Hi Everyone!

On this page you will find some material about Lesson 37. Read through the material below, watch the videos, and follow up with your instructor if you have questions.

**Lesson 37: Logarithmic and Exponential Equations**

Table of Contents

### Resources

*In this section you will find some important information about the specific resources related to this lesson: *

*the learning outcomes,**the section in the textbook,**the WeBWorK homework sets,**a link to the pdf of the lesson notes,**a link to a video lesson.*

**Learning Outcomes.**

- Solve a logarithmic equation.
- Understand the need of checking the solution(s) of a logarithmic equation.
- Solve an exponential equation.
- Write an exponential expression in logarithmic form.

**Topic**. This lesson covers Section 8.7: Logarithmic and Exponential Equations and Applications.

**WeBWorK**. There are two WeBWorK assignments on today’s material:

ExponentialEquations

ExponentialEquations-Calc

**Lesson Notes.**

These notes are used in Lessons 36 and 37. For today’s lesson, you can consider all pages, which will include a brief review on Compound Interest.

**Video Lesson. **

Video Lesson 37 (based on Lesson 37 Notes)

This video is used in Lessons 36 and 37. For today’s lesson, watch the whole video.

### Warmup Questions

*These are questions on fundamental concepts that you need to know before you can embark on this lesson. Don’t skip them! Take your time to do them, and check your answer by clicking on the “Show Answer” tab.*

#### Warmup Question 1

Solve $\log_5x=-3$.

#### Show Answer 1

$x=5^{-3}=\dfrac{1}{5^3}=\dfrac{1}{125}$

**Check**: $\log_5(5^{-3}) = -3$ $\quad\checkmark$

The solution set is $\left\{\dfrac{1}{125}\right\}$.

#### Warmup Question 2

Solve $3^x=9$

#### Show Answer 2

$$3^x=9$$

$$3^x=3^2$$

$$x=2$$

The solution set is $\left\{2\right\}$.

#### Warmup Question 3

Solve $5^{x+1}=1$.

#### Show Answer 3

$$5^{x+1}=1$$

$$5^{x+1}=5^0$$

$$x+1=0$$

$$x=-1$$

The solution set is $\left\{-1\right\}$.

### Review

*If you are not comfortable with the Warmup Questions, don’t give up! Click on the indicated lesson for a quick catchup. A brief review will help you boost your confidence to start the new lesson, and that’s perfectly fine.*

### Quick Intro

*This is like a mini-lesson with an overview of the main objects of study. It will often contain a list of key words, definitions and properties – all that is new in this lesson. We will use this opportunity to make connections with other concepts. It can be also used as a review of the lesson.*

**A Quick Intro to Logarithmic and Exponential Equations**

**Key Words.** Logarithmic equation, exponential form, checking the solution, exponential equation.

**Logarithmic Equations**

$\bigstar$ In the Warmup Question 1, we solved $\log_5x=-3$ by writing it in **exponential form**:

$$x=5^{-3}.$$

This is the basic idea in solving **logarithmic equations**. We may be given a more complicated equation, but if we leave it in the same form as $\log_5x=-3$, then you can solve it by rewriting it in exponential form. What is special about this form is that:

**there is a single logarithm on the left side**. If the equation has more logarithm terms with the same base, you can try and combine them by using properties such as

$$\log_bx +\log_b y = \log_b(xy), and$$

$$\log_bx-\log_by = \log_b\left(\dfrac{x}{y}\right).$$

**on the other side there is a number**. If you have another logarithm term on the other side, then you can use the following:

$$\log_bx=\log_by\quad\text{ implies }\quad x=y.$$

$\bigstar$ Attention: when we take the logarithm of $x$, we need to make sure $x$ is positive. So you need to check your answer.

**Exponential Equations**

$\bigstar$ In the Warmup Question 2 , we solved $3^x=9$ by writing $9$ as $3^2$ so that $3^x=3^2$ and deduce that $x=2$. In the last step we obtained $x=2$ by comparing the exponents in $3^x=3^2$. This is true because $3^x$ and $3^2$ are powers with the same base $3$. In general,

$$b^x=b^y\quad\text{ implies }\quad x=y.$$

When solving an **exponential equation**, try to rewrite it in a way so that you have a single exponential term on each side where both bases are the same. If the bases are not the same like

$$3^{x+1}=5^{x},$$

take the logarithm (with any base) of both sides:

$$\ln(3^{x-1})=\ln(5^x)$$

Using the logarithm properties, we obtain

$$(x-1)\ln 3 = x\ln 5$$

$$x\ln 3 – \ln 3 = x\ln 5$$

$$x\ln 3 – x\ln 5 = \ln 3$$

$$x(\ln 3 – \ln 5) = \ln 3$$

$$x = \dfrac{\ln 3}{\ln 3 – \ln 5}$$

### Video Lesson

*Many times the mini-lesson will not be enough for you to start working on the problems. You need to see someone explaining the material to you. In the video you will find a variety of examples, solved step-by-step – starting from a simple one to a more complex one. Feel free to play them as many times as you need. Pause, rewind, replay, stop… follow your pace!*

**Video Lesson**

**A description of the video**

This video is used in Lessons 36 and 37. For today’s lesson, watch the whole video.

In this video you will see the equations

- $2^x=8$
- $2^x=7$
- $2^{5x-1}=8$
- $2^{3x-1}=5$
- $7^x=3^{x-2}$

You will also see the following applications that require solving an exponential equation.

- How long does it take to have \$1,600 if you invest \$800 at a rate of $3\%$ compounded quarterly?
- How long must you invest \$1,000 to have \$1500 with a rate of $2\%$ annually compounded continuously?

### Try Questions

*Now that you have read the material and watched the video, it is your turn to put in practice what you have learned. We encourage you to try the Try Questions on your own. When you are done, click on the “Show answer” tab to see if you got the correct answer.*

#### Try Question 1

Solve in exact form.

$$5^x=3$$

#### Show Answer 1

$$5^x=3$$

$$x=\log_53$$

The solution set is $\left\{\log_53\right\}$.

#### Try Question 2

Solve.

$$6^{x+1}=216$$

#### Show Answer 2

$$6^{x+1}=216$$

$$6^{x+1}=6^3$$

$$x+1=3$$

$$x=2$$

The solution set is $\left\{2\right\}$.

#### Try Question 3

Solve in exact form. $$9^{2x-1}=7$$

#### Show Answer 3

$$9^{2x-1}=7$$

$$9^{2x-1}=9^{\log_97}$$

$$2x-1=\log_97$$

$$2x=\log_97+1$$

$$x=\dfrac{\log_97+1}{2}$$

The solution set is $\left\{\dfrac{\log_97+1}{2}\right\}$.

#### Try Question 4

For how long do we have to invest \$1000 at a rate of $2.5\%$ compounded continuously to obtain a final amount of \$1100?

#### Show Answer 4

$$Pe^{rt} = A$$

$$1000e^{0.025 t} = 1100 $$

$$e^{0.025 t} = \dfrac{1100}{1000} $$

$$e^{0.025 t} = 1.1 $$

$$ \ln\left(e^{0.025 t}\right) =\ln(1.1)$$

$$0.025 t= \ln(1.1) $$

$$t = \dfrac{\ln(1.1)}{ 0.025} $$

$$ t \approx 3.82124$$

It will take about 3.8 years to obtain a final amount of \$ 1100.

### WeBWorK

*You should now be ready to start working on the WeBWorK problems. Doing the homework is an essential part of learning. It will help you practice the lesson and reinforce your knowledge.*

**WeBWorK**

It is time to do the homework on WeBWork:

ExponentialEquations

ExponentialEquations-Calc

When you are done, come back to this page for the Exit Questions.

### Exit Questions

*After doing the WeBWorK problems, come back to this page. The Exit Questions include vocabulary checking and conceptual questions. Knowing the vocabulary accurately is important for us to communicate. You will also find one last problem. All these questions will give you an idea as to whether or not you have mastered the material. Remember: the “Show Answer” tab is there for you to check your work!*

#### Exit Questions

- What does it mean to solve an exponential equation?
- What about the equation $4^x=5$ tells us we should use the logarithm? How do we know if we should use the $\ln$ or the $\log$ to solve an equation?
- Is it necessary to check your answer to an exponential equation even if you know you have not made an error?

$\bigstar$ (a) Solve $4^{2x+7}=16.$

$\bigstar$ (b) Solve $2^{t+1}= 5$ in exact form.

#### Show Answer

(a)

$$4^{2x+7}=16$$

$$4^{2x+7}=4^2$$

$$2x+7=2$$

$$2x=-5$$

$$x=-\dfrac{5}{2}$$

The solution set is $\left\{-\dfrac{5}{2}\right\}$.

(b)

$$2^{t+1}= 5$$

$$\log(2^{t+1}) =\log 5$$

$$(t+1)\log 2 = \log 5$$

$$t\log 2 + \log 2 = \log 5$$

$$ t\log 2 = \log 5 – \log 2$$

$$t = \dfrac{\log 5 – \log 2}{\log 2}$$

The solution set is $\left\{\dfrac{\log 5 – \log 2}{\log 2}\right\}$.

### Need more help?

*Don’t wait too long to do the following.*

- Watch the additional video resources.

- Talk to your instructor.
- Form a study group.
- Visit a tutor. For more information, check the tutoring page.