Course: 11.11 溶解对数等数

Section outline

Select section General

Collapse Expand
General

Collapse all Expand all
- Select activity 新闻通告
  
  新闻通告 Forum
Select section 解析对数等量

Collapse Expand
解析对数等量
If you number the keys on a piano from 1 to 88 left to right, you can relate t he key number to the frequency or pitch of the key with the logarithmic function $\begin{aligned} n = 12 \log_{2} (\frac{f}{440}) + 49, \end{aligned}$ where $\begin{aligned} n \end{aligned}$ is the key number, and $\begin{aligned} f \end{aligned}$ is the frequency in Hertz. A middle C note is produced by the 40th key ¹ . What frequency is a middle C?
::如果您将钢琴上的键数从1到 88 到右, 您可以用对数函数 n=12log2( f440)+49, 键数为 n, 键数为 Hertz , 频率为 Hertz 。第40 个键生成了一个中间的 C 音符。 1 中间的 C 频率是多少?

To find out, we need to solve a logarithmic equation. In this section, we learn the techniques necessary to do that.
::要找出答案,我们需要解决对数方程。在本节中,我们学习了这样做的必要技巧。

Solving Logarithmic Equations
::解析对数等量

A logarithmic equation has the variable within the log arithm. We can solve these equations by converting to exponential form.
::对数方程式在对数内有变量。我们可以转换成指数形式来解析这些方程式。

How To Solve Logarithmic Equations
::如何解析对数等

1. Isolate the term with the logarithm. If there is more than one term with a logarithm, use logarithmic properties to create a single logarithm.
::1. 用对数分隔术语。如果对数有一个以上的术语,则使用对数属性来创建单一的对数。

EXCEPTION: If $\begin{aligned} \log_{b} u = \log_{b} v \end{aligned}$ , then $\begin{aligned} u = v \end{aligned}$ .
::例外:如果logbu=logbv,那么u=v。

2. Convert to exponential form. Solve the resulting equation.
::2. 转换成指数形式,解决所产生的方程式。

3. Check for extraneous solutions.
::3. 寻找不相干的解决办法。

Let's consider the exception 1st.
::让我们考虑一下例外1号

Example 1
::例1

Solve $\begin{aligned} \log_{4} (2 x - 3) = \log_{4} 8 x \end{aligned}$ .
::解析对数4( 2x- 3) = log48x 。

Solution: If the logarithms have the same base as they do here, then the arguments must be equal.
::解答: 如果对数与这里的对数有相同的基数, 那么参数必须是相等的。

$\begin{aligned} \log_{4} (2 x - 3) & = \log_{4} 8 x \\ 2 x - 3 & = 8 x \\ - 3 & = 6 x \\ - \frac{1}{2} & = x \end{aligned}$

::对数 4( 2x- 3) = log48x2x-3=8x-3=6x-12=x

Next, we need to check if our solution is extraneous—that is, solutions that our algebraic steps created, but do not actually make both sides of the equation equal. In this case, we particularly need to look out for situations where we try to take the log of a negative number or 0.
::接下来,我们需要检查我们的解决方案是否不相干 — — 也就是说,我们代数步骤创造的解决方案,但实际上并不使等式的两面都相等。在这种情况下,我们尤其需要关注我们试图将负数或负数记录到零的情况。

$\begin{aligned} \log_{4} (2 (- \frac{1}{2}) + 3) & = \log_{4} 8 (- \frac{1}{2}) \\ \log_{4} (- 2) & = \log_{4} - 4 \end{aligned}$

::log4( 2( -12)+3) = log48( -12) log4( 2-2) = log4-4

We cannot calculate the logarithm on the right side of the equation, so the solution is extraneous. There is no solution to this equation.
::我们无法计算方程右侧的对数, 因此解答是外在的。此方程没有解答。

by CK-12 shows an example of the exception.
::CK-12 显示例外示例。

Example 2
::例2

Solve $\begin{aligned} \log_{2} (x + 5) = 9 \end{aligned}$ .
::解析对数2(x+5)=9。

Solution: Since the term with the logarithm is already isolated, we can convert to exponential form. Here, 2 is the base, 9 is the exponent, and $\begin{aligned} x + 5 \end{aligned}$ is the result.
::解答: 由于对数术语已经孤立, 我们可以转换成指数形式。这里, 2 是基数, 9 是指数, 9 是结果 x+ 5 。

$\begin{aligned} \log_{2} (x + 5) & = 9 \\ 2^{9} & = x + 5 \\ 512 & = x + 5 \\ 507 & = x \end{aligned}$

::log2(x+5)=929=x+5512=x+5507=x

Checking our solution yields $\begin{aligned} \log_{2} (507 + 5) = 9 \to \log_{2} 512 = 9. \end{aligned}$
::正在检查我们的溶液产值 log2( 507+5) = 9log2512=9。

Note there is an alternate approach to solving logarithmic equations. The 2nd way to solve this equation is to express both sides as exponents with a base of 2, and then use the inverse property.
::注意有解决对数方程式的另一种方法。解决此方程式的第二种方法是将两边作为标语方块以 2 为基数表示, 然后使用反向属性。

$\begin{aligned} 2^{\log_{2} (x + 5)} & = 2^{9} \\ x + 5 & = 512 \\ x & = 507 \end{aligned}$

::2log2(x+5)=29x+5=512x=507

Example 3
::例3

Solve $\begin{aligned} 9 + 2 \log_{3} x = 23. \end{aligned}$
::9+2log3=x=23 解决9+2log3=x=23。

Solution: Isolate the log and then convert to exponential form.
::溶液: 分离日志, 然后转换成指数形式。

$\begin{aligned} 9 + 2 \log_{3} x & = 23 \\ 2 \log_{3} x & = 14 \\ \log_{3} x & = 7 \\ x & = 3^{7} = 2, 187 \end{aligned}$

::9+2log3x=232log3x=14log3x=7x=37=2,187

2,187 is positive, so when we check it we will have a valid solution.
::2,187是积极的,所以当我们检查它时,我们将有一个有效的解决办法。

Example 4
::例4

Solve $\begin{aligned} 3 \ln (- x) - 5 = 10 \end{aligned}$ .
::解决 3ln(-x)- 5=10 。

Solution: First, add 5 to both sides and then divide by 3 to isolate the natural log.
::解决办法:首先,在两边加5,然后除以3,分离自然原木。

$\begin{aligned} 3 \ln (- x) - 5 & = 10 \\ 3 \ln (- x) & = 15 \\ \ln (- x) & = 5 \end{aligned}$

::3ln(-x) -5=103ln(-x)=15ln(-x)=5

Convert to exponential form with $\begin{aligned} e \end{aligned}$ as the base.
::转换成以 e 为基数的指数形式。

$\begin{aligned} - x & = e^{5} \\ x & = - e^{5} \approx - 148.41 \end{aligned}$

::-x=e5x=-e5-148.41

Checking the solution, we have $\begin{aligned} 3 \ln (- (- e^{5})) - 5 = 10 \to 3 \ln e^{5} - 5 = 10 \to 3 \cdot 5 - 5 = 10 \end{aligned}$
::在检查解决方案时,我们有 3ln(-(-)-(-))-5=103ln5-5-5=1035-5=1035-5=10

WARNING
::警告

A negative solution or a solution of 0 is not automatically an extraneous solution. You need to check whether the argument of the logarithm is a negative number or 0.

Example 5
::例5

You can relate the key number to the frequency or pitch of the key with the logarithmic function $\begin{aligned} n = 12 \log_{2} (\frac{f}{440}) + 49, \end{aligned}$ where $\begin{aligned} n \end{aligned}$ is the key number and $\begin{aligned} f \end{aligned}$ is the frequency in Hertz. A middle C note is produced by the 40th key ¹ . What frequency is a middle C?
::您可以用对数函数 n=12log2(f440)+49 将键的频率或方位与键的频率或方位连接, n 是键码, f 是 Hertz 的频率。第40个键生成了一个中间的C注。 1 中间的C 频率是多少?

Solution: Substitute 40 for $\begin{aligned} n \end{aligned}$ and solve.
::溶液: n 和溶解替代 40 。

$\begin{aligned} 40 & = 12 \log_{2} (\frac{f}{440}) + 49 \\ - 9 & = 12 \log_{2} (\frac{f}{440}) \\ - \frac{3}{4} & = \log_{2} (\frac{f}{440}) \\ 2^{- \frac{3}{4}} & = \frac{f}{440} \\ 440 \cdot 2^{- \frac{3}{4}} & = f \\ 261.63 & = f \end{aligned}$

::40=12log2(f440)+49-9=12log2(f440)-34=log2(f440)2-34=f440440440}2-34=f261.63=f

The frequency of a middle C is about 262 Hertz.
::中间C的频率约为262赫兹。

by Mathispower4u shows several examples of solving logarithmic equations.
::Mathispower4u 表示解析对数方程式的几个例子。

Example 6
::例6

Solve $\begin{aligned} \log 5 x + \log (x - 1) = 2 \end{aligned}$ .
::解析对数5x+log(x-1)=2。

Solution: Simplify the left-hand side by using the product rule .
::解决方案:使用产品规则简化左手侧。

$\begin{aligned} \log 5 x + \log (x - 1) = 2 \\ \log [5 x (x - 1)] = 2 \\ \log (5 x^{2} - 5 x) = 2 \end{aligned}$

::对数5x+log(x-1)=2log[5x(x-1)]=2log(5x2-5x)=2

Now that there is a single log isolated on one side, we can convert to exponential form.
::现在单面有一个单日志, 我们可以转换成指数形。

$\begin{aligned} 5 x^{2} - 5 x & = 10^{2} \\ 5 x^{2} - 5 x & = 100 \\ x^{2} - x - 20 & = 0 \\ (x - 5) (x + 4) & = 0 \\ x & = 5, - 4 \end{aligned}$

::5x2-5x=1025x2-5x=100x2-x-20=0(x-5)(x+4)=0x=5、4

Now, check both solutions.
::现在,检查两种解决方案。

$\begin{aligned} \log 5 (5) + \log (5 - 1) & = 2 \log 5 (- 4) + \log ((- 4) - 1) = 2 \\ \log 25 + \log 4 & = 2 \log (- 20) + \log (- 5) = 2 \\ \log 100 & = 2 \end{aligned}$

::5(5)+log(5-1)=2log5(4)+log((4)-1)=2log25+log4=2log(20)+log(5)=2log100=2

-4 is an extraneous solution. In the step $\begin{aligned} \log (- 20) + \log (- 5) = 2 \end{aligned}$ , we cannot take the log of a negative number, therefore -4 is not a solution. 5 is the only solution.
::4 是一个不相干的解决办法。在步数 log( 20) +log( 5) = 2 中, 我们无法使用负数的日志, 因此 - 4 不是一个解决办法。 5 是唯一的解决办法。

Example 7
::例7

Solve $\begin{aligned} \ln (x - 1) - \ln (x + 1) = 8 \end{aligned}$
::解决内(x- 1)- ln(x+1)=8

Solution: Simplify the left-hand side using the quotient rule and then convert to exponential form.
::解决方案: 使用商数规则简化左手侧, 然后转换成指数形式。

$\begin{aligned} \ln (x - 1) - \ln (x + 1) & = 8 \\ \ln (\frac{x - 1}{x + 1}) & = 8 \\ \frac{x - 1}{x + 1} & = e^{8} \\ x - 1 & = (x + 1) (e^{8}) \\ x - 1 & = x e^{8} + e^{8} \\ x - x e^{8} & = 1 + e^{8} \\ x (1 - e^{8}) & = 1 + e^{8} \\ x & = \frac{1 + e^{8}}{1 - e^{8}} \approx - 1.00067 \end{aligned}$

::In(x- 1)- ln(x+1)=8ln(x-1x+1)=8x-1x+1=e8x-1=(x+1)(e8)x-1=x8+e8x-xe8=1+e8x(1-e8)=1+e8x=1+e8x=1+e8x=1+e8x=1+e81-e81-e8*_1=-1.0067

Checking our solution, we get $\begin{aligned} \ln (- 1.0067 - 1) - \ln (- 1.0067 + 1) = 8 \end{aligned}$ , which does not work because the logarithms are of negative numbers. Therefore, there is no solution for this equation.
::检查我们的解决方案后, 我们得到 In( -1.0067- 1)- ln( -1. 0067+1) =8 , 因为对数为负数, 无效。因此, 此方程式没有解决方案。

by CK-12 shows an example of using both the logarithmic properties and the exception to solve a logarithmic equation.
::使用 CK-12 来显示一个使用对数属性和例外来解析对数方程式的示例。

Feature: Shaky Ground
::地貌变暗

by Meredith Beaton
::梅雷迪思·贝顿(签名)

An earthquake is the result of two plates of Earth's crust suddenly slipping past each other as they travel along the molten lava ocean underneath our planet's crust. The epicenter of the earthquake is the location on Earth's surface above where the earthquake started. Not all earthquakes are big enough to be felt. In fact, thousands of earthquakes occur every year but are so small we just can't feel them!
::地震是地球地壳的两个板块突然滑过地球地壳的两块板块造成的。当它们沿着熔岩熔岩海洋在我们星球地壳下行走时,地震的震中点是地震起源地上方的地球表面。并非所有地震都足够大,可以感觉到。事实上,每年发生的地震数以万计,但规模如此之小,我们无法感觉到它们!

News You Can Use
::您可以使用新闻新闻

Often, we only feel earthquakes that have a magnitude of at least 3.5. But what is the magnitude of an earthquake, and what does it really tell us? Dr. Charles Richter was a geologist who studied earthquakes. He discovered that the seismic waves (shown on a seismograph) that radiated from the epicenter of the earthquake provide good estimates for the magnitude of an earthquake, or how big the earthquake is. The bigger the seismic waves, the bigger the earthquake.
::通常,我们只感受到至少3.5级的地震。但地震的规模是什么?它告诉我们什么?查尔斯·里希特博士是一位地质学家,研究地震。他发现地震震中辐射的地震波(在地震地震仪上出现)提供了地震规模或地震规模的准确估计。地震波越大,地震就越大。

Richter's formula for calculating the magnitude of an earthquake takes the logarithm of the height of the seismic wave and scales it according to the distance that the seismic wave was felt to the epicenter of the earthquake:
::Richter计算地震规模的公式是地震波高度的对数,并根据地震震中感受到地震波的距离大小计算:

$\begin{aligned} Magnitude = \log [A] [(m^{2})] + 3 \log (8 (Δ t)) - 2.92 \end{aligned}$

::磁度=log[A][(m2)+3log(8)()-2.92]

$\begin{aligned} A = \end{aligned}$ the height (or amplitude) of the seismic waves
::A=地震波的高度(或振幅)

$\begin{aligned} m = \end{aligned}$ the distance (in meters) from where the seismic waves are felt and the epicenter
::m= 地震波感知和震中距离(米)

$\begin{aligned} Δ t = \end{aligned}$ the amount of time between when the earthquake happened at the epicenter and when the seismic waves are felt
::* t=地震在震中发生和地震波感受到之间的时间

The Richter scale measures the magnitude of all earthquakes and ranks them from 0.1 to 9.9, where 9.9 are the worst earthquakes we have ever experienced. This calculation tells geologists and government agencies a lot about the impact of an earthquake and its expected destruction.
::里氏级测量所有地震的规模,并将其从0.1到9.9级,其中9.9级是有史以来最严重的地震。这一计算让地质学家和政府机构大量了解地震的影响及其预期的破坏。

For example, the devastating earthquake on April 25, 2015, in Nepal registered a magnitude of 7.8. An earthquake with a magnitude of 7.8 may not sound much bigger than an earthquake with a magnitude of about 4, just a few number lower. However, instead of being almost double in magnitude, this more powerful earthquake causes the Earth to shake about 6,310 times as much.
::例如,2015年4月25日尼泊尔的毁灭性地震规模为7.8级。规模为7.8级的地震听起来可能不会比规模为4级的地震大得多,只是少了几个数字。然而,这次更强大的地震不是几乎是规模的两倍,而是震动地球的6 310倍。

by National Geographic describes earthquakes and the Richter scale.
::国家地理表示地震和里氏级。

Summary
::摘要

To solve logarithmic equations, isolate the term with the logarithm and convert to exponential form.
::要解析对数方程式, 请用对数分离术语, 并转换成指数形式。

Make sure to check for extraneous solutions.
::确保检查不相干的解决办法。

Review
::回顾

S olve the following equations for $\begin{aligned} x \end{aligned}$ . Round answers to three decimal places and check for extraneous solutions.
::解析 x. 的以下方程式。小数点后三个位的圆形答案, 并检查不相干解决方案。

1. $\begin{aligned} \log_{2} x = 15 \end{aligned}$
::1. log2x=15

2. $\begin{aligned} \log_{12} x = 2.5 \end{aligned}$
::2. log12x=2.5

3. $\begin{aligned} \log_{9} (x - 5) = 2 \end{aligned}$
::3. log9(x-5)=2

4. $\begin{aligned} \log_{7} (2 x + 3) = 3 \end{aligned}$
::4. log7( 2x+3)=3

5. $\begin{aligned} 8 \ln (3 - x) = 5 \end{aligned}$
::5. 8ln(3-x)=5

6. $\begin{aligned} 4 \log_{3} 3 x - \log_{3} x = 5 \end{aligned}$
::6. 4log33x-log3x=5

7. $\begin{aligned} \log (x + 5) + \log x = \log 14 \end{aligned}$
::7.(x+5)+logx=log14

8. $\begin{aligned} 2 \ln x - \ln x = 0 \end{aligned}$
::8. 2lnxx-lnx=0

9. $\begin{aligned} 3 \log_{3} (x - 5) = 3 \end{aligned}$
::9. 3log3(x-5)=3

10. $\begin{aligned} \frac{2}{3} \log_{3} x = 2 \end{aligned}$
::10. 23log3x=2

11. $\begin{aligned} 5 \log \frac{x}{2} - 3 \log \frac{1}{x} = \log 8 \end{aligned}$
::11. 5log_x2 - 3log_1x=log_8

12. $\begin{aligned} 2 \ln x^{e + 2} - \ln x = 10 \end{aligned}$
::12. 2lnxxe+2-lnx=10

13. $\begin{aligned} 2 \log_{6} x + 1 = \log_{6} (5 x + 4) \end{aligned}$
::13. 2log6x+1=log6(5x+4)

14. $\begin{aligned} 2 \log_{\frac{1}{2}} x + 2 = \log_{\frac{1}{2}} (x + 10) \end{aligned}$
::14. 2log12x+2=log12(x+10)

15. $\begin{aligned} 3 \log_{\frac{2}{3}} x - \log_{\frac{2}{3}} 27 = \log_{\frac{2}{3}} 8 \end{aligned}$
::15. 3log23x-log2327=log238

Explore More
::探索更多

1. Soils with a pH above 5.5 are best for plants. If $\begin{aligned} pH = - \log H^{+}, \end{aligned}$ where $\begin{aligned} H^{+} \end{aligned}$ is the hydrogen ion concentration ² , what is the hydrogen ion concentration when the pH is 5.5?
::1.pH值高于5.5的土壤最适合植物。如果pH=-logH+,H+是氢离子浓度2,pH值为5.5时氢离子浓度是什么?

2. The Beer-Lambert Law relates the ability of a light to penetrate a material to the concentration of a material. A version of the formula is $\begin{aligned} A = \log (\frac{1}{T}), \end{aligned}$ where $\begin{aligned} T \end{aligned}$ is the transmittance, a measure of the light that transmitted through a material, and $\begin{aligned} A \end{aligned}$ is the absorbance, a measure of the light absorbed ³ . If the absorbance is 0.5, what is the transmittance?
::2. 《啤酒-伯特法》涉及光能穿透材料使其浓缩的能力,公式的版本是A=log(1T),即传输的信号,即通过材料传输的光的量度,A是吸收的量度,是吸收的光的量度。3如果吸收为0.5,什么是传输?

3. In astronomy, the limiting magnitude is a scale that measures the faintest objects that can be measured with a telescope. If you have a 10-inch telescope, the equation for the limiting magnitude is $\begin{aligned} M_{T} = M_{E} + 4 + 2.5 \log P, \end{aligned}$ where $\begin{aligned} M_{T} \end{aligned}$ is the limiting magnitude looking through the telescope, $\begin{aligned} M_{E} \end{aligned}$ is the limiting magnitude when looking at the night sky with the naked eye, and $\begin{aligned} P \end{aligned}$ is the power or magnification of the telescope. Find the magnification if the limiting magnitude through the telescope is 15, and the limiting magnitude with the naked eye is 5. In urban areas, the limiting magnitude with the naked eye is reduced due to light pollution ⁴ . Find the magnification when the limiting magnitude with the naked eye is 3.
::3. 在天文学中,极限星系是测量可以用望远镜测量的最微小天体的尺度,如果有10英寸望远镜,极限星系的方程式是MT=ME+4+2.5logP,其中MT是通过望远镜观察的极限星度,ME是用肉眼观察夜间天空的极限星度,Pis是望远镜的功率或放大度。如果通过望远镜观测的极限星系是15,而肉眼的极限星系是5,则在城市地区,由于光污染,肉眼的极限星系缩小了4。在以肉眼观察的极限星系是3时,找到放大度。

Answers for Review and Explore More Problems
::回顾和探讨更多问题的答复

Please see the Appendix.
::请参看附录。

References
::参考参考资料

1. "Piano Key Frequencies," last edited April 3, 2017,
::1. "钢琴键盘", 上次编辑于2017年4月3日, 2017年4月3日,

2. "Efficient Fertilizer Use Guide Soil pH | Mosaic Crop Nutrition,"
::2. "高效肥料使用指南土壤pH 蚊虫作物营养 " ,

3. "Beer-Lambert Law," last edited May 16, 2017,
::2017年5月16日,

4. "Limiting Magnitude," last edited May 16, 2017,
::4. 2017年5月16日上次编辑的"缩小磁度"

Section outline

General

解析对数等量

Solving Logarithmic Equations ::解析对数等量

How To Solve Logarithmic Equations ::如何解析对数等

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

WARNING ::警告

Example 5 ::例5

Example 6 ::例6

Example 7 ::例7

Feature: Shaky Ground ::地貌变暗

News You Can Use ::您可以使用新闻新闻

Summary ::摘要

Review ::回顾

Explore More ::探索更多

Answers for Review and Explore More Problems ::回顾和探讨更多问题的答复

References ::参考参考资料

Solving Logarithmic Equations
::解析对数等量

How To Solve Logarithmic Equations
::如何解析对数等

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

WARNING
::警告

Example 5
::例5

Example 6
::例6

Example 7
::例7

Feature: Shaky Ground
::地貌变暗

News You Can Use
::您可以使用新闻新闻

Summary
::摘要

Review
::回顾

Explore More
::探索更多

Answers for Review and Explore More Problems
::回顾和探讨更多问题的答复

References
::参考参考资料