Section: 指数等量 | 3.6 指数等量

Section outline

When you were first learning equations, you learned the rule that whatever you do to one side of an equation , you must also do to the other side so that the equation stays in balance. The basic techniques of adding, subtracting, multiplying and dividing both sides of an equation worked to solve almost all equations up until now. With logarithms, you have more tools to isolate a variable . Consider the following equation and ask yourself: why is $\begin{aligned} x = 3 \end{aligned}$ ? Logically it makes sense that if the bases match, then the exponents must match as well, but how can it be shown for examples like this one?
::当你刚开始学习方程式时,你学会了一条规则,即无论你对方程式的一边做什么,你也必须对另一方做什么,以使方程式保持平衡。增加、减、乘和分割一个方程式的两边的基本技术一直有助于解决几乎所有方程式,直到现在为止。有了对数,你有更多的工具来孤立变量。想想下面的方程式,然后问问你自己:为什么是x=3?逻辑上讲,如果基准匹配,那么出价者也必须匹配,但是如何用这个例子来显示?

\begin{aligned} {1.79898}^{2 x} = {1.79898}^{6} \end{aligned}

::1.798982x=1.798986

Solving Exponential Equations
::解决指数等号

A common technique for solving equations with unknown variables in exponents is to take the log of the desired base of both sides of the equation. Then, you can use properties of logs to simplify and solve the equation.
::用于解析方程式中未知变量的常见技术是将方程式两侧的预期基数进行日志。然后,您可以使用日志的属性来简化和解析方程式。

Take the following equation. To solve for $\begin{aligned} t \end{aligned}$ , you should first simplify the expression as much as possible and then take the natural log of both sides.
::取下以下方程。要解决 t, 您应该首先尽可能简化表达式, 然后使用双方的自然日志。

$\begin{aligned} 9, 000 = 300 \cdot \frac{(1.06)^{t} - 1}{0.06} \end{aligned}$
::9,000=300(1.06)t-10.06

\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

It does not matter what base you use in this situation as long as you use the same base on both sides. Choosing natural log allows you to use a calculator to finish the problem.
::只要在两侧使用相同的基数,在这种情况下使用什么基数并不重要。选择自然日志可以使您使用计算器来完成问题。

Note that this type of equation is common in financial mathematics. The equation above represents the unknown amount of time it will take you to save $9,000 in a savings account if you save $300 at the end of each year in an account that earns 6% annual compound interest.
::请注意, 此等式在金融数学中很常见。以上等式代表未知的时间, 如果您在每年年底在一个每年赚取6%年复利的账户中节省300美元, 节省在储蓄账户中9000美元。

The other good base to use is base 10. When solving the following equation for $\begin{aligned} x \end{aligned}$ : $\begin{aligned} 16^{x} = 25 \end{aligned}$ , you will need to use a calculator to get the final answer and your calculator can handle base 10 as well. First take the log of both sides. Then, use log properties and your calculator to help.
::另一个良好的使用基础是基数 10 。当解析 x 的下列方程式时: 16x=25, 您需要使用计算器来获得最后答案, 您的计算器也可以同时处理 10 基数。首先使用两边的日志。然后, 使用日志属性和计算器来帮助。

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

::16x=25log_16x=log_25xlog_16=log_25x=log_25x=log_25x=25log_16x=1.16

Examples
::实例

Example 1
::例1

Earlier, you were asked how to show that if the bases match in an equation, the exponents should match. In the equation, logs can be used to reduce the equation to $\begin{aligned} 2 x = 6 \end{aligned}$ .
::早些时候, 有人询问您如何显示如果基数符合方程式, 指数应该匹配。在方程式中, 日志可以用来将方程式减为 2x=6 。

$\begin{aligned} {1.79898}^{2 x} = {1.79898}^{6} \end{aligned}$
::1.798982x=1.798986

Take the log of both sides and use the property of exponentiation of logs to bring the exponent out front.
::把两边的日志都取下来使用日志表的属性将日志表带出来

\begin{aligned} \log {1.79898}^{2 x} & = \log {1.79898}^{6} \\ 2 x \cdot \log 1.79898 & = 6 \cdot \log 1.79898 \\ 2 x & = 6 \\ x & = 3 \end{aligned}

::log1. 798982x=log1.7989862xlog_1.7989898=6_log@1. 798982x=6x=3

Example 2
::例2

Solve the following equation for all possible values of $\begin{aligned} x \end{aligned}$ : $\begin{aligned} (x + 1)^{x - 4} - 1 = 0 \end{aligned}$
::为 x 的所有可能值解决下列方程x+1)x-4-1=0)

$\begin{aligned} (x + 1)^{x - 4} - 1 = 0 \end{aligned}$
:x+1)x-4-1=0

$\begin{aligned} (x + 1)^{x - 4} = 1 \end{aligned}$
:x+1)x-4=1

Case 1 is that $\begin{aligned} x + 1 \end{aligned}$ is positive in which case you can take the log of both sides.
::例1是 x+1 是正数, 在此情况下, 您可以使用双方的日志。

\begin{aligned} \log {(x + 1)}^{(x - 4)} & = \log 1 \\ (x - 4) \cdot \log (x + 1) & = 0 \\ x - 4 = 0 & o r \log (x + 1) = 0 \\ x = 4 & o r (x + 1) = 10^{0} = 1 \\ x & = 4, 0 \end{aligned}

::log* (x+1) (x- 4) =log*%1(x- 4) =(x+1) =0x-4=0 orlog* (x+1) =0x=4 或(x+1) =100=1x=4,0

Note that $\begin{aligned} \log 1 = 0 \end{aligned}$
::注意对数1=0

Case 2 is that $\begin{aligned} (x + 1) \end{aligned}$ is negative 1 and raised to an even power. This happens when $\begin{aligned} x = - 2 \end{aligned}$ .
::案例2是 (x+1) 负 1 , 并升至偶数。当 x\\\\ 2 时会发生这种情况。

\begin{aligned} {(x + 1)}^{(x - 4)} & = 1 \\ {(- 2 + 1)}^{(- 2 - 4)} - 1 & = {(- 1)}^{- 6} - 1 \\ = \frac{1}{{(- 1)}^{6}} - 1 \\ = 0 \end{aligned}

x+1)(x-4)=1(-2+1)(-2-4)-1=(-1)-6-1(1)-1(1(-1)-6-1(-1)6-1=0)

The reason why this exercise is included is because you should not fall into the habit of assuming that you can take the log of both sides of an equation. It is only valid when the argument is strictly positive. For example, $\begin{aligned} \log {(- 2 + 1)}^{(- 2 - 4)} = \log (- 1) \end{aligned}$ is not possible.
::之所以将这一练习包括在内,是因为您不应该习惯于假设您可以使用方程两侧的日志, 只有当参数绝对肯定时才有效。例如, log *% (-2+1)(-2- 4) =log * (-1) 是不可能的。

Example 3
::例3

Light intensity, measured in lumens, can be described by the relationship between $\begin{aligned} i \end{aligned}$ for intensity and $\begin{aligned} d \end{aligned}$ for depth in feet as it travels at specific depths of water in a swimming pool . What is the intensity of light at 10 feet?
::以润滑度测量的光强度可以描述为强度i与深度d之间的关系,因为它在游泳池中特定水深处行走。 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

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 \end{aligned}

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

Example 4
::例4

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

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

::- exe- x3=14

First solve for $\begin{aligned} e^{x} \end{aligned}$ ,
::前妻的第一个解决方案,

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

::-e-x3=14ex(ex-e-x)=(42)exex2x-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

Note that the negative result is extraneous because $\begin{aligned} e^{x} \end{aligned}$ must be greater than zero, so you only proceed in solving for $\begin{aligned} x \end{aligned}$ for one result.
::请注意,负结果是不相干,因为前一结果必须大于零,所以您只能为一个结果解决 x 。

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

::-42.023796x 42.0237963.738

Example 5
::例5

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- log2x712 ) 解决下列方程式。

In calculus it is common to use a small substitution to simplify the problem and then substitute back later. In this case let $\begin{aligned} u = \log_{2} x \end{aligned}$ . Notice that this is a quadratic problem.
::在微积分中,通常使用一个小的替代来简化问题,然后再重新替换。在此情况下,请使用 u =log2 x。注意这是一个二次问题。

\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)

Now substitute back and solve for $\begin{aligned} x \end{aligned}$ in each case.
::现在换回并解决每个情况中的 x 。

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

::对等2\\ x=3\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ x=8log2\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ x=16

Summary

Solving exponential equations often involves taking the log of the desired base on both sides of the equation. This will isolate the missing variable. Then the equation can be solved with properties of logs.
::解析指数方程式时, 通常需要在方程式的两侧取出理想基数的日志。这将分离缺失的变量。然后方程式可以用日志的属性解析。

Review
::回顾

Solve each equation for $\begin{aligned} x \end{aligned}$ . Round each answer to three decimal places.
::x 的每个方程式都解答。将每个方程式的答案回合到小数点后的三个位数。

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

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

3. $\begin{aligned} 12^{4 x} = 1020 \end{aligned}$
::3. 124x=1020

4. $\begin{aligned} 7^{3 x} = 2400 \end{aligned}$
::4. 73x=2400

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.
::使用日志属性和重写成指数方程式,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)

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

Solving Exponential Equations ::解决指数等号

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Review ::回顾

Review (Answers) ::回顾(答复)