Hi Everyone!
On this page you will find some material about Lesson 25. Read through the material below, watch the videos, and follow up with your instructor if you have questions.
Lesson 25: Exponential Expressions
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.
- Recognize an exponential expression.
- Approximate exponential expressions with non-integer exponents.
- Distinguish an exponential growth expression from an exponential decay expression.
- Solve simple exponential growth problems.
Topic. This lesson covers
Section 8.3.1: Definition of an Exponential Function,
Section 8.3.2: Approximating Exponential Expressions with a Calculator, and
Section 8.3.4: Applications of Exponential Functions.
WeBWorK. There is one WeBWorK assignment on today’s material:
ExponentialFunctions
Lesson Notes.
These notes are used in Lessons 25-27. For today’s lesson, only consider page 1.
Video Lesson.
Video Lesson 25 (based on Lesson 25 Notes)
This video is used in Lessons 25-27. For today’s lesson, watch from [0:0] to [4:36].
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
(a) In the expression $x^3$, what are the base and the exponent? Find its value when $x=2$.
(b) In the expression $3^x$, what are the base and the exponent? Find its value when $x=2$.
Show Answer 1
(a) In the expression $x^3$, the base is $x$ and the exponent is $3$. When $x=2$, we have
$$x^3=2^3=8.$$
(b) In the expression $3^x$, the base is $3$ and the exponent is $x$. When $x=2$, we have
$$3^x=3^2 =9.$$
Warmup Question 2
Evaluate $27^{2/3}$ without using a calculator.
Show Answer 2
$$27^{2/3} = (27^{1/3})^2 = (\sqrt[3]{27})^2$$
$$= (\sqrt[3]{3^3})^2=3^2=9$$
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 Exponential Expressions
Key Words. Exponential expression, exponential growth, exponential decay, population growth.
In the Warmup Question 1 we saw two expressions,
$$x^3\quad\text{ and }\quad 3^x.$$
$\bullet$ $x^3$ is a polynomial (here $x$ is the base; polynomial will be further studied in MAT 1375)
$\bullet$ $3^x$ is an exponential expression (here $x$ is the exponent)
$\bigstar$ In general, an exponential expression looks like
$$b^x$$
where $b$ is any real number such that $b>0$ and $b\neq 1$.
Examples of exponential expressions: $5^x$, $\left(\dfrac{1}{4}\right)^x$, $(\sqrt 7)^x$ etc.
$\bigstar$ Returning to $3^x$, the exponent $x$ can take any real value. For example, what if $x=\sqrt 3$? Then we have
$$3^{\sqrt 3}.$$
Now use a calculator to find what this number is.
$$3^{\sqrt 3} \approx 6.70499185.$$
The symbol $\approx$ reads “is approximately.”
Exponential expressions appear in many real world problems.
$\bigstar$ If $b>1$, then $b^x$ is an exponential growth expression.
$\bigstar$ If $0<b<1$, then $b^x$ is an exponential decay expression.
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 25-27. For today’s lesson, watch from [0:0] to [4:36].
In this video you will see the exponential expression $2^x$.
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
The half-life of a radioactive substance is the amount of time that it takes for half of the original amount of the substance to change. An isotope of radium called Radium 226 has a half-life of 1620 years. The amount of Radium 226 present after $t$ years in grams is given by
$$A(t) = \left(\dfrac{1}{2}\right)^{t/1620}.$$
Find the amount of Radium 226 present after 3500 years.
Show Answer 1
$$A(3500) = \left(\dfrac{1}{2}\right)^{3500/1620}\approx 0.22$$
After 3500 years, about 0.22 g remains.
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:
ExponentialFunctions
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 is the difference between $y=x^2$ and $y=2^x$?
- What are some applications of exponential equations?
$\bigstar$ Population growth model
Suppose that the initial population is $P_0=5,000$. Let $r=0.025$ be the annual rate of increase of the population. After $t$ years, the population is $P(t) = P_0(1+r)^t,$ that is, $$P(t) = 5000(1+0.025)^t.$$
Find the population after 7 years.
Show Answer
$$P(7) = 5000(1+0.025)^7 = 5000(1.025)^7\approx 5,943$$
After 7 years, the population will be approximately 5,943.
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.
