (x+1)-3\log y\end{array}$
  ::11. 12log(x+1)-3logy

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  12.  $\begin{array}{r}4{\log }_{2}y+\frac{1}{3}{\log }_2{x}^3\end{array}$
  ::12. 4log2y+13log2x3

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  13.  $\begin{array}{r}\frac{1}{5}\left[10{\log }_2(x-3)+{\log }_{2}32-{\log }_{2}y\right]\end{array}$
  ::13. 15[10log2(x-3)+log232-log2

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  14.  $\begin{array}{r}4\left[\frac{1}{2}{\log }_{3}y-\frac{1}{3}{\log }_{3}x-{\log }_{3}z\right]\end{array}$
  ::14. 4[12log3]-13log3]-x-log3z]

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  Answers for Review Problems
  ::回顾问题的答复

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  Please see the Appendix.
  ::请参看附录。

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  References
  ::参考参考资料

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  1. "Decibel," last edited May 16, 2017,
  ::1. 2017年5月16日编辑的《Decibel》