8.4: The Number e
The interest on a sum of money that compounds continuously can be calculated with the formula
Guidance
There are many special numbers in mathematics:
From the previous concept, we learned that the formula for compound interest is
Investigation: Finding the values of (1+1n)n as n gets larger
1. Copy the table below and fill in the blanks. Round each entry to the nearest 4 decimal places.

1  2  3  4  5  6  7  8 




2. Does it seem like the numbers in the table are approaching a certain value? What do you think the number is?
3. Find
4. Fill in the blanks: As
We define
Example A
Graph
Solution: As you would expect, the graph of
The asymptote is
Example B
Simplify
Solution: The bases are the same, so you can just add the exponents. The answer is
Example C
Gianna opens a savings account with $1000 and it accrues interest continuously at a rate of 5%. What is the balance in the account after 6 years?
Solution: In the previous concept, the word problems dealt with interest that compounded monthly, quarterly, annually, etc. In this example, the interest compounds continuously. The equation changes slightly, from
Intro Problem Revisit Plug the given values into the equation
Therefore, at the end of 4 years, you will have earned $105.20 in interest.
Guided Practice
1. Determine if the following functions are exponential growth, decay, or neither.
a)
b)
c)
d)
2. Simplify the following expressions with
a)
b)
3. The rate of radioactive decay of radium is modeled by
Answers
1. Recall to be exponential growth, the base must be greater than one. To be exponential decay, the base must be between zero and one.
a) Exponential growth; \begin{align*}e>1\end{align*}
b) Neither; \begin{align*}a<0\end{align*}
c) Exponential decay; \begin{align*}e^{x}= \left(\frac{1}{e}\right)^x\end{align*} and \begin{align*}0< \frac{1}{e}<1\end{align*}
d) Exponential growth; \begin{align*}\left(\frac{1}{e}\right)^{x}=e^x\end{align*}
2. a) \begin{align*}2e^{3} \cdot e^2 = 2e^{1}\end{align*} or \begin{align*}\frac{2}{e}\end{align*}
b) \begin{align*}\frac{4e^6}{16e^2} = \frac{e^4}{4}\end{align*}
3. Use the formula given in the problem and solve for what you don’t know.
\begin{align*}R &= Pe^{0.00043t} \\ 698.9 &= Pe^{0.00043(5000)} \\ 698.9 &= P(0.11648) \\ 6000 &= P\end{align*}
There was about 6000 grams of radium to start with.
Vocabulary
 Natural Number (Euler Number)
 The number \begin{align*}e\end{align*}, such that as \begin{align*}n \rightarrow \infty, \left(1+ \frac{1}{n}\right)^n \rightarrow e\end{align*}. \begin{align*}e \approx 2.71828\end{align*}.
Practice
Determine if the following functions are exponential growth, decay or neither. Give a reason for your answer.
 \begin{align*}y=\frac{4}{3} e^x\end{align*}
 \begin{align*}y=e^{x}+3\end{align*}
 \begin{align*}y= \left(\frac{1}{e}\right)^x +2\end{align*}
 \begin{align*}y= \left(\frac{3}{e}\right)^{x} 5\end{align*}
Simplify the following expressions with \begin{align*}e\end{align*}.
 \begin{align*}e^{3} \cdot e^{12}\end{align*}
 \begin{align*}\frac{5e^{4}}{e^3}\end{align*}
 \begin{align*}6e^5e^{4}\end{align*}
 \begin{align*}\left(\frac{4e^4}{3e^{2}e^3}\right)^{2}\end{align*}
Solve the following word problems.
The population of Springfield is growing exponentially. The growth can be modeled by the function \begin{align*}P=Ie^{0.055t}\end{align*}, where \begin{align*}P\end{align*} represents the projected population, \begin{align*}I\end{align*} represents the current population of 100,000 in 2012 and \begin{align*}t\end{align*} represents the number of years after 2012.
 To the nearest person, what will the population be in 2022?
 In what year will the population double in size if this growth rate continues?
The value of Steve’s car decreases in value according to the exponential decay function: \begin{align*}V=Pe^{0.12t}\end{align*}, where \begin{align*}V\end{align*} is the current value of the vehicle, \begin{align*}t\end{align*} is the number of years Steve has owned the car and \begin{align*}P\end{align*} is the purchase price of the car, $25,000.
 To the nearest dollar, what will the value of Steve’s car be in 2 years?
 To the nearest dollar, what will the value be in 10 years?
Naya invests $7500 in an account which accrues interest continuously at a rate of 4.5%.
 Write an exponential growth function to model the value of her investment after \begin{align*}t\end{align*} years.
 How much interest does Naya earn in the first six months to the nearest dollar?
 How much money, to the nearest dollar, is in the account after 8 years?
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Here you'll use the natural number, @$\begin{align*}e\end{align*}@$, in exponential functions and reallife situations.