Subsection 10.3.1 The Natural Exponential Function
The natural exponential function is the function \(f(x) = e^x\text{.}\) Values for \(e^x\) can be obtained with a calculator using the \(\boxed{e^x}\) key ( 2nd LN on most calculators). For example, you can evaluate \(e^1\) by pressing
\(\qquad\qquad\)2nd LN \(1\)
to confirm the value of \(e\) given above.
Checkpoint 10.3.1. Practice 1.
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 \(y = 3^x\text{.}\) Compare the tables of values and the graphs of the three functions below. You can verify the table and graphs on your caclulator.
| \(x\) |
\(y=2^x\) |
\(y=e^x\) |
\(y=3^x\) |
| \(-3\) |
\(0.125\) |
\(0.050\) |
\(0.037\) |
| \(-2\) |
\(0.250\) |
\(0.135\) |
\(0.111\) |
| \(-1\) |
\(0.500\) |
\(0.368\) |
\(0.333\) |
| \(0\) |
\(1\) |
\(1\) |
\(1\) |
| \(1\) |
\(2\) |
\(2.718\) |
\(3\) |
| \(2\) |
\(4\) |
\(7.389\) |
\(9\) |
| \(3\) |
\(8\) |
\(20.086\) |
\(27\) |
Checkpoint 10.3.2. QuickCheck 1.
Example 10.3.3.
Graph each function. How does each graph differ from the graph of \(y = e^x\text{?}\)
\(\displaystyle g(x) = e^{x+2}\)
\(\displaystyle h(x) = e^x + 2\)
Solution.
The graph of \(g\) is shifted \(2\) units to the left of \(y = e^x\text{.}\) The graph of \(h\) is shifted \(2\) units up from \(y = e^x\text{.}\) The graphs are shown above.
Subsection 10.3.2 The Natural Logarithmic Function
The base \(e\) logarithm of a number \(x\text{,}\) or \(\log_ e x\text{,}\) is called the natural logarithm of \(x\) and is denoted by \(\ln x\text{.}\)
The Natural Logarithm.
The natural logarithm is the logarithm base \(e\text{.}\)
\begin{equation*}
\blert{\ln x = \log_{e}{x}, ~~~~ x\gt 0}
\end{equation*}
The natural logarithm of \(x\) is the exponent to which \(e\) must be raised to produce \(x\text{.}\) For example, the natural logarithm of \(10\text{,}\) or \(\ln 10\text{,}\) is the solution of the equation
\begin{equation*}
e^y = 10
\end{equation*}
You can verify on your calculator that
\begin{equation*}
e^{\alert{2.3}} ~~ \approx 10 ~~ \text{ or } ~~ \ln 10 \approx \alert{2.3}
\end{equation*}
In general, natural logs obey the same conversion formulas that work for logs to other bases.
Conversion Formulas for Natural Logs.
\begin{equation*}
\blert{y= \ln x~~~~~~\text{if and only if}~~~~~~e^y=x}
\end{equation*}
Checkpoint 10.3.4. QuickCheck 2.
In particular,
\begin{align*}
\ln e \amp = 1~~~~ \text{ because }~~~~ e^1 = e \\
\ln 1 \amp = 0~~~~ \text{ because }~~~~ e^0 = 1
\end{align*}
As is the case with exponential and log functions with other bases, the natural log function, and the natural exponential function, \(f(x) = e^x\text{,}\) "undo" each other, so they are inverse functions.
Example 10.3.5.
Graph \(f(x) = e^x\) and \(g(x) = \ln x\) on the same grid.
Solution.
We can make a table of values for \(g(x) = \ln x\) by interchanging the columns in the table for \(f(x) = e^x\text{.}\) Plotting the points gives us the graph below.
| \(x\) |
\(y=\ln x\) |
| \(0.050\) |
\(-3\) |
| \(0.135\) |
\(-2\) |
| \(0.368\) |
\(-1\) |
| \(1\) |
\(0\) |
| \(2.718\) |
\(1\) |
| \(7.389\) |
\(2\) |
| \(20.086\) |
\(3\) |
You can see from the graph that the natural log function has only positive numbers as input values. The natural logs of negative numbers and zero are undefined. You can also see that the natural log of a number greater than 1 is positive, while the logs of fractions between 0 and 1 are negative.
Checkpoint 10.3.6. Practice 2.
The three properties of logarithms also apply to base \(e\) logarithms.
Properties of Natural Logarithms.
If \(x, y \gt 0\text{,}\) then
\(\displaystyle \ln(xy) = \ln x + \ln y\)
\(\displaystyle \ln {\dfrac{x}{y}} = \ln x - \ln y\)
\(\displaystyle \ln {x^n} = n \ln x\)
If \(x \le 0\text{,}\) \(\ln x\) is undefined.
And because the functions \(y=e^x\) and \(y= \ln x\) are inverse functions, the following properties are also true.
The Natural Log and \(e^x\).
\begin{equation*}
\blert{\ln {e^x} =x}~~~~~~\text{and}~~~~~~\blert{e^{\ln {x}} = x}
\end{equation*}
Example 10.3.7.
Simplify each expression.
\(\displaystyle \ln e^{0.3x}\)
\(\displaystyle e^{2 \ln(x+3)}\)
Solution.
The natural log is the log base \(e\text{,}\) and hence the inverse of \(e^x\text{.}\) Therefore,
\begin{equation*}
\ln {e^{0.3x}} = 0.3x
\end{equation*}
First, we simplify the exponent using the third property of logs to get
\begin{equation*}
2 \ln(x + 3) = \ln(x + 3)^2
\end{equation*}
Then \(e^{2 \ln(x+3)} = e^{\ln(x+3)^2} = (x + 3)^2\text{.}\)
Checkpoint 10.3.8. Practice 3.
Subsection 10.3.4 Exponential Growth and Decay
Recall that functions of the form
\begin{equation*}
P(t) = P_0\cdot b^t
\end{equation*}
describe exponential growth when \(b \gt 1\) and exponential decay when \(0 \lt b \lt 1\text{.}\) Exponential growth and decay can also be modeled by functions of the form
\begin{equation*}
\blert{P(t) = P_0 \cdot e^{kt}}
\end{equation*}
where we have substituted \(e^k\) for the growth factor \(b\text{,}\) so that
\begin{align*}
P(t) \amp = P_0 \cdot b^t\\
\amp = P_0 \cdot \left(e^k\right)^t = P_0 \cdot e^{kt}
\end{align*}
We can find the value of \(k\) by solving the equation \(b = e^k\) for \(k\text{,}\) to get \(k = \ln b\text{.}\)
For instance, consider a colony of bacteria that grows according to the formula
\begin{equation*}
P(t) = 100 \cdot \alert{3^t}
\end{equation*}
We can express this function in the form \(P(t) = 100 \cdot \alert{e^{kt}}\) if we set
\begin{equation*}
3 = e^k ~ \text{ or } ~ k = \ln 3 \approx 1.0986
\end{equation*}
Thus, the growth law for the colony of bacteria can be written
\begin{equation*}
P(t) \approx 100 \cdot e^{1.0986t}
\end{equation*}
By graphing both functions on your calculator, you can verify that
\begin{equation*}
P(t) = 100 \cdot 3^t~~~\text{and}~~~P(t) = 100 \cdot e^{1.0986t}
\end{equation*}
are just two ways of writing the same function.
Example 10.3.17.
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 \(e\text{.}\)
Solution.
The growth factor was \(b = 1 + r = 1.064\text{.}\) The population \(t\) years later was
\begin{equation*}
P(t) = 768,000 (1.064)^t
\end{equation*}
We use the formula \(P(t) = P_0 \cdot e^{kt}\text{,}\) where \(e^k = 1.064\text{.}\) Solving for \(k\text{,}\) we find
\begin{equation*}
k = \ln 1.064 = 0.062
\end{equation*}
so \(P(t) = 768,000 e^{0.062t}\text{.}\)
Checkpoint 10.3.18. Practice 7.
If \(k\) is negative, then \(e^k\) is a fraction less than \(1\text{.}\) For example, if \(k = -2\text{,}\)
\begin{equation*}
e^{-2} = \frac{1}{e^2} \approx \frac{1}{7.3891} \approx 0.1353
\end{equation*}
Thus, for negative values of \(k\text{,}\) the function \(P(t) = P_0 e^{kt}\) describes exponential decay.
Exponential Growth and Decay.
The function
\begin{equation*}
P(t) = P_0 e^{kt}
\end{equation*}
describes exponential growth if \(k \gt 0\text{,}\) and exponential decay if \(k \lt 0\text{.}\)
Checkpoint 10.3.19. QuickCheck 4.
Example 10.3.20.
Express the decay law \(N(t) = 60 (0.8)^t\) in the form \(N(t) = N_0 e^{kt}\text{.}\)
Solution.
For this decay law, \(N_0 = 60\) and \(b = 0.8\text{.}\) We would like to find a value for \(k\) so that \(e^k = b = 0.8\text{,}\) that is, we must solve the equation
\begin{align*}
e^k \amp = 0.8\amp\amp \blert{\text{Take natural log of both sides.}}\\
\ln e^k\amp = \ln 0.8\amp\amp \blert{\text{Simplify.}}\\
k\amp = \ln 0.8 \approx -0.2231
\end{align*}
Replacing \(b\) with \(e^k\text{,}\) we find that the decay law is
\begin{equation*}
N(t) \approx 60e^{-0.2231t}
\end{equation*}
Checkpoint 10.3.21. Practice 8.
