<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# The Number e

## Practice exponential functions including the Euler number (e)

Estimated7 minsto complete
%
Progress
Practice The Number e
Progress
Estimated7 minsto complete
%
The Number e

Previously, the word problems you have dealt with involve interest that compounded monthly, quarterly, annually, etc. In this example, the interest compounds continuously. The equation changes slightly, from \begin{align*}A=P \left(1+ \frac{r}{n}\right)^{nt}\end{align*} to \begin{align*}A=Pe^{rt}\end{align*}, without \begin{align*}n\end{align*}, because there is no longer any interval. Therefore, the equation for this problem is \begin{align*}A=1000e^{0.05(6)}\end{align*} and the account will have 1349.86 in it. Compare this to daily accrued interest, which would be \begin{align*}A=1000 \left(1+ \frac{0.05}{365}\right)^{365(6)}=1349.83\end{align*}. ### Examples #### Example 1 Earlier, you were asked to find how much money you will earn after 4 years. Plug the given values into the equation \begin{align*}I=Pe^{rt}\end{align*} and solve for I. \begin{align*}I=Pe^{rt} - P\\ I = 1000\cdot e^{0.025\cdot 4} - 1000\\ I = 1000 \cdot e^{0.1} - 1000\\ I = 1000 \cdot 1.1052 - 1000\\ I = 1105.20 - 1000 = 105.20\end{align*} Therefore, at the end of 4 years, you will have earned105.20 in interest.

Determine if Examples 2-5 are exponential growth, decay, or neither.

Recall to be exponential growth, the base must be greater than one. To be exponential decay, the base must be between zero and one.

#### Example 2

\begin{align*}y=\frac{1}{2}e^x\end{align*}

Exponential growth; \begin{align*}e>1\end{align*}

#### Example 3

\begin{align*}y=-4e^x\end{align*}

Neither; \begin{align*}a<0\end{align*}

#### Example 4

\begin{align*}y=e^{-x}\end{align*}

Exponential decay; \begin{align*}e^{-x}= \left(\frac{1}{e}\right)^x\end{align*} and \begin{align*}0< \frac{1}{e}<1\end{align*}

#### Example 5

\begin{align*}y=2 \left(\frac{1}{e}\right)^{-x}\end{align*}

Exponential growth; \begin{align*}\left(\frac{1}{e}\right)^{-x}=e^x\end{align*}

#### Example 6

Simplify the following expression with \begin{align*}e\end{align*}.

\begin{align*}2e^{-3} \cdot e^2\end{align*}

\begin{align*}2e^{-3} \cdot e^2 = 2e^{-1}\end{align*} or \begin{align*}\frac{2}{e}\end{align*}

#### Example 7

Simplify the following expression with \begin{align*}e\end{align*}.

\begin{align*}\frac{4e^6}{16e^2}\end{align*}

\begin{align*}\frac{4e^6}{16e^2} = \frac{e^4}{4}\end{align*}

#### Example 8

The rate of radioactive decay of radium is modeled by \begin{align*}R=Pe^{-0.00043t}\end{align*}, where \begin{align*}R\end{align*} is the amount (in grams) of radium present after \begin{align*}t\end{align*} years and \begin{align*}P\end{align*} is the initial amount (also in grams). If there is 698.9 grams of radium present after 5,000 years, what was the initial amount?

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*}

### Review

Determine if the following functions are exponential growth, decay or neither. Give a reason for your answer.

1. \begin{align*}y=\frac{4}{3} e^x\end{align*}
2. \begin{align*}y=-e^{-x}+3\end{align*}
3. \begin{align*}y= \left(\frac{1}{e}\right)^x +2\end{align*}
4. \begin{align*}y= \left(\frac{3}{e}\right)^{-x} -5\end{align*}

Simplify the following expressions with \begin{align*}e\end{align*}.

1. \begin{align*}e^{-3} \cdot e^{12}\end{align*}
2. \begin{align*}\frac{5e^{-4}}{e^3}\end{align*}
3. \begin{align*}6e^5e^{-4}\end{align*}
4. \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.

1. To the nearest person, what will the population be in 2022?
2. 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. 1. To the nearest dollar, what will the value of Steve’s car be in 2 years? 2. 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%.

1. Write an exponential growth function to model the value of her investment after \begin{align*}t\end{align*} years.
2. How much interest does Naya earn in the first six months to the nearest dollar?
3. How much money, to the nearest dollar, is in the account after 8 years?

To see the Review answers, open this PDF file and look for section 8.4.

### Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes

### Vocabulary Language: English

$e$

$e$ is an irrational number that is approximately equal to 2.71828. As $n \rightarrow \infty, \left(1+ \frac{1}{n}\right)^n \rightarrow e$.

e

$e$ is an irrational number that is approximately equal to 2.71828. As $n \rightarrow \infty, \left(1+ \frac{1}{n}\right)^n \rightarrow e$.

Euler number

The Euler number is the irrational number $e$, such that as $n \rightarrow \infty, \left(1+ \frac{1}{n}\right)^n \rightarrow e$. $e \approx 2.71828$.