Course: 3.10 对数模型 | The Way To Learn

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对数模型

In prior lessons, you used an exponential model to predict the population of a town based on a constant growth rate such as 6% per year.
::在以往的课程中,你使用指数模型,根据每年6%的持续增长率预测一个城镇的人口。

In the real world however, populations often do not just grow continuously and without limit. A town originally founded near a convenient water source may grow very quickly at first, but the expansion will slow dramatically as houses and businesses run out of room near the water source, and need to begin transporting water further and further away.
::然而,在现实世界中,人口往往不仅仅持续和无限制地增长。最初在方便水源附近建立的城镇一开始可能增长很快,但随着房屋和企业在水源附近耗尽房房房,扩张将大大放缓,需要开始越来越远的运水。

How can a situation like this be modeled with an equation?
::怎样才能用方程式来模拟这种情形呢?

Logarithmic Models
::对数模型

In a prior lesson, we considered the solutions of simple log equations. Now we return to that topic and explore some more complex examples. Solving more complicated log equations can be less difficult than you might think, by using our knowledge of log properties.
::在先前的教训中,我们考虑了简单的日志方程式的解决方案。现在我们回到这个话题, 探索一些更复杂的例子。通过使用我们对日志属性的知识, 解决更复杂的日志方程式可能比您想象的要困难得多。

For example, consider the equation log ₂ ( x ) + log ₂ ( x - 2) = 3. We can solve this equation using a log property.
::例如,考虑方程式log2(x)+log2(x-2)=3。我们可以使用日志属性解析此方程式。

log ₂ ( x ) + log ₂ ( x - 2) = 3
log ₂ ( x ( x - 2)) = 3	log _b x + log _b y = log _b ( xy )
log ₂ ( x ² - 2 x ) = 3 $\begin{aligned} \Rightarrow \end{aligned}$	write the equation in exponential form.
2 ³ = x ² - 2 x
x ² - 2 x - 8 = 0	Solve the resulting quadratic
( x - 4) ( x + 2) = 0
x = -2, 4

The resulting quadratic has two solutions. However, only x = 4 is a solution to our original equation, as log ₂ (-2) is undefined. We refer to x = -2 as an extraneous solution.
::所产生的二次方程式有两个解决方案。然而, 只有 x = 4 是我们原始方程式的解决方案, 因为log2( 2-2) 尚未定义。我们称 x = - 2 是一个外部解决方案。

Examples
::实例

Example 1
::例1

Earlier, you were asked how a situation could be modeled with an equation.
::早些时候,有人问你,如何用方程式模拟局势。

A population that increases continuously at a constant rate may be modeled with an exponential function.
::以恒定速率持续增长的人口可模拟成指数函数。

A population that increases rapidly and then levels off may be modeled with a logarithmic function.
::人口迅速增长,然后达到水平,可以模拟对数功能。

Example 2
::例2

Solve each equation.
::解决每个方程式

log ( x + 2) + log 3 = 2
::日对数 (x+2) + 日对数 3 = 2

log (3( x + 2)) = 2	log _b x + log _b y = log _b ( xy )
log (3 x + 6) = 2	Simplify the expression 3( x +2)
10 ² = 3 x + 6	Write the log expression in exponential form
100 = 3 x + 6
3 x = 94	Solve the linear equation
x = 94/3

ln ( x + 2) - ln ( x ) = 1
::内(x+2) - 内(x)=1

$\begin{aligned} l n (\frac{x + 2}{x}) = 1 \end{aligned}$	$\begin{aligned} l o g_{b} x - l o g_{b} y = l o g_{b} (\frac{x}{y}) \end{aligned}$
$\begin{aligned} e^{1} = \frac{x + 2}{x} \end{aligned}$	Write the log expression in exponential form.
$\begin{aligned} e x = x + 2 \end{aligned}$	Multiply both sides by $\begin{aligned} x \end{aligned}$ .
$\begin{aligned} e x - x = 2 \end{aligned}$	Factor out $\begin{aligned} x \end{aligned}$ .
$\begin{aligned} x (e - 1) = 2 \end{aligned}$	Isolate $\begin{aligned} x \end{aligned}$ .
$\begin{aligned} x = \frac{2}{e - 1} \end{aligned}$

The solution above is an exact solution. If we want a decimal approximation, we can use a calculator to find that x ≈ 1.16. We can also use a graphing calculator to find an approximate solution. Consider again the equation ln ( x + 2) - ln ( x ) = 1. We can solve this equation by solving a system:
::上面的解决方案是一个精确的解决方案。如果我们想要一个小数近似值, 我们可以使用计算器来找到 x = = 1. 16。我们还可以使用图形计算器来找到一个大致的解决方案。再考虑一下 in (x+ 2) - in (x) = 1 的方程。我们可以通过解决一个系统来解析这个方程 :

$\begin{aligned} {\begin{cases} y = l n (x + 2) - l n (x) \\ y = 1 \end{cases} \end{aligned}$
::{y=ln(x+2)-ln(x)y=1

If you graph the system on your graphing calculator, you should see that the curve and the horizontal line intersection at one point. Using the INTERSECT function on the CALC menu (press <2nd>[CALC]), you should find that the x coordinate of the intersection point is approximately 1.16. This method will allow you to find approximate solutions for more complicated log equations.
::如果您在图形化计算器上绘制系统图, 您应该看到曲线和水平线交叉点在一个点上。使用 CALC 菜单上的 InterSECT 函数 (press < 2nd> [CALC]) , 您应该发现交叉点的 x 坐标约为 1. 16 。这个方法可以让您找到比较复杂的日志方程式的大致解决方案。

Example 3
::例3

Use a graphing calculator to solve each equation:
::使用图形计算计算器解析每个方程式 :

log(5 - x ) + 1 = log x
::log( 5 - x) + 1 = log x

The graphs of y = log (5 - x ) + 1 and y = log x intersect at x ≈ 4.5454545.
::y 的图形 = 日志 (5 - x) + 1 和 y = 日志 x 在x = = 4.5454545 时交叉的对数 x 。

Therefore the solution of the equation is x ≈ 4.54.
::因此,方程式的解决方案是 x 4.54。

log ₂ (3 x + 8) + 1 = log ₃ (10 - x )
::log2 (3x+8) + 1 = log3 (10 - x)

First, in order to graph the equations, you must rewrite them in terms of a common log or a natural log. The resulting equations are: $\begin{aligned} y = \frac{l o g (3 x + 8)}{l o g 2} + 1 \end{aligned}$ and $\begin{aligned} y = \frac{l o g (10 - x)}{l o g 3} \end{aligned}$ .
::首先,要绘制方程图, 您必须用普通日志或自然日志重写它们。由此产生的方程是 y=log( 3x+8) log2+1 和 y=log( 10- x) log3 。

The graphs of these functions intersect at x ≈ -1.87. This value is the approximate solution to the equation.
::这些函数的图形在 x \\ \ - 1. 87 时交叉。这个值是方程的大致解决方案。

Example 4
::例4

Consider population growth:
::考虑人口增长:

Year	Population
1	2000
5	4200
10	6500
20	8800
30	10500
40	12500

If we plot this data, we see that the growth is not quite linear, and it is not exponential either.
::如果我们绘制这些数据,我们就会发现增长不是线性增长,也不是指数性增长。

We can find a logarithmic function to model this data. First enter the data in the table in L1 and L2. Then press STAT to get to the CALC menu. This time choose option 9. You should get the function y = 930.4954615 + 2780.218173 ln x. If you view the graph and the data points together, as described in the Technology Note above, you will see that the graph of the function does not touch the data points, but models the general trend of the data.
::我们可以找到一个对数函数来模拟此数据。首先在 L1 和 L2 的表格中输入数据, 然后按 STAT 键以获取 CALC 菜单。这次选择选项 9 。您应该获得 y = 930.4954615 + 2780.211873 in x 。如果您同时查看上述技术注释中描述的图形和数据点, 您可以看到该函数的图形不会触动数据点, 而是模拟数据的一般趋势。

Note about technology: you can also do this using an Excel spreadsheet. Enter the data in a worksheet, and create a scatterplot by inserting a chart. After you create the chart, from the chart menu, choose “add trendline.” You will then be able to choose the type of function. Note that if you want to use a logarithmic function, the domain of your data set must be positive numbers. The chart menu will actually not allow you to choose a logarithmic trendline if your data include zero or negative x values. See below:
::有关技术的注释 : 您也可以使用 Excel 电子表格这样做。在工作表中输入数据, 并通过插入图表创建散射图。在您创建图表后, 从图表菜单中选择“ 添加趋势线 ” 。然后您可以选择函数类型。请注意, 如果您想要使用对数函数, 您数据集的域必须是正数。如果您的数据包含 0 或负 x 值, 图表菜单实际上不允许您选择对数趋势线。见 :

Example 5
::例5

Solve for $\begin{aligned} x \end{aligned}$ : $\begin{aligned} l o g_{2} x - l o g_{2} (x - 4) = 12 \end{aligned}$ .
::x: log2x- log2(x- 4)=12 的溶解。

To solve $\begin{aligned} l o g_{2} x - l o g_{2} (x - 4) = 12 \end{aligned}$ :
::要解析对数2x-log2(x-4)=12:

$\begin{aligned} l o g_{2} \frac{x}{x - 4} = 12 \end{aligned}$ : Using $\begin{aligned} l o g_{x} y - l o g_{x} z = l o g_{x} \frac{y}{z} \end{aligned}$
::log2xx- 4= 12 : 使用 logxy- logxz = logxyz

$\begin{aligned} 2^{12} = \frac{x}{x - 4} \end{aligned}$ : Write in exponential form
::212=xx-4:以指数形式写字

$\begin{aligned} 4, 096 = \frac{x}{x - 4} \end{aligned}$ : With a calculator
::4,096=xx-4:使用计算器

$\begin{aligned} 4, 096 x - 16384 = x \end{aligned}$ : Multiply both sides by $\begin{aligned} x - 4 \end{aligned}$
::4 096x- 16384=x: 将两边乘以 x-4

$\begin{aligned} 4, 095 x = 16, 384 \end{aligned}$ : Simplify
::4 095x=16 384:简化

$\begin{aligned} x = 4 \end{aligned}$ : Divide
::x=4 : 除法

Example 6
::例6

Biologists use the formula $\begin{aligned} n = k \cdot l o g A \end{aligned}$ to estimate the number of species $\begin{aligned} n \end{aligned}$ that live in a given area $\begin{aligned} A \end{aligned}$ by multiplying by a constant $\begin{aligned} k \end{aligned}$ which changes by location. If a particular rain forest has a constant $\begin{aligned} k \end{aligned}$ of $\begin{aligned} 943 \end{aligned}$ how many species would be estimated to live in an area of $\begin{aligned} 950 k m^{2} \end{aligned}$ ?
::生物学家使用计算公式 n=klogA 来估计生活在给定区域A 中的 n 种物种数量, 计算方法是乘以一个常数 k , 按位置变化。如果某个雨林的常数 k 为 943 , 估计有多少种物种生活在 950 km2 区域 ?

To find the number of species in an area of 950km ² :
::在950公里区域寻找物种数量:

$\begin{aligned} n = 943 \cdot l o g 950 \end{aligned}$ : Substitute the given $\begin{aligned} k \end{aligned}$ and A values
::n=943 log950: 替代给定 k 和 A 值

$\begin{aligned} n = 943 \cdot 2.977 \end{aligned}$ : With a calculator
::n=9432.977:使用计算器

$\begin{aligned} n = 2, 807 \end{aligned}$
::n=2,807

Therefore 2,807 species would likely live in the area.
::因此,2 807种物种很可能生活在这一地区。

Review
::回顾

Express 1-7 in exponential form:
::指数式1-7快递:

$\begin{aligned} l o g_{12} \frac{1}{1728} = - 3 \end{aligned}$
::log 12117283

$\begin{aligned} l o g_{216} 6 = \frac{1}{3} \end{aligned}$
::log2166=13

$\begin{aligned} l o g_{\frac{1}{3}} \frac{1}{9} = 3 \end{aligned}$
::日志1319=3

$\begin{aligned} l o g_{\frac{1}{4}} \frac{1}{16} = 2 \end{aligned}$
::log14116=2

$\begin{aligned} l o g_{5} 125 = 3 \end{aligned}$
::log5125=3 log5125=3

$\begin{aligned} l o g_{15} 225 = 2 \end{aligned}$
::15225=2 log 15225=2

$\begin{aligned} l o g_{25} 5 = \frac{1}{2} \end{aligned}$
::log255=12

For questions 8-13, solve for $\begin{aligned} x \end{aligned}$ .
::问题8-13, 解答 x。

$\begin{aligned} l o g_{x} 64 = 2 \end{aligned}$
::logx64=2

$\begin{aligned} l o g_{3} 6561 = x \end{aligned}$
::log36561=x

$\begin{aligned} l o g_{5} x = 4 \end{aligned}$
::log5x=4

$\begin{aligned} l o g_{x} 27 = 3 \end{aligned}$
::logx27=3

$\begin{aligned} l o g_{2} x = 6 \end{aligned}$
::log2x=6 log2x=6

$\begin{aligned} l o g_{4} 64 = x \end{aligned}$
::log464=x

For questions 14-19, solve for the variable.
::问题14-19,解决变量。

$\begin{aligned} 4 l o g (\frac{x}{5}) + l o g (\frac{625}{4}) = 2 l o g x \end{aligned}$
::4log( x5) +log( 6254) = 2logx

$\begin{aligned} l o g_{5} x + \frac{l o g_{5} 125}{l o g_{5} x} = \frac{7}{2} \end{aligned}$
::log5x+log5125log5x=72

$\begin{aligned} l o g p = \frac{2 - l o g p}{l o g p} \end{aligned}$
::logp=2 - loglogp

$\begin{aligned} 2 l o g x - 2 l o g (x + 1) = 0 \end{aligned}$
::2logx-2log( x+1)=0

$\begin{aligned} l o g (25 - z^{3}) - 3 l o g (4 - z) = 0 \end{aligned}$
::log( 25 - z3) - 3log( 4 - z) =0

$\begin{aligned} \frac{l o g (35 - y^{3})}{l o g (5 - y)} = 3 \end{aligned}$
::log( 35 - Y3) log( 5 - Y) = 3

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。

Section outline

General

对数模型

Logarithmic Models ::对数模型

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Example 6 ::例6

Review ::回顾

Review (Answers) ::回顾(答复)