5.3 The Natural Base

We have looked at logarithms with various bases, and in particular we studied the common or base 10 logarithms, which often appear in applications. There is another base for logarithms and exponential functions that is often used in applications. This base is an irrational number called $e$ , where

$e approximately equals 2.71828182845$

The number $e$ is essential for many advanced topics, and it is often called the natural base. It may seem strange to use an irrational number as the base for exponential functions, but just as the irrational number $π$ arises naturally in geometry, so does $e$ arise in calculus and its applications. At the end of this section we'll look at a specific case of how base $e$ occurs, and how its use is connected to the ideas and techniques of calculus.

The Natural Exponential Function

The natural exponential function is the function

$f (x) = e^{x}$

Values for $e^{x}$ can be obtained with a calculator using the $e^{x}$ key ( 2nd LN on most calculators). For example, you can evaluate $e to the power 1$ by pressing

2nd LN $1$

to confirm the value of $e$ given above. (We'll explain why we use the 2nd LN key a little later.)

Try a few more calculations to become familiar with base $e$ .

Use your calculator to evaluate the following powers.

$e^{2} \approx$ _____
$e^{3.5} \approx$ _____
$e^{- 0.5} \approx$ _____

$e squared approximately equals 7.389$
$e^{3.5} \approx 33.115$
$e^{- 0.5} \approx 0.6065$

Use your calculator to evaluate the following powers. Round to four decimal places.

$e^{2}$
$e to the power 3.5$
$e^{- 0.5}$

$e squared approximately equals 7.3891$
$e^{3.5} \approx 33.1155$
$e^{- 0.5} \approx 0.6065$

What about the graph of this new exponential function? Because $e$ is a number between $2$ and $3$ , the graph of $f (x) = e^{x}$ lies between the graphs of $y = 2^{x}$ and $y = 3^{x}$ . Compare the tables of values and the graphs of the three functions below. For example, note that for $x equals 2$ , the value of $e squared equals 7.389$ is between $2 squared equals 4$ and $3 squared equals 9$ . You can verify the table and graphs on your calculator.

$x$	$y = 2^{x}$	$y = e^{x}$	$y = 3^{x}$
$minus 3$	$0.125$	$0.050$	$0.037$
$minus 2$	$0.250$	$0.135$	$0.111$
$minus 1$	$0.500$	$0.368$	$0.333$
$0$	$1$	$1$	$1$
$1$	$2$	$2.718$	$3$
$2$	$4$	$7.389$	$9$
$3$	$8$	$20.086$	$27$

The value of $e^{2}$ is closest to

_____

The value of e 2 is closest to which of these?
1. $3$
2. $5$
3. $7$
4. $9$
The value of e − 1 is closest to which of these?
1. $minus 3$
2. $\frac{1}{2}$
3. $\frac{1}{3}$
4. $2$

Variations on the natural exponential function occur in many disciplines. For example, the graph in the figure below is called a "bell curve." It is the graph of the normal distribution in statistics.

Example 1

The normal distribution above shows the heights of men in the US. The average or mean height for men is 70 inches, and the formula for the graph is

$f (x) = 0.13 e^{- (x - 70)^{2} / 18}$

The probability that a given man has a height between two values is the area under the curve between those values. Evaluate the function for $x equals 70$ and for $x equals 67$ .

To evaluate the function we follow the order of operations. For $x equals 70$ we have

$\begin{aligned} f (70) & = 0.13 e^{- (70 - 70)^{2} / 18} & Simplify the exponent first. \\ = 0.13 e^{0} = 0.13 \end{aligned}$

For $x equals 67$ we have

$\begin{aligned} f (67) & = 0.13 e^{- (67 - 70)^{2} / 18} & - (67 - 70)^{2} / 18 = - 0.5 \\ = 0.13 e^{- 0.5} = 0.079 \end{aligned}$

This next graph is an example of a logistic function, which models population growth with an upper bound.

The logistic function shown above models the spread of Covid in China during the 2020 epidemic. It gives the number of infections $N (t)$ , in thousands, reported $t$ days after January 21, 2020. The equation for this model is

$N (t) = \frac{83.5}{1 + 267 e^{- 0.214 t}}$

According to the model, how many cases of Covid were reported on February 20 (day $t equals 30$ )?

58,194 cases

The Natural Logarithmic Function

Recall that each exponential function with base $b$ has an inverse function, the logarithmic function with the same base. For example, the function $y equals log sub 2 ⁡ open parenthesis x close parenthesis$ is the inverse of the function $y = 2^{x}$ . It gives the exponent needed on 2 to give $x$ , so that, for instance, $log sub 2 ⁡ open parenthesis 16 close parenthesis equals 4$ , because $2^{4} = 16$ .

The base $e$ logarithm of a number $x$ , or $\log_{e} (x)$ , is called the natural logarithm of $x$ and is denoted by $\ln (x)$ or $\ln x$ . (Why “ln” and not “nl”? The natural logarithm is denoted by “ln” because it stands for “logarithmus naturalis,” which is the Latin for “natural logarithm.”) Here is its official definition.

The natural logarithm of $x$ is the exponent to which $e$ must be raised to produce $x$ . For example, the natural logarithm of $10$ , or $ln ⁡ open parenthesis 10 close parenthesis$ , is the solution of the equation

$e^{y} = 10$

You can verify on your calculator that

$e^{2.3} \approx 10 or \ln (10) \approx 2.3$

As is the case with exponential and log functions with other bases, the natural log function, $y = \ln (x)$ , and the natural exponential function, $f (x) = e^{x}$ , “undo” each other, so they are inverse functions. (This is why many calculators use 2nd LN to indicate $e^{x}$ .)

Use your calculator to evaluate each logarithm. Round your answers to four decimal places.

$ln ⁡ 100 approximately equals$ _____
$ln ⁡ 0.01 approximately equals$ _____
$\ln e^{3} \approx$ _____

$ln ⁡ 100 approximately equals 4.6052$
$\ln 0.01 \approx - 4.6052$
$ln ⁡ e cubed equals 3$

Use your calculator to evaluate each logarithm. Round your answers to four decimal places.

$ln ⁡ open parenthesis 10 close parenthesis$
$ln ⁡ open parenthesis 0.1 close parenthesis$
$ln ⁡ open parenthesis 100 close parenthesis$
$ln ⁡ open parenthesis 0.01 close parenthesis$

$ln ⁡ open parenthesis 10 close parenthesis approximately equals 2.3026$
$\ln (0.1) \approx - 2.3026$
$ln ⁡ open parenthesis 100 close parenthesis approximately equals 4.6052$
$\ln (0.01) \approx - 4.6052$

As usual, we can gain a better understanding of a new function by looking at its graph.

Example 3

Graph $f (x) = e^{x}$ and $g (x) = \ln (x)$ on the same grid.
Give the domain and range of the natural log function.

Earlier we made a table of values to graph the fuction $f (x) = e^{x}$ , and we can make a table of values for $g (x) = \ln (x)$ by interchanging the columns in that table. Plotting the points gives us the graph below.

$x$	$y = \ln (x)$
$0.050$	$minus 3$
$0.135$	$minus 2$
$0.368$	$minus 1$
$1$	$0$
$2.718$	$1$
$7.389$	$2$
$20.086$	$3$

You can see that the graph of $g (x) = \ln (x)$ is the reflection of the graph of $f (x) = e^{x}$ about the line $y = x$ . In particular, notice that while the graph of $f (x) = e^{x}$ has no points with negative $y$ -values, the graph of $g (x) = \ln (x)$ has no points with negative $x$ -values. Thus, the domain of the natural log function is the same as the range of $y = e^{x}$ , or all positive numbers. The range of $y = \ln (x)$ is the same as the domain of $y = e^{x}$ , or all real numbers.

From the graph of $g (x) = \ln (x)$ you can make the following observations.

The natural log function has only positive numbers as input values.
The natural logs of negative numbers and zero are undefined.
The natural log of a number greater than 1 is positive, while the logs of numbers between 0 and 1 are negative.

Properties of the Natural Logarithm

Natural logs obey the same conversion formulas that work for logs to other bases.

The conversion formulas are just another way of saying the the the natural log function, $g (x) = \ln (x)$ , and the natural exponential function, $f (x) = e^{x}$ , are inverse functions.

In particular,

$\begin{aligned} \ln (e) & = 1 because e^{1} = e \\ \ln (1) & = 0 because e^{0} = 1 \end{aligned}$

Which of the following is equivalent to $e^{x} = k$ ?

_____

$\ln k = x$

Which of the following is equivalent to $e^{x} = k$ ?

$e^{k} = x$
$\ln (e) = x$
$\ln (k) = x$
$k^{x} = e$

We use natural logarithms in the same way that we use logs to other bases. The properties of logarithms that we studied in Properties of Logarithms also apply to logarithms base $e$ .

Because the functions $y = e^{x}$ and $y = \ln x$ are inverse functions, the following properties are also true.

Example 4

Simplify each expression.

$ln ⁡ open parenthesis e to the power 0.3 x close parenthesis$
$e^{2 \ln (x + 3)}$

The natural log is the log base $e$ , and hence the inverse of $e^{x}$ . Therefore,
$\ln (e^{0.3 x}) = 0.3 x$
First, we simplify the exponent using the third property of logs to get
$2 \ln (x + 3) = \ln ((x + 3)^{2})$
Then $e^{2 \ln (x + 3)} = e^{\ln ((x + 3)^{2})} = (x + 3)^{2}$ .

Simplify each expression. Use "sqrt(x)" to get $\sqrt{x}$ .

$e^{(\ln x) / 2} =$ _____
$\ln (\frac{1}{e^{4 x}}) =$ _____

$\sqrt{x}$
$minus 4 x$

Simplify each expression.

$e to the power open parenthesis ln ⁡ open parenthesis x close parenthesis / 2 close parenthesis$
$\ln (\frac{1}{e^{4 x}})$

$\sqrt{x}$ or $x to the power 1 / 2$
$minus 4 x$

Explain why $\ln (\frac{1}{e^{3}}) = - 3$ .

_____

Explain why $\ln (\frac{1}{e^{3}}) = - 3$ .

Solving Equations

We use the natural logarithm to solve exponential equations with base $e$ . The techniques we've learned for solving other exponential equations also apply to equations with base $e$ .

Solve each equation. Round your answers to four decimal places.

$\ln x = - 0.2$
$x \approx$ _____
$e^{x} = 8$
$x \approx$ _____

$0.8187$
$2.0794$

Solve each equation. Round your answers to four decimal places.

$\ln (x) = - 0.2$
$e^{x} = 8$

$0.8187$
$2.0794$

Why is the equation $e^{x} = 6.5$ easier to solve than $8^{x} = 6.5$ ?

_____

There is a button for log base on the calculator, but not a button for log base 8.

Which statement below explains why the equation $e^{x} = 6.5$ is easier to solve than $8^{x} = 6.5$ ?

8 is larger than 6.5.
$k$ is a constant.
There is a button for log base $e$ on the calculator, but not a button for log base 8.
Because $e$ is an irrational number.

To solve more complicated exponential equations, we isolate the power on one side of the equation before converting to logarithmic form.

Example 6

Solve $200 = 60 + 20 e^{0.4 x}$

First, we isolate the power.

$\begin{aligned} 200 & = 60 + 20 e^{0.4 x} & Subtract 60 from both sides. \\ 140 & = 20 e^{0.4 x} & Divide both sides by 20. \\ 7 & = e^{0.4 x} \end{aligned}$

Then we convert the equation to logarithmic form.

$\begin{aligned} 0.4 x & = \ln (7) & Divide both sides by 0.4. \\ x & = \frac{\ln (7)}{0.4} \end{aligned}$

Rounded to four decimal places, $x approximately equals 4.8648$ .

Note

We can also solve the equation in Example,

$7 = e^{0.4 x}$

by taking the natural logarithm of both sides. This gives us

$\begin{aligned} \ln (7) & = \ln e^{0.4 x} & Simplify the right side. \\ \ln (7) & = 0.4 x \end{aligned}$

because $\ln (e^{a}) = a$ for any number $a$ . We then proceed with the solution as before.

Solve

$80 - 16 e^{- 0.2 x} = 70.3$

$x =$ _____

$x = - 5 \ln (\frac{9.7}{16}) \approx 2.5023$

Solve

$80 - 16 e^{- 0.2 x} = 70.3$

Isolate the power, take the natural log of both sides, and solve as usual to find $x = - 5 \ln (\frac{9.7}{16}) \approx 2.5023$

Example 7

The logistic function

$P = \frac{20}{1 + 5 e^{- 0.4 t}}$

has its upper bound at $P equals 20$ . Find the value of $t$ for which $P equals 10$ .

We must solve the equation $10 = \frac{20}{1 + 5 e^{- 0.4 t}}$ for $t$ . To begin, we multiply both sides of the equation by the denominator, $1 + 5 e^{- 0.4 t}$ , to get

$1 + 5 e^{- 0.4 t} = \frac{20}{10} = 2$

Then we isolate the power, $e^{- k t}$ , as follows:

$\begin{aligned} 1 + 5 e^{- 0.4 t} & = 2 & Subtract 1 from both sides, then divide by 5. \\ e^{- 0.4 k t} & = 0.2 \end{aligned}$

Next, we take the natural logarithm of both sides to get

$\ln (e^{- 0.4 t}) = \ln (0.2)$

and recall that $\ln (e^{x}) = x$ to simplify the left side.

$- 0.4 t = \ln (0.2)$

Finally, we divide both sides by $minus 0.4$ to solve for $t$ .

$t = \frac{- \ln (0.2)}{0.4} \approx 4.02$

Solve $N = A e^{- k t}$ for $k$ .

$k =$ _____

$Divide both sides by A .$

$Take the natural log of both sides.$

$Divide both sides by - t .$

$k = \frac{- \ln (N / A)}{t}$

Solve $N = A e^{- k t}$ for $k$ .

$k = \frac{- \ln (N / A)}{t}$

Delbert says that he will begin solving the equation $100 e^{0.6 t} = 40$ by computing $e to the power 0.6 t$ . Is this a good strategy? Why or why not?

_____

Delbert says that he will begin solving the equation $100 e^{0.6 t} = 40$ by computing $e to the power 0.6 t$ . Is this a good strategy? Why or why not?

Exponential Growth and Decay

In Exponential Growth and Decay, we considered functions of the form

$P (t) = P_{0} \cdot b^{t}$

which describe exponential growth when $b greater than 1$ and exponential decay when $0 less than b less than 1$ . Exponential growth and decay can also be modeled by functions of the form

$P (t) = P_{0} \cdot e^{k t}$

where we have substituted $e^{k}$ for the growth factor $b$ , so that

$\begin{aligned} P (t) & = P_{0} \cdot b^{t} \\ = P_{0} \cdot {(e^{k})}^{t} = P_{0} \cdot e^{k t} \end{aligned}$

We can find the value of $k$ by solving the equation $b = e^{k}$ for $k$ , to get $k = \ln b$ .

For instance, consider a colony of bacteria grows according to the formula

$P (t) = 100 \cdot 3^{t}$

We can express this function in the form $P (t) = 100 \cdot e^{k t}$ if we set

$3 = e^{k} or k = \ln (3) \approx 1.0986$

Thus, the growth law for the colony of bacteria can be written

$P (t) \approx 100 \cdot e^{1.0986 t}$

By graphing both functions on your calculator, you can verify that

$\begin{matrix} P (t) = 100 \cdot 3^{t} \\ and \\ P (t) = 100 \cdot e^{1.0986 t} \end{matrix}$

are just two ways of writing the same function.

Sometimes exponential growth is given as a percentage, so for example we might say “prices rose by 5% annually.” In this case the growth factor is $b = 1 + r$ , where $r$ is the percentage rate in decimal form. For a percent decrease, $b = 1 - r$ .

Example 8

From 1990 to 2000, the population of Clark County, Nevada, grew by $6.4 %$ per year.

What was the growth factor for the population of Clark County from 1990 to 2000? If the population of Clark County was $768, 000$ in 1990, write a formula for the population $t$ years later.
Write a growth formula for Clark County using base $e$ .

The growth factor was $b = 1 + r = 1.064$ . The population $t$ years later was
$P (t) = 768, 000 (1.064)^{t}$
We use the formula $P (t) = P_{0} \cdot e^{k t}$ , where $e^{k} = 1.064$ . Solving for $k$ , we find
$k = \ln (1.064) = 0.062$
so $P (t) = 768, 000 e^{0.062 t}$ .

From 1994 to 1998, the number of personal computers connected to the Internet grew according to the formula $N (t) = 2.8 e^{0.85 t}$ , where $t equals 0$ in 1994 and $N$ is in millions. (Source: Los Angeles Times, September 6, 1999)

Evaluate $N open parenthesis 1 close parenthesis equals$ _____. By what percent did the number of Internet users grow in one year?
About _____%
Express the growth law in the form $N (t) = N_{0} (1 + r)^{t}$ .
$N (t) =$ _____

$e^{k} = 1 + r$

$N open parenthesis 1 close parenthesis approximately equals 6.55$ , $134 %$
$N (t) \approx 2.8 (2.3396)^{t}$

From 1994 to 1998, the number of personal computers connected to the Internet grew according to the formula

$N (t) = 2.8 e^{0.85 t}$

where $t equals 0$ in 1994 and $N$ is in millions. (Source: Los Angeles Times, September 6, 1999)

Evaluate $N open parenthesis 1 close parenthesis$ . By what percent did the number of Internet users grow in one year?
Express the growth law in the form $N (t) = N_{0} (1 + r)^{t}$ . (Hint: $e^{k} = 1 + r$ .)

$N open parenthesis 1 close parenthesis approximately equals 6.55$ , $134 %$
$N (t) \approx 2.8 (2.3396)^{t}$

Now, what about exponential decay, where the the decay factor $b$ is a number less than 1? If $k$ is negative, then $e^{k}$ is a number less than $1$ . For example, if $k = - 2$ ,

$e^{- 2} = \frac{1}{e^{2}} \approx \frac{1}{7.3891} \approx 0.1353$

Thus, for negative values of $k$ , the function $P (t) = P_{0} e^{k t}$ describes exponential decay.

The natural log of a fraction between 0 and 1 is

_____

negative.

The natural log of a number between 0 and 1 is

positive.
negative.
undefined.
between $e to the power 0$ and $e to the power 1$ .

Example 9

Express the decay law $N (t) = 60 (0.8)^{t}$ in the form $N (t) = N_{0} e^{k t}$ .

For this decay law, $N sub 0 equals 60$ and $b equals 0.8$ . We would like to find a value for $k$ so that $e^{k} = b = 0.8$ , that is, we must solve the equation

$\begin{aligned} e^{k} & = 0.8 & Take natural log of both sides. \\ \ln e^{k} & = \ln (0.8) & Simplify. \\ k & = \ln (0.8) \approx - 0.2231 \end{aligned}$

Replacing $b$ with $e^{k}$ , we find that the decay law is

$N (t) \approx 60 e^{- 0.2231 t}$

A scientist isolates $25$ grams of krypton-91, which decays according to the formula

$N (t) = 25 e^{- 0.07 t}$

, where $t$ is in seconds.

Complete the table of values showing the amount of krypton-91 left at $10$ -second intervals over the first minute.
$t$ $0$ $10$ $20$ $30$
$N (t)$ _____ _____ _____ _____
$t$ $40$ $50$ $60$
$N (t)$ _____ _____ _____
Use the table to choose a suitable window and graph the function $N (t)$ .
Write and solve an equation to answer the question: How long does it take for 60% of the krypton-91 to decay?
_____
Answer: $t =$ _____ seconds

$t$	$0$	$10$	$20$	$30$
$N (t)$	_____	_____	_____	_____

$t$	$40$	$50$	$60$
$N (t)$	_____	_____	_____

If $60 %$ of the krypton-91 has decayed, $40 %$ of the original $25$ grams remains.

$t$ $0$ $10$ $20$ $30$ $40$ $50$ $60$
$N (t)$ $25$ $12.41$ $6.16$ $3.06$ $1.52$ $0.75$ $0.37$
A graph is below.
$25 e^{- 0.07 t} = 0.40 (25)$ ; $t = \frac{\ln (0.4)}{- 0.07} \approx 13.09$ seconds

$t$	$0$	$10$	$20$	$30$	$40$	$50$	$60$
$N (t)$	$25$	$12.41$	$6.16$	$3.06$	$1.52$	$0.75$	$0.37$

Graph for part (b):

A scientist isolates $25$ grams of krypton-91, which decays according to the formula

$N (t) = 25 e^{- 0.07 t}$

, where $t$ is in seconds.

Complete the table of values showing the amount of krypton-91 left at

10

-second intervals over the first minute.

$t$	$0$	$10$	$20$	$30$	$40$	$50$	$60$
$N (t)$

Use the table to choose a suitable window and graph the function $N (t)$ .
Write and solve an equation to answer the question: How long does it take for 60% of the krypton-91 to decay?
Hint: If $60 %$ of the krypton-91 has decayed, $40 %$ of the original $25$ grams remains.

$t$ $0$ $10$ $20$ $30$ $40$ $50$ $60$
$N (t)$ $25$ $12.41$ $6.16$ $3.06$ $1.52$ $0.75$ $0.37$
$25 e^{- 0.07 t} = 0.40 (25)$ ; $t = \frac{\ln (0.4)}{- 0.07} \approx 13.09$ seconds

$t$	$0$	$10$	$20$	$30$	$40$	$50$	$60$
$N (t)$	$25$	$12.41$	$6.16$	$3.06$	$1.52$	$0.75$	$0.37$

Explain how to rewrite $P (t) = P_{0} b^{t}$ with the natural base.

_____

Explain how to rewrite $P (t) = P_{0} b^{t}$ with the natural base.

Continuous Compounding

In Section 2.1 we looked at a formula for savings accounts on which the interest is compounded $n$ times per year, and we saw that the amount $A (t)$ on such an account increased when $n$ increased. But there is a limit or upper bound to the amount, no matter how large the value of $n$ . At this upper bound we say that the interest is compounded continuously, and the amount is given by the function

$A (t) = P e^{r t}$

where $P$ is the principal invested and $r$ is the interest rate.

Example 10

Suppose you invest $ $500$ in an account that pays $8 %$ interest compounded continuously. You leave the money in the account without making any additional deposits or withdrawals.

Write a formula that gives the value of your account $A (t)$ after $t$ years.
Make a table of values showing $A (t)$ for the first $5$ years.
Graph the function $A (t)$ .
How much will the account be worth after $10$ years?
How long will it be before the account is worth $ $1000$ ?

We substitute $500$ for $P$ , and $0.08$ for $r$ to find
$A (t) = 500 e^{0.08 t}$
We evaluate the formula for $A (t)$ to obtain a table.
$t$ $A (t)$
$0$ $500$
$1$ $541.64$
$2$ $586.76$
$3$ $635.62$
$4$ $688.56$
$5$ $745.91$
The graph of $A (t)$ is shown above.
We evaluate $A (t)$ for $t equals 10$ .
$\begin{aligned} A (10) & = 500 e^{0.08 (10)} \\ = 500 e^{0.8} \\ \approx 500 (2.2255) = 1112.77 \end{aligned}$
The account will be worth $ $1112.77$ after $10$ years.
We substitute $1000$ for $A (t)$ and solve the equation.
$\begin{aligned} 1000 & = 500 e^{0.08 t} & Divide both sides by 500. \\ 2 & = e^{0.08 t} & Take natural log of both sides. \\ \ln (2) & = \ln (e^{0.08 t}) = 0.08 t & Divide both sides by 0.08. \\ t & = \frac{\ln (2)}{0.08} \approx 8.6643 \end{aligned}$
The account will be worth $ $1000$ after approximately $8.7$ years.

Zelda invested $1000 in an account that pays 4.5% interest compounded continuously. How long will it be before the account is worth $2000?

Answer: about _____ years.

About 15.4 years

Zelda invested $1000 in an account that pays 4.5% interest compounded continuously. How long will it be before the account is worth $2000?

About 15.4 years

Explain why solving exponential equations in base $e$ is no harder than solving exponential equations in base 10.

_____

Explain why solving exponential equations in base $e$ is no harder than solving exponential equations in base 10.

Section Summary

Vocabulary

Look up the definitions of new terms in the Glossary.

Natural exponential function
Natural logarithm
Continuous compounding

CONCEPTS

The natural base is an irrational number called $e$ , where
$e approximately equals 2.71828182845$
The natural exponential function is the function $f (x) = e^{x}$ . The natural log function is the function $g (x) = \ln x = \log_{e} x$ .
Note
Conversion Formulas for Natural Logs
$y = \ln (x) if and only if e^{y} = x$
Note
Properties of Natural Logarithms
If $x, y greater than 0$ , then
1. $\ln (x y) = \ln (x) + \ln (y)$
2. $\ln (\frac{x}{y}) = \ln (x) - \ln (y)$
3. $\ln (x^{k}) = k \ln (x)$
We use the natural logarithm to solve exponential equations with base $e$ .
Note
Exponential Growth and Decay
The function
$P (t) = P_{0} e^{k t}$
describes exponential growth if $k greater than 0$ , and exponential decay if $k less than 0$ .
Continuous compounding: The amount accumulated in an account after $t$ years at interest rate $r$ compounded continuously is given by
$A (t) = P e^{r t}$
where $P$ is the principal invested.

STUDY QUESTIONS

State the value of $e$ to $3$ decimal places. Memorize this value.
Explain why $\ln e^{x} = x$ .
State the formula for exponential growth using base $e$ .
How is the formula for exponential decay in base $e$ different from the formula for exponential growth?

SKILLS

Practice each skill in the Homework problems listed.

Graph exponential functions base $e$ : #1–4
Simplify expressions: #5 and 6
Solve exponential and log equations base $e$ : #7–10, 23–30
Use the properties of logs and exponents with the natural base: #19–22, 37–40
Use the natural exponential function in applications: #11–14, 47–58
Convert between $P (t) = P_{0} (1 + r)^{t}$ and $P (t) = P_{0} e^{k t}$ : #15–18, 41–46

Homework 5.3

For Problems 1–4, use your calculator to complete the table for each function. Then choose a suitable window and graph the function.

$x$	$minus 10$	$minus 5$	$0$	$5$	$10$	$15$	$20$
$f (x)$

$f (x) = e^{0.2 x}$

$x$	$minus 10$	$minus 5$	$0$	$5$	$10$	$15$	$20$
$f (x)$	$0.135$	$0.368$	$1$	$2.718$	$7.389$	$20.086$	$54.598$

$f (x) = e^{0.6 x}$

$f (x) = e^{- 0.3 x}$

$x$	$minus 10$	$minus 5$	$0$	$5$	$10$	$15$	$20$
$f (x)$	$20.086$	$4.482$	$1$	$0.223$	$0.05$	$0.011$	$0.00248$

$f (x) = e^{- 0.1 x}$

For Problems 5–6, simplify.

$\ln (e^{2})$
$e to the power ln ⁡ open parenthesis 5 t close parenthesis$
$e^{- \ln (x)}$
$\ln (\sqrt{e})$

$2$
$5 t$
$\frac{1}{x}$
$\frac{1}{2}$

$\ln (e^{x^{4}})$
$e to the power 3 ln ⁡ open parenthesis x close parenthesis$
$e^{\ln (x) - \ln (y)}$
$\ln (\frac{1}{e^{2 t}})$

For Problems 7–10, solve for $x$ . Round your answers to two decimal places.

$e^{x} = 1.9$
$e^{x} = 45$
$e^{x} = 0.3$

$0.64$
$3.81$
$minus 1.20$

$e^{x} = 2.1$
$e^{x} = 60$
$e^{x} = 0.9$

$ln ⁡ open parenthesis x close parenthesis equals 1.42$
$ln ⁡ open parenthesis x close parenthesis equals 0.63$
$\ln (x) = - 2.6$

$4.14$
$1.88$
$0.07$

$ln ⁡ open parenthesis x close parenthesis equals 2.03$
$ln ⁡ open parenthesis x close parenthesis equals 0.59$
$\ln (x) = - 3.4$

The number of bacteria in a culture grows according to the function

$N (t) = N_{0} e^{0.04 t}$

where $N sub 0$ is the number of bacteria present at time $t equals 0$ and $t$ is the time in hours.

Write a growth law for a sample in which $6000$ bacteria were present initially.
Make a table of values for $N (t)$ in $5$ -hour intervals over the first $30$ hours.
Graph $N (t)$ .
How many bacteria were present at $t equals 24$ hours?
How much time must elapse (to the nearest tenth of an hour) for the original $6000$ bacteria to increase to $100, 000$ ?

$N (t) = 6000 e^{0.04 t}$
$t$ $0$ $5$ $10$ $15$ $20$ $25$ $30$
$N (t)$ $6000$ $7328$ $8951$ $10, 933$ $13, 353$ $16, 310$ $19, 921$
$15, 670$
$70.3$ hrs

$t$	$0$	$5$	$10$	$15$	$20$	$25$	$30$
$N (t)$	$6000$	$7328$	$8951$	$10, 933$	$13, 353$	$16, 310$	$19, 921$

Hope invests $$ 2000$ in a savings account that pays $5 \frac{1}{2} %$ annual interest compounded continuously.

Write a formula that gives the amount of money $A (t)$ in Hope’s account after $t$ years.
Make a table of values for $A (t)$ in $2$ -year intervals over the first $10$ years.
Graph $A (t)$ .
How much will Hope's account be worth after $7$ years?
How long will it take for the account to grow to $$ 5000$ ?

The intensity, $I$ (in lumens), of a light beam after passing through $t$ centimeters of a filter having an absorption coefficient of $0.1$ is given by the function

$I (t) = 1000 e^{- 0.1 t}$

Graph $I (t)$ .
What is the intensity (to the nearest tenth of a lumen) of a light beam that has passed through $0.6$ centimeter of the filter?
How many centimeters (to the nearest tenth) of the filter will reduce the illumination to $800$ lumens?

$941.8$ lumens
$2.2$ cm

X-rays can be absorbed by a lead plate so that

$I (t) = I_{0} e^{- 1.88 t}$

where $I sub 0$ is the X-ray count at the source and $I (t)$ is the X-ray count behind a lead plate of thickness $t$ inches.

Graph $I (t)$ .
What percent of an X-ray beam will penetrate a lead plate $\frac{1}{2}$ inch thick?
How thick should the lead plate be in order to screen out $70 %$ of the X-rays?

For problems 15–18, express each exponential function in the form $P (t) = P_{0} b^{t}$ . Is the function increasing or decreasing? What is its initial value?

$P (t) = 20 e^{0.4 t}$

$P (t) = 20 {(e^{0.4})}^{t} \approx 20 \cdot {1.492}^{t}$ ; increasing; initial value $20$

$P (t) = 0.8 e^{1.3 t}$

$P (t) = 6500 e^{- 2.5 t}$

$P (t) = 6500 {(e^{- 2.5})}^{t} \approx 6500 \cdot {0.082}^{t}$ ; decreasing; initial value $6500$

$P (t) = 1.7 e^{- 0.02 t}$

Fill in the table, rounding your answers to four decimal places.

$x$	$0$	$0.5$	$1$	$1.5$	$2$	$2.5$
$e^{x}$

Compute the ratio of each function value to the previous one. Explain the result.

$x$ $0$ $0.5$ $1$ $1.5$ $2$ $2.5$
$e^{x}$ $1$ $1.6487$ $2.7183$ $4.4817$ $7.3891$ $12.1825$
Each ratio is $e^{0.5} \approx 1.6487$ : Increasing $x$ -values by a constant $Δ x equals 0.5$ corresponds to multiplying the $y$ -values of the exponential function by a constant factor of $e^{Δ x}$ .

$x$	$0$	$0.5$	$1$	$1.5$	$2$	$2.5$
$e^{x}$	$1$	$1.6487$	$2.7183$	$4.4817$	$7.3891$	$12.1825$

Fill in the table, rounding your answers to four decimal places.

$x$	$0$	$2$	$4$	$6$	$8$	$10$
$e^{x}$

Compute the ratio of each function value to the previous one. Explain the result.

Fill in the table, rounding your answers to the nearest integer.

$x$	$0$	$0.6931$	$1.3863$	$2.0794$	$2.7726$	$3.4657$	$4.1589$
$e^{x}$

Subtract each $x$ -value from the next one. Explain the result.

$x$ $0$ $0.6931$ $1.3863$ $2.0794$ $2.7726$ $3.4657$ $4.1589$
$e^{x}$ $1$ $2$ $4$ $8$ $16$ $32$ $64$
Each difference in $x$ -values is approximately $ln ⁡ open parenthesis 2 close parenthesis approximately equals 0.6931$ : Increasing $x$ -values by a constant $Δ x equals ln ⁡ open parenthesis 2 close parenthesis$ corresponds to multiplying the $y$ -values of the exponential function by a constant factor of $e^{Δ x} = e^{\ln (2)} = 2$ . That is, each function value is approximately equal to double the previous one.

$x$	$0$	$0.6931$	$1.3863$	$2.0794$	$2.7726$	$3.4657$	$4.1589$
$e^{x}$	$1$	$2$	$4$	$8$	$16$	$32$	$64$

Fill in the table, rounding your answers to the nearest integer.

$x$	$0$	$1.0986$	$2.1972$	$3.2958$	$4.3944$	$5.4931$	$6.5917$
$e^{x}$

Subtract each $x$ -value from the next one. Explain the result.

For Problems 23–30, solve. Round your answers to two decimal places.

$6.21 = 2.3 e^{1.2 x}$

$0.8277$

$22.26 = 5.3 e^{0.4 x}$

$6.4 = 20 e^{0.3 x} - 1.8$

$minus 2.9720$

$4.5 = 4 e^{2.1 x} + 3.3$

$46.52 = 3.1 e^{1.2 x} + 24.2$

$1.6451$

$1.23 = 1.3 e^{2.1 x} - 17.1$

$16.24 = 0.7 e^{- 1.3 x} - 21.7$

$minus 3.0713$

$55.68 = 0.6 e^{- 0.7 x} + 23.1$

For Problems 31–36, solve the equation for the specified variable.

$y = e^{k t},$ for $t$

$t = \frac{1}{k} \ln (y)$

$\frac{T}{R} = e^{t / 2},$ for $t$

$y = k (1 - e^{- t}),$ for $t$

$t = \ln (\frac{k}{k - y})$

$B - 2 = (A + 3) e^{- t / 3},$ for $t$

$T = T_{0} \ln (k + 10),$ for $k$

$k = e^{T / T_{0}} - 10$

$P = P_{0} + \ln (10 k),$ for $k$

Fill in the table, rounding your answers to three decimal places.

$n$	$0.39$	$3.9$	$39$	$390$
$\ln (n)$

Subtract each natural logarithm in your table from the next one. (For example, compute $ln ⁡ open parenthesis 3.9 close parenthesis minus ln ⁡ open parenthesis 0.39 close parenthesis$ .) Explain the result.

$n$ $0.39$ $3.9$ $39$ $390$
$\ln n$ $minus 0.942$ $1.361$ $3.664$ $5.966$
Each difference in function values is approximately $ln ⁡ open parenthesis 10 close parenthesis approximately equals 2.303$ : Multiplying $x$ -values by a constant factor of $10$ corresponds to adding a constant value of ln (10) to the $y$ -values of the natural log function.

$n$	$0.39$	$3.9$	$39$	$390$
$\ln n$	$minus 0.942$	$1.361$	$3.664$	$5.966$

Fill in the table, rounding your answers to three decimal places.

$n$	$0.64$	$6.4$	$64$	$640$
$\ln (n)$

Subtract each natural logarithm in your table from the next one. (For example, compute $ln ⁡ open parenthesis 6.4 close parenthesis minus ln ⁡ open parenthesis 0.64 close parenthesis$ .) Explain the result.

Fill in the table, rounding your answers to three decimal places.

$n$	$2$	$4$	$8$	$16$
$\ln (n)$

Divide each natural logarithm in your table by $ln ⁡ 2$ . Explain the result.

$n$ $2$ $4$ $8$ $16$
$\ln (n)$ $0.693$ $1.386$ $2.079$ $2.773$
Each quotient equals $k$ , where $n = 2^{k}$ . Because $\ln (n) = \ln (2^{k}) = k \cdot \ln (2)$ , $k = \frac{\ln (n)}{\ln (2)}$ .

$n$	$2$	$4$	$8$	$16$
$\ln (n)$	$0.693$	$1.386$	$2.079$	$2.773$

Fill in the table, rounding your answers to three decimal places.

$n$	$5$	$25$	$125$	$625$
$\ln (n)$

Divide each natural logarithm in your table by $ln ⁡ open parenthesis 5 close parenthesis$ . Explain the result.

For Problems 41–46,

Express each growth or decay law in the form $N (t) = N_{0} e^{k t}$ .
Check your answer by graphing both forms of the function on the same axes. Do they have the same graph?

$N (t) = 100 \cdot 2^{t}$

$N (t) = 100 e^{(\ln (2)) t} \approx 100 e^{0.6931 t}$

$N (t) = 50 \cdot 3^{t}$

$N (t) = 1200 (0.6)^{t}$

$N (t) = 1200 e^{\ln (0.6) t} \approx 1200 e^{- 0.5108 t}$

$N (t) = 300 (0.8)^{t}$

$N (t) = 10 (1.15)^{t}$

$N (t) = 10 e^{\ln (1.15) t} \approx 10 e^{0.1398 t}$

$N (t) = 1000 (1.04)^{t}$

The population of Citrus Valley was $20, 000$ in $2000$ . In $2010$ , it was $35, 000$ .

What is $P sub 0$ if $t equals 0$ in $2000$ ?
Use the population in $2010$ to find the growth factor $e^{k}$ .
Write a growth law of the form $P (t) = P_{0} e^{k t}$ for the population of Citrus Valley.
If it continues at the same rate of growth, what will the population be in $2030$ ?

$20, 000$
${(\frac{35, 000}{20, 000})}^{1 / 10} \approx e^{0.056}$
$P (t) = 20, 000 e^{0.056 t}$
$107, 188$

A copy of Time magazine cost $ $1.50$ in $1981.$ In $1988$ , the cover price had increased to $ $2.00$ .

What is $P sub 0$ if $t equals 0$ in $1981$ ?
Use the price in $1988$ to find the growth factor $e^{k}$ .
Find a growth law of the form $P (t) = P_{0} e^{k t}$ for the price of Time.
In $1999$ , a copy of Time cost $ $3.50$ . Did the price of the magazine continue to grow at the same rate from $1981$ to $1999$ ?

Cobalt-60 is a radioactive isotope used in the treatment of cancer. A $500$ -milligram sample of cobalt-60 decays to $385$ milligrams after $2$ years.

Using $P sub 0 equals 500$ , find the decay factor $e^{k}$ for cobalt-60.
Write a decay law $N (t) = N_{0} e^{k t}$ for cobalt-60.
How much of the original sample will be left after $10$ years?

${(\frac{385}{500})}^{1 / 2} \approx e^{- 0.1307}$
$N (t) = 500 e^{- 0.1307 t}$
$135.3$ mg

Weed seeds can survive for a number of years in the soil. An experiment on cultivated land found $155$ million weed seeds per acre, and in the following years the experimenters prevented the seeds from coming to maturity and producing new weeds. Four years later, there were $13.6$ million seeds per acre. (Source: Burton, 1998)

Find the annual decay factor $e^{k}$ for the number of weed seeds in the soil.
Write an exponential formula with base $e$ for the number of weed seeds that survived after $t$ years.

Problems 51–58 are about doubling time and half-life.

Delbert invests $ $500$ in an account that pays $9.5 %$ interest compounded continuously.

Write a formula for $A (t)$ that gives the amount of money in Delbert's account after $t$ years.
How long will it take Delbert's investment to double to $ $1000$ ?
How long will it take Delbert's money to double again, to $ $2000$ ?
Graph $A (t)$ and illustrate the doubling time on your graph.
Choose any point $open parenthesis t sub 1, A sub 1 close parenthesis$ on the graph, then find the point on the graph with vertical coordinate $2 A sub 1$ . Verify that the difference in the $t$ -coordinates of the two points is the doubling time.

$A (t) = 500 e^{0.095 t}$
$7.3$ years
$7.3$ years

d–e

The growth of plant populations can be measured by the amount of pollen they produce. The pollen from a population of pine trees that lived more than $9500$ years ago in Norfolk, England, was deposited in the layers of sediment in a lake basin and dated with radiocarbon techniques.

The figure shows the rate of pollen accumulation plotted against time, and the fitted curve $P (t) = 650 e^{0.00932 t}$ . (Source: Burton, 1998)

What was the annual rate of growth in pollen accumulation?
Find the doubling time for the pollen accumulation, that is, the time it took for the accumulation rate to double.
By what factor did the pollen accumulation rate increase over a period of $500$ years?

Technetium-99m (Tc-99m) is an artificially produced radionuclide used as a tracer for producing images of internal organs such as the heart, liver, and thyroid. A solution of Tc-99m with initial radioactivity of $10, 000$ becquerels (Bq) decays according to the formula

$N (t) = 10, 000 e^{- 0.1155 t}$

where $t$ is in hours.

How long will it take the radioactivity to fall to half its initial value, or $5000$ Bq?
How long will it take the radioactivity to be halved again?
Graph $N (t)$ and illustrate the half-life on your graph.
Choose any point $open parenthesis t sub 1, N sub 1 close parenthesis$ on the graph, then find the point on the graph with vertical coordinate $0.5 N sub 1$ . Verify that the difference in the $t$ -coordinates of the two points is the half-life.

$6$ hours
$6$ hours

All living things contain a certain amount of the isotope carbon-14. When an organism dies, the carbon-14 decays according to the formula

$N (t) = N_{0} e^{- 0.000124 t}$

where $t$ is measured in years. Scientists can estimate the age of an organic object by measuring the amount of carbon-14 remaining.

When the Dead Sea scrolls were discovered in 1947, they had $78.8 %$ of their original carbon-14. How old were the Dead Sea scrolls then?
What is the half-life of carbon-14, that is, how long does it take for half of an object's carbon-14 to decay?

The half-life of iodine-131 is approximately $8$ days.

If a sample initially contains $N sub 0$ grams of iodine-131, how much will it contain after $8$ days? How much will it contain after $16$ days? After $32$ days?
Use your answers to part (a) to sketch a graph of $N (t)$ , the amount of iodine-131 remaining, versus time. (Choose an arbitrary height for $N sub 0$ on the vertical axis.)
Calculate $k$ , and hence find a decay law of the form $N (t) = N_{0} e^{k t}$ , where $k less than 0$ , for iodine-131.

$\frac{1}{2} N_{0}$ , $\frac{1}{4} N_{0}$ , $\frac{1}{16} N_{0}$
$N (t) = N_{0} e^{- 0.0866 t}$

The half-life of hydrogen-3 is $12.5$ years.

If a sample initially contains $N sub 0$ grams of hydrogen-3, how much will it contain after $12.5$ years? How much will it contain after $25$ years?
Use your answers to part (a) to sketch a graph of $N (t)$ , the amount of hydrogen-3 remaining, versus time. (Choose an arbitrary height for $N sub 0$ on the vertical axis.)
Calculate $k$ , and hence find a decay law of the form $N (t) = N_{0} e^{k t}$ , where $k less than 0$ , for hydrogen-3.

A Geiger counter measures the amount of radioactive material present in a substance. The table shows the count rate for a sample of iodine-128 as a function of time. (Source: Hunt and Sykes, 1984)

Time (min)	$0$	$10$	$20$	$30$	$40$	$50$	$60$	$70$	$80$	$90$
Counts/sec	$120$	$90$	$69$	$54$	$42$	$33$	$25$	$19$	$15$	$13$

Graph the data and use your calculator's exponential regression feature to fit a curve to them.
Write your equation in the form $G (t) = G_{0} e^{k t}$ .
Calculate the half-life of iodine-128.

$y = 116 (0.975)^{t}$
$G (t) = 116 e^{- 0.025 t}$
$28$ minutes

The table shows the count rate for sodium-24 registered by a Geiger counter as a function of time. (Source: Hunt and Sykes, 1984)

Time (min)	$0$	$10$	$20$	$30$	$40$	$50$	$60$	$70$	$80$	$90$
Counts/sec	$180$	$112$	$71$	$45$	$28$	$18$	$11$	$7$	$4$	$3$

Graph the data and use your calculator's exponential regression feature to fit a curve to them.
Write your equation in the form $G (t) = G_{0} e^{k t}$ .
Calculate the half-life of sodium-24.

Modeling, Functions, and Graphs by Katherine Yoshiwara (yoshiwarabooks.org), GNU Free Documentation License 1.2 or later. Adapted for the XYZ HTML edition with the authors' permission (recorded 2026-07-04). License: GFDL-1.2-or-later.

$t$	$A (t)$
$0$	$500$
$1$	$541.64$
$2$	$586.76$
$3$	$635.62$
$4$	$688.56$
$5$	$745.91$