Checkpoint 10.3.1. Practice 1.
Use your calculator to evaluate the following powers. Round to four decimal places.
-
\(\displaystyle e^2\)
-
\(\displaystyle e^{3.5}\)
-
\(\displaystyle e^{-0.5}\)
Section 10.3 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 used in scientific applications. \(~\alert{\text{[TK]}}~~\) This base is an irrational number called \(e\text{,}\) where
\begin{equation*}
\blert{e \approx 2.71828182845}
\end{equation*}
Aside
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 \(\pi\) 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.
Subsection 10.3.1 The Natural Exponential Function
The natural exponential function is the function
\begin{equation*}
f(x) = e^x
\end{equation*}
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
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 \(e\text{.}\)
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 \(y = 3^x\text{.}\) Compare the table of values and the graphs of the three functions below. For example, note that for \(x=2\text{,}\) the value of \(e^2 = 7.389\) is between \(2^2 = 4\) and \(3^2 = 9\text{.}\) You can verify the table and the graphs on your calculator.
| \(x\) | \(y=2^x\) | \(y=e^x\) | \(y=3^x\) |
| \(-3\) | \(0.125\) | \(0.050\) | \(0.037\) |
| \(-2\) | \(0.250\) | \(0.135\) | \(0.111\) |
| \(-1\) | \(0.500\) | \(0.368\) | \(0.333\) |
| \(0\) | \(1\) | \(1\) | \(1\) |
| \(1\) | \(2\) | \(2.718\) | \(3\) |
| \(2\) | \(4\) | \(7.389\) | \(9\) |
| \(3\) | \(8\) | \(20.086\) | \(27\) |
Checkpoint 10.3.2. QuickCheck 1.
-
The value of \(e^2\) is closest to which of these?
-
The value of \(e^{-1}\) is closest to which of these?
-
\(\displaystyle -3\)
-
\(\displaystyle \dfrac{1}{2}\)
-
\(\displaystyle \dfrac{1}{3}\)
-
\(\displaystyle 2\)
-
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 10.3.3.
The normal distribution above shows the heights of men in the US. \(\alert{\text{[TK]}}~~\) The average or mean height for men is 70 inches, and the formula for the graph is
\begin{equation*}
f(x) = 0.13~ e^{-(x-70)^2/18}
\end{equation*}
Evaluate the function for \(x=70\) and for \(x=67\text{.}\)
Aside
Solution.
To evaluate the function we follow the order of operations. For \(x=70\) we have
\begin{align*}
f(70) \amp = 0.13~ e^{-(70-70)^2/18} \amp \amp \blert{\text{Simplify the exponent first.}}\\
\amp = 0.13~ e^0 = 0.13
\end{align*}
For \(x=67\) we have
\begin{align*}
f(67) \amp = 0.13~ e^{-(67-70)^2/18} \amp \amp \blert{-(67-70)^2/18 = -0.5} \\
\amp = 0.13~ e^{-0.5} = 0.079
\end{align*}
This next graph is an example of a logistic function, which models population growth with an upper bound.
Checkpoint 10.3.4. Practice 2.
The logistic function shown above models the spread of Covid in China during the 2020 epidemic. It gives the number of infections \(N(t)\text{,}\) in thousands, reported \(t\) days after January 21, 2020. The equation for this model is
\begin{equation*}
N(t) = \dfrac{83.5}{1+267~ e^{-0.214t}}
\end{equation*}
According to the model, how many cases of Covid were reported on February 20 (day \(t=30\))?
To use the natural base for exponential growth and decay, we will need to look at functions of the form \(~y=e^{kx}~\) for various values of \(k\text{.}\) Here are some examples.
Example 10.3.5.
Use technology to graph each function. How does the coefficient of \(x\) make each graph different from the graph of \(y = e^x\text{?}\)
-
\(\displaystyle g(x) = e^{0.5x}\)
-
\(\displaystyle h(x) = e^{-0.5x}\)
Solution.
-
The graph of \(y = e^x\) is shown in red. For positive \(x\text{,}\) the coefficient \(0.5\) makes the graph of \(g\) grow more slowly than \(y = e^x\text{,}\) because \(e^{0.5x}\) is smaller than \(e^x\text{.}\) So for positive \(x\) the graph of \(g\) lies below the graph of \(y = e^x\text{.}\)
-
Because \(e^{-0.5x} = \dfrac{1}{e^{0.5x}}\text{,}\) the graph of \(h\) is decreasing, and in fact it is the reflection of the graph of \(g\) about the \(y\)-axis.
Subsection 10.3.2 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 \(~y=\log_2{(x)}~\) is the inverse of the function \(~y=2^x\text{.}\) It gives the exponent needed on 2 to give \(x\text{,}\) so that, for instance, \(~\log_2{(16)} = 4~\text{,}\) because \(~2^4 = 16\text{.}\)\(\alert{\text{[TK]}}~~\)
Aside
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 \(\ln {(x)}\text{.}\) (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.
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*}
As is the case with exponential and log functions with other bases, the natural log function, \(y= \ln {(x)}\text{,}\) and the natural exponential function, \(f(x) = e^x\text{,}\) “undo” each other, so they are inverse functions. (This is why many calculators use 2nd LN to indicate \(e^x\text{.}\))
Checkpoint 10.3.6. Practice 3.
Use your calculator to evaluate each logarithm. Round your answers to four decimal places.
-
\(\displaystyle \ln {(10)}\)
-
\(\displaystyle \ln {(0.1)}\)
-
\(\displaystyle \ln {(100)}\)
-
\(\displaystyle \ln {(0.01)}\)
As usual, we can gain a better understanding of a new function by looking at its graph.
Example 10.3.7.
Solution.
Earlier we made a table of values to graph the fuction \(f(x) = e^x\text{,}\) 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.
| \(x\) | \(y=\ln {(x)}\) |
| \(0.050\) | \(-3\) |
| \(0.135\) | \(-2\) |
| \(0.368\) | \(-1\) |
| \(1\) | \(0\) |
| \(2.718\) | \(1\) |
| \(7.389\) | \(2\) |
| \(20.086\) | \(3\) |
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 \(y=x\text{.}\) 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.
From the graph of \(g(x) = \ln {(x)}\) you can make the following observations. \(~\alert{\text{[TK]}}~~\)
-
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.
Aside
Subsection 10.3.3 Properties of the Natural Logarithm
In general, natural logs obey the same conversion equations that work for logs to other bases. In fact, all the techniques you have already learned for working with exponential and log functions still work for base \(e\text{.}\)
Conversion Equations for Natural Logs.
\begin{equation*}
\blert{y= \ln {(x)}~~~~~~\text{if and only if}~~~~~~e^y=x}
\end{equation*}
Checkpoint 10.3.8. QuickCheck 2.
Which of the following is equivalent to \(e^x=k\) ?
-
\(\displaystyle e^k=x\)
-
\(\displaystyle \ln{(e)} = x\)
-
\(\displaystyle \ln{(k)} = x\)
-
\(\displaystyle k^x = e\)
In particular,
\begin{align*}
\ln {(e)} \amp = 1~~~~ \text{ because }~~~~ e^1 = e \\
\ln {(1)} \amp = 0~~~~ \text{ because }~~~~ e^0 = 1
\end{align*}
Properties of Natural Logarithms.
If \(x, y \gt 0\text{,}\) then
-
\(\displaystyle \ln{(xy)} = \ln {(x)} + \ln {(y)}\)
-
\(\displaystyle \ln {\left(\dfrac{x}{y}\right)} = \ln {(x)} - \ln {(y)}\)
-
\(\displaystyle \ln {(x^n)} = n \ln {(x)}\)
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 10.3.9.
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 10.3.10. Practice 4.
Simplify each expression.
Subsection 10.3.4 Solving Equations
We use the natural logarithm to solve exponential equations with base \(e\text{.}\) The techniques we’ve learned for solving other exponential equations also apply to equations with base \(e\text{.}\) \(~\alert{\text{[TK]}}\)
Aside
Example 10.3.11.
Solve each equation for \(x\text{.}\)
-
\(\displaystyle e^x = 0.24\)
-
\(\displaystyle \ln {(x)} = 3.5\)
Solution.
-
We convert the equation to logarithmic form and evaluate using a calculator.\begin{equation*} x = \ln {(0.24)} \approx -1.427 \end{equation*}
-
We convert the equation to exponential form and evaluate.\begin{equation*} x = e^{3.5} \approx 33.1155 \end{equation*}
Checkpoint 10.3.12. Practice 5.
Solve each equation. Round your answers to four decimal places.
Checkpoint 10.3.13. QuickCheck 3.
-
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 10.3.14.
Solve \(~200 = 60 + 20 e^{0.4x}\)
Solution.
First, we isolate the power.
\begin{align*}
200 \amp = 60 + 20 e^{0.4x} \amp \amp \blert{\text{Subtract 60 from both sides.}}\\
140 \amp = 20 e^{0.4x} \amp \amp \blert{\text{Divide both sides by 20.}}\\
7 \amp = e^{0.4x}
\end{align*}
Then we convert the equation to logarithmic form.
\begin{align*}
0.4x \amp = \ln {(7)} \amp\amp \blert{\text{Divide both sides by 0.4.}}\\
x \amp= \frac{\ln {(7)}}{0.4}
\end{align*}
Rounded to four decimal places, \(x \approx 4.8648\text{.}\)
Note 10.3.15.
When solving the equation \(7 = e^{0.4x}\) in the Example above, we can think of taking the natural logarithm of both sides. This gives us
\begin{align*}
\ln {(7)} \amp = \ln {(e^{0.4x})}\amp\amp \blert{\text{Simplify the right side.}}\\
\ln {(7)} \amp = 0.4x
\end{align*}
because \(\ln{(e^a)} = a\) for any number \(a\text{.}\) We then proceed with the solution as before.
Checkpoint 10.3.16. Practice 6.
Example 10.3.17.
The logistic function
\begin{equation*}
P = \dfrac{20}{1 + 5e^{-0.4t}}
\end{equation*}
has its upper bound at \(~P=20\text{.}\) Find the value of \(t\) for which \(~P=10\text{.}\)
Solution.
We must solve the equation \(~10 = \dfrac{20}{1 + 5e^{-0.4t}}~\) for \(t\text{.}\) To begin, we multiply both sides of the equation by the denominator, \(1 + 5e^{-0.4t}\text{,}\) to get
\begin{equation*}
1 + 5e^{-0.4t} = \dfrac{20}{10} = 2
\end{equation*}
Then we isolate the power, \(e^{-kt}\text{,}\) as follows:
\begin{align*}
1 + 5e^{-0.4t} \amp = 2 \amp \amp \blert{\text{Subtract 1 from both sides, then divide by 5.}}\\
e^{-0.4kt} \amp = 0.2
\end{align*}
Next, we take the natural logarithm of both sides to get
\begin{equation*}
\ln {(e^{-0.4t})} = \ln{(0.2)}
\end{equation*}
and recall that \(\ln {(e^x)} = x\) to simplify the left side.
\begin{equation*}
-0.4t = \ln{(0.2)}
\end{equation*}
Finally, we divide both sides by \(-0.4\) to solve for \(t\text{.}\)
\begin{equation*}
t =\frac{-\ln{(0.2)}}{0.4} \approx 4.02
\end{equation*}
Checkpoint 10.3.18. Practice 7.
Subsection 10.3.5 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 \(0 \lt b \lt 1\text{.}\) \(~\alert{\text{[TK]}}\)
Aside
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{gather*}
P(t) = 100 \cdot 3^t\\
~~~\text{and}~~~\\
P(t) = 100 \cdot e^{1.0986t}
\end{gather*}
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 \(~b=1+r~\text{,}\) where \(r\) is the percentage rate in decimal form. For a percent decrease, \(~b=1-r~\text{.}\) \(\alert{\text{[TK]}}~~\)
Aside
Example 10.3.19.
From 1990 to 2000, the population of Clark County, Nevada, grew by \(6.4\%\) per year. \(~\alert{\text{[TK]}}~~\)
-
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 \(P(t) = 768,000 e^{0.062t}\text{.}\)
Aside
Checkpoint 10.3.20. Practice 8.
From 1994 to 1998, the number of personal computers connected to the Internet grew according to the formula
\begin{equation*}
N(t) = 2.8e^{0.85t}
\end{equation*}
where \(t = 0\) in 1994 and \(N\) is in millions. (Source: Los Angeles Times, September 6, 1999)
-
Evaluate \(N(1)\text{.}\) By what percent did the number of Internet users grow in one year?
-
Express the growth law in the form \(~N(t) = N_0 (1 + r)^t\text{.}\) (Hint: \(e^k = 1 + r\text{.}\))
Now, what about exponential decay, where the the decay fator \(b\) is a number less than 1? If \(k\) is negative, then \(e^k\) is a number less than 1. For example, if \(k = -2\text{,}\)
\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 10.3.21. QuickCheck 4.
The natural log of a number between 0 and 1 is
Example 10.3.22.
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 10.3.23. Practice 9.
A scientist isolates \(25\) grams of krypton-91, which decays according to the formula
\begin{equation*}
N(t) = 25e^{-0.07t}\text{,}
\end{equation*}
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.
\(t\) \(0\) \(10\) \(20\) \(30\) \(40\) \(50\) \(60\) \(N(t)\) \(\hphantom{000}\) \(\hphantom{000}\) \(\hphantom{000}\) \(\hphantom{000}\) \(\hphantom{000}\) \(\hphantom{000}\) \(\hphantom{000}\) -
Use the table to choose a suitable window and graph the function \(N(t)\text{.}\)
-
Write and solve an equation to answer the question: How long does it take for 60% of the krypton-91 to decay?
Subsection 10.3.6 Continuous Compounding
In Section 7.5.4 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 \(n\text{.}\) At this upper bound we say that the interest is compounded continuously, and the amount is given by the function
\begin{equation*}
A(t) = Pe^{rt}
\end{equation*}
where \(P\) is the principal invested and \(r\) is the interest rate. \(~\alert{\text{[TK]}}\)
Aside
Example 10.3.24.
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 \(A(t)\text{.}\)
-
How much will the account be worth after 10 years?
-
How long will it be before the account is worth $1000?
Solution.
-
We substitute 500 for \(P\text{,}\) and 0.08 for \(r\) to find\begin{equation*} A(t) = 500e^{0.08t} \end{equation*}
-
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 \(t = 10\text{.}\)\begin{align*} A(10) \amp= 500 e^{0.08(10)}\\ \amp = 500 e^{0.8}\\ \amp \approx 500(2.2255) = 1112.77 \end{align*}The account will be worth $1112.77 after 10 years.
-
We substitute 1000 for \(A(t)\) and solve the equation.\begin{align*} 1000 \amp= 500 e^{0.08t}\amp\amp \blert{\text{Divide both sides by 500.}}\\ 2 \amp = e^{0.08t}\amp\amp \blert{\text{Take natural log of both sides.}}\\ \ln {(2)} \amp= \ln {(e^{0.08t})} = 0.08t\amp\amp \blert{\text{Divide both sides by 0.08.}}\\ t \amp= \frac{\ln {(2)}}{0.08}\approx 8.6643 \end{align*}The account will be worth $1000 after approximately 8.7 years.
Checkpoint 10.3.25. Practice 10.
Zelda invested $1000 in an account that pays 4.5% interest compounded continuously. How long will it be before the account is worth $2000?
In the next Investigation we’ll see how the formula for interest compounded continuously, \(~A(t) = Pe^{rt}~\text{,}\) arises when \(n\text{,}\) the number of compounding periods, increases.
Investigation 10.3.1. Interest Compounded Continuously.
We learned in Section 7.5.4 that the amount, \(A\) (principal plus interest), accumulated in an account with interest compounded \(n\) times annually is
\begin{equation*}
A=P\left(1+\dfrac{r}{n} \right)^{nt}
\end{equation*}
where \(P\) is the principal invested, \(r\) is the interest rate, and \(t\) is the time period, in years.
-
Suppose you keep \(\$1000\) in an account that pays 8% interest. How much is the amount \(A\) after 1 year if the interest is compounded twice a year? Four times a year?\begin{gather*} n = \alert{2}: A = 1000\left(1 + \frac{0.08}{\alert{2}} \right)^{\alert{2}(1)}= \\ n = \alert{4}: A = 1000\left(1 + \frac{0.08}{\alert{4}} \right)^{\alert{4}(1)}= \end{gather*}
-
What happens to \(A\) as we increase \(n\text{,}\) the number of compounding periods per year? Fill in the table showing the amount in the account for different values of \(n\text{.}\)
\(n\) \(~~~A~~~\) \(1\) (annually) \(1080\) \(2\) (semiannually) \(\) \(4\) (quarterly) \(\) \(6\) (bimonthly) \(\) \(12\) (monthly) \(\) \(365\) (daily) \(\) \(1000\) \(\) \(10,000\) \(\) 
-
Plot the values in the table from \(n = 1\) to \(n = 12\text{,}\) and connect them with a smooth curve. Describe the curve: What is happening to the value of \(A\text{?}\)
-
In part (2), as you increased the value of \(n\text{,}\) the other parameters in the formula stayed the same. In other words, \(A\) is a function of \(n\text{,}\) given by \(A = 1000 \left(1 + \dfrac{0.08}{n} \right)^n\text{.}\) Use your calculator to graph A on successively larger intervals:
-
\(\displaystyle \text{Xmin} = 0, \text{Xmax} = 12; \text{Ymin} = 1080, \text{Ymax} = 1084\)
-
\(\displaystyle \text{Xmin} = 0, \text{Xmax} = 50; \text{Ymin} = 1080, \text{Ymax} = 1084\)
-
\(\displaystyle \text{Xmin} = 0, \text{Xmax} = 365; \text{Ymin} = 1080, \text{Ymax} = 1084\)
-
-
Use the Trace feature or the Table feature to evaluate \(A\) for very large values of \(n\text{.}\) Rounded to the nearest penny, what is the largest value of \(A\) that you can find?
-
As \(n\) increases, the values of \(A\) approach a limiting value. Although \(A\) continues to increase, it does so by smaller and smaller increments and will never exceed $1083.29. When the number of compounding periods increases without bound, we call the limiting result continuous compounding.
-
Is there an easier way to compute \(A\) under continues compounding? Yes! Compute \(1000e^{0.08}\) on your calculator. (Press 2nd LN to enter \(e^x\text{.}\)) Compare the value to your answer in part (5) for the limiting value.
-
Repeat your calculations for two other interest rates, 15% and (an extremely unrealistic) 100%, again for an investment of $1000 for 1 year. In each case, compare the limiting value of \(A\text{,}\) and compare to the value of \(1000e^r\text{.}\)
-
\(r=0.15\) \(n\) \(A\) \(1\) \(115\) \(2\) \(\hphantom{00000}\) \(4\) \(\) \(6\) \(\) \(12\) \(\) \(3652\) \(\) \(1000\) \(\) \(10,000\) \(\) \(~~1000e^{0.15}= \) -
\(r=1\) \(n\) \(A\) \(1\) \(200\) \(2\) \(\hphantom{00000}\) \(4\) \(\) \(6\) \(\) \(12\) \(\) \(3652\) \(\) \(1000\) \(\) \(10,000\) \(\) \(~~1000e^{1}= \)
-
-
In part (8b), you have computed an approximation for \(1000e\text{.}\) What is the value of \(e\text{,}\) rounded to 5 decimal places?
-
Complete the table of values. What does \(\left(1 + \dfrac{1}{n} \right)^n\) appear to approach as \(n\) increases?
\(n\) \(100\) \(1000\) \(10,000\) \(100,000\) \(\left(1+\dfrac{1}{n} \right)^n \)
In the Investigation, we saw that the number \(e\) arises as the limiting value of the expression \(\left(1 + \dfrac{1}{n} \right)^n\) as \(n\) increases without bound. The limiting process is one of the fundamental methods of calculus, and helps us understand and analyze other continuous phenomena, such as rates of growth.
Exercises 10.3.7 Homework 10.3
Skills Practice
Exercise Group.
For Problems 1-4, use your calculator to complete the table for each function. Then choose a suitable window and graph the function.
| \(x\) | \(-10\) | \(-5\) | \(0\) | \(5\) | \(10\) | \(15\) | \(20\) |
| \(f(x)\) | \(\hphantom{0000}\) | \(\hphantom{0000}\) | \(\hphantom{0000}\) | \(\hphantom{0000}\) | \(\hphantom{0000}\) | \(\hphantom{0000}\) | \(\hphantom{0000}\) |
Exercise Group.
For Problems 5 and 6, simplify.
5.
-
\(\displaystyle \ln {(e^2)}\)
-
\(\displaystyle e^{\ln {(5t)}} \)
-
\(\displaystyle e^{-\ln {(x)}} \)
-
\(\displaystyle \ln {(\sqrt{e})} \)
6.
-
\(\displaystyle \ln {(e^{x^4})}\)
-
\(\displaystyle e^{3 \ln {(x)}} \)
-
\(\displaystyle e^{\ln {(x)} -\ln {(y)}} \)
-
\(\displaystyle \ln \left(\dfrac{1}{e^{2t}} \right) \)
Exercise Group.
For Problems 7–10, solve for \(x\text{.}\) Round your answers to two decimal places.
7.
8.
9.
-
\(\displaystyle \ln {(x)} = 1.42\)
-
\(\displaystyle \ln {(x)} = 0.63\)
-
\(\displaystyle \ln {(x)} = -2.6\)
10.
-
\(\displaystyle \ln {(x)} = 2.03\)
-
\(\displaystyle \ln {(x)} = 0.59\)
-
\(\displaystyle \ln {(x)} = -3.4\)
Exercise Group.
For Problems 11-14, express each exponential function in the form \(P(t) = P_0b^t\text{.}\) Is the function increasing or decreasing? What is its initial value?
11.
\(P(t) = 20e^{0.4t}\)
12.
\(P(t)=0.8 e^{1.3t} \)
13.
\(P(t) = 6500e^{-2.5t}\)
14.
\(P(t)=1.7 e^{-0.02t} \)
15.
-
Fill in the table, rounding your answers to four decimal places.
\(x\) \(0\) \(0.5\) \(1\) \(1.5\) \(2\) \(2.5\) \(e^x\) \(\phantom{000} \) \(\phantom{000}\) \(\phantom{000}\) \(\phantom{000}\) \(\phantom{000}\) \(\phantom{000}\) -
Compute the ratio of each function value to the previous one. Explain the result.
16.
-
Fill in the table, rounding your answers to four decimal places.
\(x\) \(0\) \(2\) \(4\) \(6\) \(8\) \(10\) \(e^x\) \(\phantom{000} \) \(\phantom{000}\) \(\phantom{000}\) \(\phantom{000}\) \(\phantom{000}\) \(\phantom{000}\) -
Compute the ratio of each function value to the previous one. Explain the result.
17.
-
Fill in the table, rounding your answers to the nearest integer.
\(x\) \(0\) \(0.6931\) \(1.3863\) \(2.0794\) \(2.7726\) \(3.4657\) \(4.1589\) \(e^x\) \(\phantom{000} \) \(\phantom{000}\) \(\phantom{000}\) \(\phantom{000}\) \(\phantom{000}\) \(\phantom{000}\) \(\phantom{000}\) -
Subtract each \(x\)-value from the next one. Explain the result.
18.
-
Fill in the table, rounding your answers to the nearest integer.
\(x\) \(0\) \(1.0986\) \(2.1972\) \(3.2958\) \(4.3944\) \(5.4931\) \(6.5917\) \(e^x\) \(\phantom{000} \) \(\phantom{000}\) \(\phantom{000}\) \(\phantom{000}\) \(\phantom{000}\) \(\phantom{000}\) \(\phantom{000}\) -
Subtract each \(x\)-value from the next one. Explain the result.
Exercise Group.
For Problems 19–26, solve. Round your answers to two decimal places.
19.
\(6.21 = 2.3e^{1.2x}\)
20.
\(22.26 = 5.3e^{0.4x}\)
21.
\(6.4 = 20e^{0.3x} - 1.8 \)
22.
\(4.5 = 4e^{2.1x} + 3.3 \)
23.
\(46.52 = 3.1e^{1.2x} + 24.2 \)
24.
\(1.23 = 1.3e^{2.1x} - 17.1 \)
25.
\(16.24 = 0.7e^{-1.3x} - 21.7 \)
26.
\(55.68 = 0.6e^{-0.7x} + 23.1 \)
Exercise Group.
For Problems 27-32, solve the equation for the specified variable.
27.
28.
29.
30.
31.
32.
33.
-
Fill in the table, rounding your answers to three decimal places.
\(n\) \(0.39\) \(3.9\) \(39\) \(390\) \(\ln {(n)}\) \(\hphantom{0000} \) \(\hphantom{0000} \) \(\hphantom{0000} \) \(\hphantom{0000} \) -
Subtract each natural logarithm in your table from the next one. (For example, compute \(\ln {(3.9)} - \ln {(0.39)}\text{.}\)) Explain the result.
34.
-
Fill in the table, rounding your answers to three decimal places.
\(n\) \(0.64\) \(6.4\) \(64\) \(640\) \(\ln n\) \(\hphantom{0000} \) \(\hphantom{0000} \) \(\hphantom{0000} \) \(\hphantom{0000} \) -
Subtract each natural logarithm in your table from the next one. (For example, compute \(\ln {(6.4)} - \ln {(0.64)}\text{.}\)) Explain the result.
35.
-
Fill in the table, rounding your answers to three decimal places.
\(n\) \(2\) \(4\) \(8\) \(16\) \(\ln {(n)}\) \(\hphantom{0000} \) \(\hphantom{0000} \) \(\hphantom{0000} \) \(\hphantom{0000} \) -
Divide each natural logarithm in your table by \(\ln 2\text{.}\) Explain the result.
36.
-
Fill in the table, rounding your answers to three decimal places.
\(n\) \(5\) \(25\) \(125\) \(625\) \(\ln {(n)}\) \(\hphantom{0000} \) \(\hphantom{0000} \) \(\hphantom{0000} \) \(\hphantom{0000} \) -
Divide each natural logarithm in your table by \(\ln {(5)}\text{.}\) Explain the result.
Exercise Group.
37.
\(N(t) = 100\cdot 2^t\)
38.
\(N(t) = 50\cdot 3^t\)
39.
\(N(t) = 1200(0.6)^t\)
40.
\(N(t) = 300(0.8)^t\)
41.
\(N(t) = 10(1.15)^t\)
42.
\(N(t) = 1000(1.04)^t\)
Applications
43.
The number of bacteria in a culture grows according to the function
\begin{equation*}
N(t) = N_0 e^{0.04t}
\end{equation*}
where \(N_0\) is the number of bacteria present at time \(t = 0\) 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 \(N(t) \text{.}\)
-
How many bacteria were present at \(t = 24\) hours?
-
How much time must elapse (to the nearest tenth of an hour) for the original 6000 bacteria to increase to 100,000?
44.
Hope invests $2000 in a savings account that pays \(5\frac{1}{2}\%\) annual interest compounded continuously.
-
Make a table of values for \(A(t)\) in 2-year intervals over the first 10 years.
-
Graph \(A(t) \text{.}\)
-
How much will Hope’s account be worth after 7 years?
-
How long will it take for the account to grow to $5000?
45.
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
\begin{equation*}
I(t) = 1000e^{-0.1t}
\end{equation*}
-
Graph \(I (t)\text{.}\)
-
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?
46.
X-rays can be absorbed by a lead plate so that
\begin{equation*}
I (t) = I_0 e^{-1.88t}
\end{equation*}
where \(I_0\) 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 \(I (t)\text{.}\)
-
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?
47.
The population of Citrus Valley was 20,000 in 2000. In 2010, it was 35,000.
-
Use the population in 2010 to find the growth factor \(e^k\text{.}\)
-
Write a growth law of the form \(P(t) = P_0 e^{kt}\) for the population of Citrus Valley.
-
If it continues at the same rate of growth, what will the population be in 2030?
48.
A copy of Time magazine cost $1.50 in 1981. In 1988, the cover price had increased to $2.00.
-
Use the price in 1988 to find the growth factor \(e^k\text{.}\)
-
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?
49.
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.
-
Write a decay law \(N(t) = N_0e^{kt}\) for cobalt-60.
-
How much of the original sample will be left after 10 years?
50.
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.
Exercise Group.
Problems 51–58 are about doubling time and half-life.
51.
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 \((t_1, A_1)\) on the graph, then find the point on the graph with vertical coordinate \(2A_1\text{.}\) Verify that the difference in the \(t\)-coordinates of the two points is the doubling time.
52.
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 \(P(t) = 650e^{0.00932t}\text{.}\) (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?
53.
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
\begin{equation*}
N(t) = 10,000e^{-0.1155t}
\end{equation*}
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 \((t_1, N_1)\) on the graph, then find the point on the graph with vertical coordinate \(0.5N_1\text{.}\) Verify that the difference in the \(t\)-coordinates of the two points is the half-life.
54.
All living things contain a certain amount of the isotope carbon-14. When an organism dies, the carbon-14 decays according to the formula
\begin{equation*}
N(t) = N_0e^{-0.000124t}
\end{equation*}
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?
55.
The half-life of iodine-131 is approximately 8 days.
-
If a sample initially contains \(N_0\) 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 \(N(t)\text{,}\) the amount of iodine-131 remaining, versus time. (Choose an arbitrary height for \(N_0\) on the vertical axis.)
-
Calculate \(k\text{,}\) and hence find a decay law of the form \(N(t) = N_0e^{kt}\text{,}\) where \(k \lt 0\text{,}\) for iodine-131.
56.
The half-life of hydrogen-3 is 12.5 years.
-
If a sample initially contains \(N_0\) 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 \(N(t)\text{,}\) the amount of hydrogen-3 remaining, versus time. (Choose an arbitrary height for \(N_0\) on the vertical axis.)
-
Calculate \(k\text{,}\) and hence find a decay law of the form \(N(t) = N_0e^{kt}\text{,}\) where \(k\lt 0\text{,}\) for hydrogen-3.
57.
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 \(G(t) = G_0e^{kt}\text{.}\)
-
Calculate the half-life of iodine-128.
58.
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 \(G(t) = G_0e^{kt}\text{.}\)
-
Calculate the half-life of sodium-24.
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