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Section 5.1 Introduction to Exponentials (EL1)

Subsection 5.1.1 Activities

Remark 5.1.1.

Linear functions have a constant rate of change - that is a constant change in output for every change in input. Let’s consider functions which do not fit this model - those which grow more rapidly and change by a varying amount for every change in input.

Activity 5.1.2.

You have two job offers on the horizon. One has offered to pay you \(\$10{,}000\) per month while the other is offering \(\$0.01\) the first month, \(\$0.02\) the second month, \(\$0.04\) the third month and doubles every month. Which job would you rather take?
(a)
Make a table representing how much money you will be paid each month for the first two years from the first job - paying \(\$10{,}000\) per month.
Answer.
month Job 1
\(1\) \(10{,}000\)
\(2\) \(10{,}000\)
\(3\) \(10{,}000\)
\(4\) \(10{,}000\)
\(5\) \(10{,}000\)
\(6\) \(10{,}000\)
\(7\) \(10{,}000\)
\(8\) \(10{,}000\)
\(9\) \(10{,}000\)
\(10\) \(10{,}000\)
\(11\) \(10{,}000\)
\(12\) \(10{,}000\)
\(13\) \(10{,}000\)
\(14\) \(10{,}000\)
\(15\) \(10{,}000\)
\(16\) \(10{,}000\)
\(17\) \(10{,}000\)
\(18\) \(10{,}000\)
\(19\) \(10{,}000\)
\(20\) \(10{,}000\)
\(21\) \(10{,}000\)
\(22\) \(10{,}000\)
\(23\) \(10{,}000\)
\(24\) \(10{,}000\)
(b)
Make a table representing how much money you will be paid each month for the first two years from the second job - paying \(\$0.01\) the first month and doubling every month after.
Answer.
month Job 2
\(1\) \(0.01\)
\(2\) \(0.02\)
\(3\) \(0.04\)
\(4\) \(0.08\)
\(5\) \(0.16\)
\(6\) \(0.32\)
\(7\) \(0.64\)
\(8\) \(1.28\)
\(9\) \(2.56\)
\(10\) \(5.12\)
\(11\) \(10.24\)
\(12\) \(20.48\)
\(13\) \(40.96\)
\(14\) \(81.92\)
\(15\) \(163.84\)
\(16\) \(327.68\)
\(17\) \(655.36\)
\(18\) \(1{,}310.72\)
\(19\) \(2{,}621.24\)
\(20\) \(5{,}242.88\)
\(21\) \(10{,}485.76\)
\(22\) \(20{,}971.52\)
\(23\) \(41{,}943.04\)
\(24\) \(83{,}886.08\)
(c)
Which job is earning more money per month after one year?
Answer.
Job 1 is earning \(\$10{,}000\) per month. Job 2 is earning \(\$20.48\) per month.
(d)
Which job is earning more money per month after 18 months?
Answer.
Job 1 is earning \(\$10{,}000\) per month. Job 2 is earning \(\$1{,}310.72\) per month.
(e)
According to your tables, does the second job ever earn more money per month than the first job?
Answer.
Yes! After 21 months, Job 1 is earning \(\$10{,}000\) per month. Job 2 is earning \(\$10{,}485.76\) per month.

Remark 5.1.3.

This idea of a function that grows very rapidly by a factor, ratio, or percent each time, like the second job in Activity 5.1.2, is considered exponential growth.

Definition 5.1.4.

Let \(a\) be a non-zero real number and \(b \neq 1\) a positive real number. An exponential function takes the form
\begin{equation*} f(x)=ab^{x} \end{equation*}
\(a\) is the initial value and \(b\) is the base.

Activity 5.1.5.

Evaluate the following exponential functions.
(a)
\(f(x)=4^{x}\) for \(f(3)\)
Answer.
\(f(3)=64\)
(b)
\(f(x)=\left( \frac{1}{3} \right)^{x}\) for \(f(3)\)
Answer.
\(f(3)=\frac{1}{27}\)
(c)
\(f(x)=3\left( 5 \right)^{x}\) for \(f(-2)\)
Answer.
\(f(-2)=\frac{3}{25}\)
(d)
\(f(x)=-2^{3x-4}\) for \(f(4)\)
Answer.
\(f(4)=-256\)

Remark 5.1.6.

Notice that in Activity 5.1.5 part (a) the ouput value is larger than the base, while in part (b) the output value is smaller than the base. This is similar to the difference between a positive and negative slope for linear functions.

Activity 5.1.7.

Consider two exponential functions \(f(x)=100(2)^{x}\) and \(g(x)=100 \left( \frac{1}{2} \right)^{x}\text{.}\)
(a)
Fill in the table of values for \(f(x)\text{.}\)
\(x\) \(f(x)\)
\(0\)
\(1\)
\(2\)
\(3\)
\(4\)
Answer.
\(x\) \(f(x)\)
\(0\) \(100\)
\(1\) \(200\)
\(2\) \(400\)
\(3\) \(800\)
\(4\) \(1600\)
(b)
Fill in the table of values for \(g(x)\text{.}\)
\(x\) \(g(x)\)
\(0\)
\(1\)
\(2\)
\(3\)
\(4\)
Answer.
\(x\) \(f(x)\)
\(0\) \(100\)
\(1\) \(50\)
\(2\) \(25\)
\(3\) \(12.5\)
\(4\) \(6.25\)
(c)
How do the values in the tables compare?
Answer.
The values of \(f(x)\) are getting larger while those of \(g(x)\) are getting smaller.

Remark 5.1.8.

In Activity 5.1.7, the only difference between the two exponential functions was the base. \(f(x)\) has a base of \(2\text{,}\) while \(g(x)\) has a base of \(\frac{1}{2}\text{.}\) Let’s use this fact to update Definition 5.1.4.

Remark 5.1.9.

An exponential function of the form \(f(x)=ab^{x}\) will grow (or increase) if \(b \gt 1\) and decay (or decrease) if \(0\lt b \lt 1\text{.}\)

Activity 5.1.10.

For each year \(t\text{,}\) the population of a certain type of tree in a forest is represented by the function \(F(t)=856(0.93)^t\text{.}\)
(a)
How many of that certain type of tree are in the forest initially?
Answer.
\(856\) trees
(b)
Is the number of trees of that type growing or decaying?
Answer.
Decaying \(b=0.93 \lt 1\)

Activity 5.1.11.

To begin creating equations for exponential functions using \(a\) and \(b\text{,}\) let’s compare a linear function and an exponential function. The tables show outputs for two different functions \(r\) and \(s\) that correspond to equally spaced input.
\(x\) \(r(x)\)
\(0\) \(12\)
\(3\) \(10\)
\(6\) \(8\)
\(9\) \(6\)
\(x\) \(s(x)\)
\(0\) \(12\)
\(3\) \(9\)
\(6\) \(6.75\)
\(9\) \(5.0625\)
(a)
Which function is linear?
Answer.
\(r(x)\) since the outputs decrease by 2 every time.
(b)
What is the initial value of the linear function?
Answer.
\(12\)
(c)
What is the slope of the linear function?
Answer.
\(-\frac{2}{3}\)
(d)
What is the initial value of the exponential function?
Answer.
\(12\)
(e)
What is the ratio of consecutive outputs in the exponential function?
  1. \(\displaystyle \frac{4}{3}\)
  2. \(\displaystyle \frac{3}{4}\)
  3. \(\displaystyle -\frac{4}{3}\)
  4. \(\displaystyle -\frac{3}{4}\)
Answer.
B

Remark 5.1.12.

In a linear function the differences are constant, while in an exponential function the ratios are constant.

Activity 5.1.13.

Find an equation for an exponential function passing through the points \((0,4)\) and \((1,6)\text{.}\)
(a)
Find the initial value.
  1. \(\displaystyle 0\)
  2. \(\displaystyle 4\)
  3. \(\displaystyle 1\)
  4. \(\displaystyle 6\)
Answer.
B
(b)
Find the common ratio.
  1. \(\displaystyle \frac{3}{2}\)
  2. \(\displaystyle 6\)
  3. \(\displaystyle \frac{2}{3}\)
  4. \(\displaystyle \frac{1}{6}\)
Answer.
A
(c)
Find the equation.
  1. \(\displaystyle f(x)=6\left( \frac{3}{2} \right)^x\)
  2. \(\displaystyle f(x)=4\left( 6 \right)^x\)
  3. \(\displaystyle f(x)=4\left( \frac{2}{3} \right)^x\)
  4. \(\displaystyle f(x)=4\left( \frac{3}{2} \right)^x\)
Answer.
D: \(f(x)=4\left( \frac{3}{2} \right)^x\)

Remark 5.1.14.

The letter \(e\) represents an irrational number and is used as a base for many real-world exponential models. To work with base \(e\text{,}\) we use the approximation, \(e \approx 2.718282\text{.}\) It can also be found on most calculators.

Activity 5.1.15.

Use a calculator to evaluate the following exponentials involving the base \(e\text{.}\)
(a)
\(f(x)=-2e^{x}-2\) for \(f(-2)\)
  1. \(\displaystyle -0.0366\)
  2. \(\displaystyle -2.2707\)
  3. \(\displaystyle -1.7293\)
  4. \(\displaystyle -16.778\)
Answer.
B
(b)
\(f(x)=\frac{1}{3}e^{x+1}\) for \(f(-1)\)
  1. \(\displaystyle 1\)
  2. \(\displaystyle 0\)
  3. \(\displaystyle 1.122\)
  4. \(\displaystyle \frac{1}{3}\)
Answer.
D

Remark 5.1.16.

Recall the negative rule of exponents which states that for any nonzero real number \(a\) and natural number \(n\)
\begin{equation*} a^{-n}=\frac{1}{a^n} \end{equation*}

Activity 5.1.17.

Let’s consider the two exponential functions \(f(x)=2^{-x}\) and \(g(x)=\left(\frac{1}{2}\right)^{x}\text{.}\)
(a)
Fill in the table of values for \(f(x)\text{.}\)
\(x\) \(f(x)\)
\(-2\)
\(-1\)
\(0\)
\(1\)
\(2\)
Answer.
\(x\) \(f(x)\)
\(-2\) \(4\)
\(-1\) \(2\)
\(0\) \(1\)
\(1\) \(\frac{1}{2}\)
\(2\) \(\frac{1}{4}\)
(b)
Fill in the table of values for \(g(x)\text{.}\)
\(x\) \(g(x)\)
\(-2\)
\(-1\)
\(0\)
\(1\)
\(2\)
Answer.
\(x\) \(g(x)\)
\(-2\) \(4\)
\(-1\) \(2\)
\(0\) \(1\)
\(1\) \(\frac{1}{2}\)
\(2\) \(\frac{1}{4}\)
(c)
What do you notice about the two functions?
Answer.
Their values are the same.
(d)
Use Remark 5.1.16 and other properties of exponents to try and rewrite \(f(x)\) as \(g(x)\text{.}\)
Answer.
\begin{equation*} f(x)=2^{-x}=\frac{1}{2^x}=\left(\frac{1}{2}\right)^x=g(x) \end{equation*}

Subsection 5.1.2 Videos

It would be great to include videos down here, like in the Calculus book!