节: 使用日志解决指数等同 | 4.5 使用日志解决指数等同

章节大纲

Introduction
::导言

A small town was established in 1950, and the population is given by
$\begin{aligned} P (t) = 2, 000 (1.05)^{t}, \end{aligned}$
where $\begin{aligned} t \end{aligned}$ is the number of years since 1950. The mayor would like to know when the population reached 20,000. In other words, a solution to the following equation is needed:
::1950年建立了一个小城镇,人口由P(t)=2,000(1.05)t给予,那里的人口是1950年以来的年数。市长想知道人口何时达到20,000。

$\begin{aligned} 20, 000 = 2, 000 (1.05)^{t} . \end{aligned}$
The difficulty is that the variable is in the exponent . We will explore algebraic techniques to handle that issue.
::20,000=2,000(1.05)t. 困难在于变量在指数中。我们将探索代数技术来解决这个问题。

Exponential Equation Solution Techniques
::指数赤道溶解技术

To solve an exponential equation:
::要解析指数方程式 :

Isolate the exponential part of the equation. If there are two exponential parts, then rewrite so there is a single exponent on each side of the equation.
::分离方程式的指数部分。如果有两个指数部分, 请重写, 这样方程式的每侧都有一个符号。

Take the logarithm of each side of the equation.
::以方程每侧的对数取出方程的对数。

Solve for the variable.
::解决变量。

Check your solution.
::检查你的解决方案。

A common technique for solving equations with variables in exponents is to take the log of both sides of the equation. The p roperties of logs can be used to simplify and solve the equation.
::用指数变量解析方程式的一个常见技术是采用方程式两侧的日志。日志的属性可用于简化和解析方程式。

Properties of Using Logarithms
::使用对数属性

$\begin{aligned} \log a^{x} & = x \cdot \log a \\ \ln e^{x} & = x \\ \log 10^{x} & = x \end{aligned}$

Examples
::实例

Example 1
::例1

Th e amount of time it will take to have $9,000 in a savings account, paying 6% annual compound interest, if $300 is deposited at the end of each year, satisfies the equation
::如果每年年底交存300美元,则储蓄账户需要9 000美元,支付6%的年度复利(如果每年年底交存300美元,则需要多少时间才能满足等式要求)

$\begin{aligned} 9, 000 = 300 \cdot \frac{(1.06)^{t} - 1}{0.06} . \end{aligned}$

::9,000=300(1.06)t-10.06。

This type of investment is called an annuity . Solve the preceding equation for $\begin{aligned} t \end{aligned}$ .
::这种类型的投资被称为年金。 t 解决前一个等式。

Solution:
::解决方案 :

$\begin{aligned} 30 & = \frac{(1.06)^{t} - 1}{0.06} \\ 1.8 & = (1.06)^{t} - 1 \\ 2.8 & = {1.06}^{t} \\ \ln 2.8 & = \ln ({1.06}^{t}) = t \cdot \ln (1.06) \\ t & = \frac{\ln 2.8}{\ln 1.06} \approx 17.67 y e a r s \end{aligned}$

::30=(1.06)t-1.006.8=(1.06t-12.8=1.06tln_2.8=ln(1.06t)=t ln(1.06t) (1.06t) =ln(2.06t) (2.8n) (1.06t) = (2.8n) (1.06) (1.06) (1.06) (17.67) 年

Example 2
::例2

Solve the following equation for $\begin{aligned} x \end{aligned}$ : $\begin{aligned} 16^{x} = 25 \end{aligned}$ .
::解析 x 的下列方程式: 16x=25 。

Solution:
::解决方案 :

Take the log of both sides. U se log properties and a calculator to approximate the solution:
::使用两边的日志。使用日志属性和计算器来接近解决方案 :

$\begin{aligned} 16^{x} & = 25 \\ \log 16^{x} & = \log 25 \\ x \log 16 & = \log 25 \\ x & = \frac{\log 25}{\log 16} \\ x & \approx 1.16 \end{aligned}$

::16x=25log_16x=log_25xlog_16=log_25x=log_25x=log_25g_16x_16x_1.16

Example 3
::例3

Solve the following equation for all possible values of $\begin{aligned} x \end{aligned}$ : $\begin{aligned} (\log_{2} x)^{2} - \log_{2} (x^{7}) = - 12 \end{aligned}$ .
::为 x 的所有可能值( log2x) 2- log2( x7) *12 ) 解决下列方程式。

Solution:
::解决方案 :

Step 1: Identify that t his is a quadratic log problem, because the logarithmic term is squared in the 1st term. Use a substitution to examine each layer of the problem.
::步骤 1: 确定这是一个二次对数问题, 因为对数术语在第一个术语中方形。使用替代来检查问题的每一层。

Step 2: Let $\begin{aligned} u = \log_{2} x \end{aligned}$ .
::步骤2:让u=log2x。

$\begin{aligned} (\log_{2} x)^{2} - 7 \log_{2} x + 12 & = 0 \\ u^{2} - 7 u + 12 & = 0 \\ (u - 3) (u - 4) & = 0 \\ u & = 3, 4 \end{aligned}$

:log2x)2--7log2x+12=0u2-7u+12=0(u-3)(u-4)=0u=3,4)

Step 3: Now, substitute back and solve for $\begin{aligned} x \end{aligned}$ in each case.
::步骤3:现在,在每种情况下以 x 替换并解决。

$\begin{aligned} \log_{2} x & = 3 ⟺ x = 2^{3} = 8 \\ \log_{2} x & = 4 ⟺ x = 2^{4} = 16 \end{aligned}$

::对数 2x=3x=23=8log2x=4x=24=16

Example 4
::例4

Return to the mayor's question from the Introduction. W hen will the small town reach a population of 20,000, as modeled by the equation $\begin{aligned} 20, 000 = 2, 000 (1.05)^{t} \end{aligned}$ ?
::回到市长在导言中提出的问题,小城镇何时才能达到以20,000=2,000(1.05)t等式为模型的20,000人口?

$\begin{aligned} 20, 000 & = 2, 000 (1.05)^{t} \\ 10 & = (1.05)^{t} \\ \log 10 & = \log (1.05)^{t} \\ \log (10) & = t \cdot \log 1.05 \\ t & = \frac{\log (10)}{\log (1.05)} \\ t & \approx 47.19 years \end{aligned}$

::20,000=2,000(1.05)t10=(1.05)tlog@10=(1.05)tlog@(10)=(1.05)tlog@1.05t=(10)g}(10) log}(1.05)t47.19年

The population will reach 20,000 in 1997, because 1997 is 47 years after 1950.
::1997年人口将达到20,000人,因为1997年是在1950年以后47年。

Example 5
::例5

List all possible values of $\begin{aligned} x \end{aligned}$ for the following equation : $\begin{aligned} (x + 1)^{x - 4} - 1 = 0. \end{aligned}$
::列出下列方程式的所有可能的 x 值 (x+1) x- 4- 1=0 。

Solution:
::解决方案 :

$\begin{aligned} (x + 1)^{x - 4} - 1 & = 0 \\ (x + 1)^{x - 4} & = 1 Add 1. \\ \log (x + 1)^{x - 4} & = \log 1 Take the log of both sides. \\ (x - 4) \cdot \log (x + 1) & = 0 Use properties of logs. \\ x - 4 = 0 or \log (x + 1) & = 0 \\ x - 4 = 0 or 10^{0} & = x + 1 \\ x - 4 = 0 or 1 - 1 & = x \\ x & = 4, 0 \end{aligned}$

:x+1) x-4- 1=0=0(x+1) x-4=1 添加 1.log(x+1) x-4=log*1 使用两侧的日志。(x- 4) log}(x+1) (x+1)=0 使用日志的属性。x- 4=0 或log (x+1)=0=0 或 100=x+1x-4=0 或 1- 1=xx=4,0 使用日志的属性。

Recall that you can only take the log of a positive argument. What if $\begin{aligned} x + 1 \end{aligned}$ is negative 1 but raised to an even power?
::回顾您只能接受正参数的日志。如果 x+1 是负 1, 但却被提升到一个偶数, 那么会怎样 ?

Notice that when $\begin{aligned} x = - 2 \end{aligned}$ ,
$\begin{aligned} (- 2 + 1)^{- 2 - 4} - 1 = (- 1)^{- 6} - 1 = \frac{1}{(- 1)^{6}} - 1 = 0, \end{aligned}$
so it is also a solution. However, $\begin{aligned} \log (- 2 + 1) = \log (- 1) \end{aligned}$ is not possible.
::当 x2, (-2+1) -2 - 4 - 1=(-1) - 6 - 1=1 (-1) - 1=0, 也是一种解决办法。但是, log *(-2+1) =log (-1) 是不可能的。

Note that you shouldn't fall into the habit of assuming you can take the log of both sides and get all the solutions . This is only true when the argument is strictly positive.
::请注意, 您不应该习惯于假设您可以使用双方的日志并获得所有解决方案。只有当争论是绝对肯定时, 才会出现这种情况。

Example 6
::例6

Light intensity as it travels at specific depths of water in a swimming pool can be described by the relationship between $\begin{aligned} i \end{aligned}$ for intensity, and $\begin{aligned} d \end{aligned}$ for depth in feet. What is the intensity of light at 10 feet?
::光强度在游泳池中特定水深中行走时的光强度可以用i与d与d与脚的深度之间的关系来描述。 10英尺的光强度是多少?

$\begin{aligned} \log (\frac{i}{12}) = - 0.0145 \cdot d \end{aligned}$
:i12) 0.0145d

Solution:
::解决方案 :

Given $\begin{aligned} d = 10 \end{aligned}$ , solve for $\begin{aligned} i \end{aligned}$ measured in lumens.
::根据 d=10, 解答我用月光测量的答案。

$\begin{aligned} \log (\frac{i}{12}) & = - 0.0145 \cdot d \\ \log (\frac{i}{12}) & = - 0.0145 \cdot 10 \\ \log (\frac{i}{12}) & = - 0.145 \\ (\frac{i}{12}) & = 10^{- 0.145} \\ i & = 12 \cdot 10^{- 0.145} \approx 8.594 lumens \end{aligned}$

:i12)0.0145(i12)0.014510log(i12)0.145(i12)=10-0.145i=1210-0.1458.594

Example 7
::例7

Solve the following equation for all possible values of $\begin{aligned} x \end{aligned}$ :
::为 x 的所有可能值解决下列方程式:

$\begin{aligned} \frac{e^{x} - e^{- x}}{3} = 14 \end{aligned}$

::- exe- x3=14

Solution:
::解决方案 :

First solve for $\begin{aligned} e^{x} \end{aligned}$ :
::ex 的第一个解答 :

$\begin{aligned} \frac{e^{x} - e^{- x}}{3} & = 14 \\ e^{x} - e^{- x} & = 42 \\ e^{x} \cdot (e^{x} - e^{- x}) & = (42) \cdot e^{x} \\ e^{2 x} - 1 & = 42 e^{x} \\ (e^{x})^{2} - 42 e^{x} - 1 & = 0 \end{aligned}$

::-e-x3=14ex-e-x=42ex__(ex-e-x)=(42__ex-e-x)=(42―ex2x-1=42ex(ex)2-42ex-1=0)

Let $\begin{aligned} u = e^{x} \end{aligned}$ .
::让u=ex。

$\begin{aligned} u^{2} - 42 u - 1 = 0 \\ u = \frac{- (- 42) \pm \sqrt{(- 42)^{2} - 4 \cdot 1 \cdot (- 1)}}{2 \cdot 1} = \frac{42 \pm \sqrt{1768}}{2} \approx 42.023796, - 0.0237960 \end{aligned}$

::u2-42u-1=0u(-42(-42-2)-(-42)-41(-1)-21=421768242.023796,-0.0237960

Since the range of the exponential function is greater than 0, $\begin{aligned} e^{x} > 0 \end{aligned}$ , then $\begin{aligned} e^{x} \approx - 0.0237960 \end{aligned}$ does not exist. Thus, $\begin{aligned} - 0.0237960 \end{aligned}$ is extraneous, so there is only one result.
::由于指数函数的范围大于 0, ex>0, 那么 ex 0.0 237960 不存在。因此, 0.0 237960 是外部的, 所以只有一个结果。

$\begin{aligned} e^{x} & \approx 42.023796 \\ x & \approx \ln 42.023796 \approx 3.738 \end{aligned}$

::-42.023796x 42.0237963.738

Summary
::摘要

To solve an exponential equation:

Isolate the exponential part of the equation. If there are two exponential parts, then rewrite so there is a single exponent on each side of the equation.
::分离方程式的指数部分。如果有两个指数部分, 请重写, 这样方程式的每侧都有一个符号。

Take the logarithm of each side of the equation.
::以方程每侧的对数取出方程的对数。

Solve for the variable.
::解决变量。

Check your solution.
::检查你的解决方案。

::要解析指数方程式 : 分离方程式的指数部分。如果有两个指数部分, 请重写, 这样方程式的每侧都有一个符号。选择方程式每一侧的对数。解决变量。请检查您的解答。

Review
::回顾

Solve each equation for $\begin{aligned} x \end{aligned}$ . If necessary, round each answer to three decimal places.
::x 的每个方程式都解答。如果需要,将每个方程式的回答按小数点后三位数进行。

1. $\begin{aligned} 4^{x} = 6 \end{aligned}$
::1. 4x=6

2. $\begin{aligned} 5^{x} = 2 \end{aligned}$
::2. 5x=2

3. $\begin{aligned} 12^{4 x} = 1, 020 \end{aligned}$
::3. 124x=1,020

4. $\begin{aligned} 7^{3 x} = 2, 400 \end{aligned}$
::4. 73x=2 400

5. $\begin{aligned} 2^{x + 1} - 5 = 22 \end{aligned}$
::5. 2x+1-5=22

6. $\begin{aligned} 5 x + 12^{x} = 5 x + 7 \end{aligned}$
::6. 5x+12x=5x+7

7. $\begin{aligned} 2^{x + 1} = 2^{2 x + 3} \end{aligned}$
::7. 2x+1=22x+3

8. $\begin{aligned} 3^{x + 3} = 9^{x + 1} \end{aligned}$
::8. 3x+3=9x+1

9. $\begin{aligned} 2^{x + 4} = 5^{x} \end{aligned}$
::9. 2x+4=5x

10. $\begin{aligned} 13 \cdot 8^{0.2 x} = 546 \end{aligned}$
::10.80.2x=546

11. $\begin{aligned} b^{x} = c + a \end{aligned}$
::11. bx=c+a

12. $\begin{aligned} 32^{x} = 0.94 - .12 \end{aligned}$
::12. 32x=0.94-12

Solve each log equation by using log properties and rewriting as an exponential equation:
::使用日志属性并重写成指数方程式来解决每个日志方程式:

13. $\begin{aligned} \log_{3} x + \log_{3} 5 = 2 \end{aligned}$
::13.3x+log3=2 对数 3x+log3=5=2

14. $\begin{aligned} 2 \log x = \log 8 + \log 5 - \log 10 \end{aligned}$
::14. 2log_x=log_8+log_5_log_10

15. $\begin{aligned} \log_{9} x = \frac{3}{2} \end{aligned}$
::15. log9x=32

Review (Answers)
::回顾(答复)

Please see the Appendix.
::请参看附录。

章节大纲

Introduction ::导言

Exponential Equation Solution Techniques ::指数赤道溶解技术

Properties of Using Logarithms ::使用对数属性

Examples ::实例

Summary ::摘要

Review ::回顾

Review (Answers) ::回顾(答复)

Introduction
::导言

Exponential Equation Solution Techniques
::指数赤道溶解技术

Properties of Using Logarithms
::使用对数属性

Examples
::实例

Summary
::摘要

Review
::回顾

Review (Answers)
::回顾(答复)