课程： 4.6 溶解对数等

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解析对数等量
Introduction
::导言

The doubling time for an investment that has earnings compounded annually is $\begin{aligned} t = \frac{\ln 2}{\ln (1 + r)}, \end{aligned}$ where $\begin{aligned} r \end{aligned}$ is the interest rate in decimal form. For interest rates close to 8%, or 0.08, this formula can be accurately approximated by the formula $\begin{aligned} t = \frac{72}{100 r} \end{aligned}$ . In other words, the doubling time is about the quotient obtained by dividing 72 by the annual interest rate, as a percent. Since it is fairly easy to divide numbers into 72, this rough rule of thumb shows up in business discussions as “the rule of 72.” For example, when does a 4% interest rate double? Since 72/4 = 18, a 4% rate should double in about 18 years. The question to be answered in this section is that, assuming an investment must double in 2 years, what would the rate of return need to be? Rather than using the approximation, the rule of 72, find the interest rate exactly. The equation to solve is $\begin{aligned} 2 = \frac{\ln 2}{\ln (1 + r)} \end{aligned}$ . Solving this problem will require us to use new techniques.
::具有年收入的双倍投资时间是 t=ln2ln( 1+r) , 它使小数点的利率翻一番。对于接近8%或0.08的利率, 这个公式可以精确地被公式 t=72100r 所接近。换句话说, 翻一番的时间是关于以年利率将72除以72乘以72而获得的商数。由于将数字分成72相当容易, 这种粗略的拇指规则在商业讨论中显示为“ 72 规则 ” 。例如, 4%利率是 72/4 = 18 , 4% 利率大约18 年应该翻一番。本节要回答的问题是, 假设投资必须在两年内翻一番, 回报率需要什么? 而不是使用近似72 规则, 找到利率的准确性。要解决的方程式是 2= ln2n2 ( 1+r) 。解决这个问题需要我们使用新技术。

Method to Solve Logarithmic Equations
::解析对数等量的方法

When solving simple exponential equations , we have used the definition $\begin{aligned} \log_{b} x = a \leftrightarrow b^{a} = x \end{aligned}$ up to this point to translate the equation into an exponential one, which is most intuitive and often easier to solve. At times, more algebraic simplification is needed, and usually this involves the properties of logs as well as the one-to-one properties:
::当解析简单的指数方程式时,我们使用了直到此点的对数logbx=aba=x的定义,将方程式转换成指数方程式,该方程式最直观,而且往往更容易解答。有时需要更多代数简化,这通常涉及日志的属性以及一对一的属性:

One-to-One Properties
::一对一属性

For $\begin{aligned} b > 0 \end{aligned}$ ,
::b>0,

Exponential functions: $\begin{aligned} A = B \end{aligned}$ if and only if $\begin{aligned} b^{A} = b^{B} \end{aligned}$
::指数函数: A=B 只有当 bA=B 时才为A=B

Logarithm functions: $\begin{aligned} A = B \end{aligned}$ if and only if $\begin{aligned} \log_{b} A = \log_{b} B \end{aligned}$
::对数函数 : A=B 仅当对数bA=logbB

Learn, Play, and Explore with Logarithmic Equations:
::用对数等量学习、玩耍和探索:

Examples
::实例

Example 1
::例1

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

Solution:

::解决方案 :

There are two different ways to solve this equation.
::解决这个方程式有两种不同的方法。

Method 1: The 1st is to use the definition of a logarithm.
::方法1:第一是使用对数定义。

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

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

Method 2: A pply the 1st one-to-one property.
::方法2:适用第一对一属性。

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

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

Example 2
::例2

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

Solution:
::解决方案 :

Step 1: A dd 5 to both sides.
::第1步:双方增加5个。

$\begin{aligned} 3 \ln (- x) - 5 & = 10 \\ 3 \ln (- x) & = 15 \end{aligned}$
Step 2: Divide by 3 to isolate the natural log.
::3ln(-x)-5=103ln(-x)=15 步骤2:除以3以分离自然日志。

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

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

Step 3: Apply the cancellation property.
::第3步:适用注销财产。

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

::eln(-x) =e5-x=e5x=e5x* e55 148.41

Example 3
::例3

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

Solution:
::解决方案 :

Step 1: Condense the lefthand side using the product property.
::第1步:利用产品特性,使左手方保持冷静。

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

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

Step 2: Apply the cancellation property.
::第2步:应用注销属性。

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

::10log( 5x2 - 5x) =1025x2 - 5x=1005x2 - 5x - 100=0x2 - x-20=0( x5) (x+4) =0x=5, - 4

Step 3: C heck to see that both answers are in the domain of the equation.
::第3步:检查是否这两个答案都属于等式领域。

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

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

In the step $\begin{aligned} \log (- 20) + \log (- 5) = 2 \end{aligned}$ , the logarithm arguments are negative. However, the domain of the log function does not include negatives. Thus, -4 is an extraneous solution, and the solution is $\begin{aligned} x = 5 \end{aligned}$ .
::在阶梯 log(- 20)+log(- 5) = 2 中,对数参数为负。但是,对数函数的域不包括负数。因此, - 4 是一个不相干的解决办法,而答案是 x= 5 。

Example 4
::例4

Return to the problem from the introduction, t he doubling time for an investment which has earnings compounded annually is $\begin{aligned} t = \frac{\ln 2}{\ln (1 + r)} \end{aligned}$ .

$\begin{aligned} 2 & = \frac{\ln 2}{\ln (1 + r)} \\ \ln (1 + r) & = \frac{\ln (2)}{2} \\ e^{\ln (1 + r)} & = e^{\frac{\ln (2)}{2}} = {(e^{\ln (2)})}^{\frac{1}{2}} \\ (1 + r) & = 2^{\frac{1}{2}} = \sqrt{2} \\ r & = \sqrt{2} - 1 \approx .414 = 41.4 % \end{aligned}$

::回到引言中的问题,每年增加收入的投资翻一番的时间是 t= ln 2ln (1+1r) 2= ln 2ln (1+r) (1+r) 1+r) = ln(2)2eln (1+r) = eln2= (eln2) 12(1+r) = 212=2-2r=1414=41.4%

Thus, the interest rate required to double the investment in 2 years is about 41.4%.
::因此,两年内投资翻一番所需的利率约为41.4%。

Example 5
::例5

Solve the following logarithmic equation:

$\begin{aligned} 9 + 2 \log_{3} x = 23. \end{aligned}$

::解析以下对数方程式: 9+2log3x=23。

Solution:
::解决方案 :

Rewrite the logarithm and apply the 1st one-to-one property.

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

::重写对数并应用第一对一属性。 9+2log3x=232log3x=14log3x=14log3x=73log3x=37x=2,187

Example 6
::例6

Solve the following logarithmic equation:
::解析以下对数方程式 :

$\begin{aligned} \ln (x - 1) - \ln (x + 1) = 8. \end{aligned}$

:x- 1) - ln (x+1)= 8 。

Solution:
::解决方案 :

Rewrite the lefthand side as a single logarithm, and then apply the definition of a logarithmic function.
::将左手侧重写为单对数,然后应用对数函数的定义。

$\begin{aligned} \ln (x - 1) - \ln (x + 1) & = 8 \\ \ln (\frac{x - 1}{x + 1}) & = 8 \\ e^{\ln (\frac{x - 1}{x + 1})} & = e^{8} \\ \frac{x - 1}{x + 1} & = e^{8} \\ x - 1 & = (x + 1) e^{8} \\ x - 1 & = x e^{8} + e^{8} \\ x - x e^{8} & = 1 + e^{8} \\ x (1 - e^{8}) & = 1 + e^{8} \\ x & = \frac{1 + e^{8}}{1 - e^{8}} \approx - 1.0007 \end{aligned}$

::In(x-1)- ln(x+1)=8ln(x-1x+1)=8eln(x-1x+1)=8eln(x-1x+1)=e8x-1=e8x-1=e8x-1=x8-e8x-xe8=1+e8x-xe8=1+e8x(1-e8)=1+e8x=1+e8x=1+e8x=1+e81-e8=_1.0007

Since a logarithm argument must be greater than 0, $\begin{aligned} \frac{x - 1}{x + 1} > 0 \end{aligned}$ and $\begin{aligned} x > 1 \end{aligned}$ . Thus, -1.0007 is not in the domain of this equation, so there is no solution.
::由于对数参数必须大于 0, x-1x+1>0 和 x>1. 因此, -1.0007 不属于此方程的范围,所以没有解决办法。

Example 7
::例7

Solve the following logarithmic equation:
::解析以下对数方程式 :

$\begin{aligned} \frac{1}{2} \log_{5} (2 x + 5) = 2. \end{aligned}$

::12log5(2x+5)=2。

Solution:

::解决方案 :

Rewrite the logarithm and apply the 1st one-to-one property.
::重写对数并应用第一对一属性。

$\begin{aligned} \frac{1}{2} \log_{5} (2 x + 5) & = 2 \\ \log_{5} (2 x + 5) & = 4 \\ 5^{\log_{5} (2 x + 5)} & = 5^{4} \\ 2 x + 5 & = 625 \\ 2 x & = 620 \\ x & = 310 \end{aligned}$

::12log5(2x+5)=2log5(2x+5)=45log5(2x+5)=542x+5=6252x=620x=310

Summary
::摘要

To solve logarithmic equations:

Use one-to-one properties:

$\begin{aligned} A = B \end{aligned}$ if and only if $\begin{aligned} b^{A} = b^{B} \end{aligned}$
::A=B 如果且仅在 bA=bB 的情况下 A=B

$\begin{aligned} A = B, \end{aligned}$ if and only if $\begin{aligned} \log_{b} A = \log_{b} B . \end{aligned}$
::A=B,如果只有对数a=logb*A=logb*B,才为A=B。

::使用一对一的属性 : A= B 仅当 bA= B A = B 仅当 bA= B 仅当logb = A = logb = B 仅当且仅在logb= A = logb = B 时才使用 A = B 。

Rewrite the equation as an exponential equation using $\begin{aligned} \log_{b} x = y ⟷ b^{y} = x . \end{aligned}$
::使用 logbx=yby=x 将方程式重写为指数方程式。

::要解析对数方程式 : 使用一对一的属性 : 如果并且只有在 bA = B A = B 时才使用 A= B, 如果并且只有在 logb\ = A = logb\ B 时才使用 A = B 。使用 logb\ x = y by=x 将方程式重写为指数方程式。

Review
::回顾

Use properties of logarithms and a calculator to solve the following equations for $\begin{aligned} x \end{aligned}$ . Round answers to three decimal places and check for extraneous solutions.
::使用对数属性和计算器解析 x. 的下列方程式。对小数点后三个位数的圆形答案, 并检查外部解决方案。

$\begin{aligned} \log_{2} x = 15 \end{aligned}$
::对数 2x=15

$\begin{aligned} \log_{12} x = 2.5 \end{aligned}$
::对数 12x=2.5

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

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

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

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

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

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

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

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

$\begin{aligned} 5 \log \frac{x}{2} - 3 \log \frac{1}{x} = \log 8 \end{aligned}$
::5log_x2 - 3log_ 1x=log_ 8

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

$\begin{aligned} 2 \log_{6} x + 1 = \log_{6} (5 x + 4) \end{aligned}$
::2log6_x+1=log6(5x+4)

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

$\begin{aligned} 3 \log_{\frac{2}{3}} x - \log_{\frac{2}{3}} 27 = \log_{\frac{2}{3}} 8 \end{aligned}$
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Review ( Answers)
::回顾(答复)

Please see the Appendix.
::请参看附录。

章节大纲

常规

解析对数等量

Introduction ::导言

Method to Solve Logarithmic Equations ::解析对数等量的方法

One-to-One Properties ::一对一属性

Examples ::实例

Summary ::摘要

Review ::回顾

Review ( Answers) ::回顾(答复)

Introduction
::导言

Method to Solve Logarithmic Equations
::解析对数等量的方法

One-to-One Properties
::一对一属性

Examples
::实例

Summary
::摘要

Review
::回顾

Review ( Answers)
::回顾(答复)