Supplemental Videos.
The main topics of this section are also presented in the following videos:
Section 4.3 Logarithms and Exponential Models
With the aid of logarithms, we can now gather more information from exponential functions. In this section we will discuss how to do this using various real world examples.
Subsection 4.3.1 Annual to Exponential Growth Rate
Recall that the annual growth rate is the value \(b-1\) in the exponential equation
\begin{equation*}
A(t)=a(b)^t.
\end{equation*}
If we want to find the continuous growth rate which is equivalent to the annual growth rate above, we first rewrite the function:
\begin{align*}
A(t)\amp =a(b)^t\\
\amp =a(e^r)^t.
\end{align*}
Next we simply need to find the value of \(r \) which keeps the above equation true. To solve for \(r \) we apply the properties of the logarithm as discussed in the previous section. Let’s take a look at a specific example of this process.
Example 4.30.
Let \(A(t)=10(.75)^t\text{.}\) Write \(A(t)\) in the form \(A(t)=ae^{rt}\) and state the continuous rate equivalent to the annual decay rate of \(25\% \text{.}\)
Solution.
We want to find values \(a,r \) such that
\begin{equation*}
10(.75)^t=ae^{rt}
\end{equation*}
for all values \(t \text{.}\) Rewriting the right hand side we get
\begin{equation*}
ae^{rt}=a(e^r)^t
\end{equation*}
which now is in the form \(a(b)^t \text{.}\) Since we want
\begin{equation*}
10(.75)^t=a(e^r)^t,
\end{equation*}
we must have \(a=10 \text{,}\) and \(e^r=.75 \text{.}\) To find the appropriate value of \(r \) we just need to solve the equation \(e^r=.75 \) for \(r \text{.}\) Applying natural log to each side we get
\begin{align*}
e^r \amp = .75\\
\ln(e^r)\amp = \ln(.75)\\
r\amp =\ln(.75)
\end{align*}
where this last equality follows due to Property \(4 \) in the previous section. So, we can rewrite \(A(t)\) as
\begin{equation*}
A(t)=10e^{\ln(.75)\cdot t}.
\end{equation*}
Furthermore, since \(\ln(.75)\approx -.2876 \text{.}\) Then a continuous decay rate of about \(28.76\%\) is equivalent to an annual decay rate of \(25\%\text{.}\)
Notice that in this previous example it was not necessary for us to use the natural logarithm. In fact, you could use a logarithm with any base you desire and still get a solution. However, since natural log has base \(e \) it is the inverse of the exponential \(e^t \) thus giving us some cancelations and hence a nicer looking solution.
Subsection 4.3.2 Double Time and Half Life
Another important piece of information we can now gather from exponentials is how to identify which input value corresponds to a given output value.
Example 4.31.
Suppose we are given the function
\begin{equation*}
A(t)=10(1.75)^t
\end{equation*}
which represents the value in dollars of an investment after \(t \) years. Identify the year in which the investment would reach $40.
Solution.
We start by setting the function equal to 40:
\begin{equation*}
40=10(1.75)^t.
\end{equation*}
Now, we divide each side by \(10 \) and apply a logarithm to each side.
\begin{align*}
40\amp = 10(1.75)^t\\
4\amp = (1.75)^t\\
\log(4)\amp = \log (1.75)^t\\
\log(4)\amp = t\cdot \log(1.75)
\end{align*}
Now, we simply divide each side by \(\log(1.75)\) to get
\begin{equation*}
t=\frac{\log(4)}{\log(1.75)}\approx 2.5\ \ \text{years}.
\end{equation*}
Caution 4.32.
In the solution for the last example, it is important that we first divided each side by the coefficient of the exponential (in this last example this coefficient was \(10 \)) before applying the logarithm. If we didn’t do this then, more often than not, we will make a calculation mistake.
This is due to the behavior of logarithms with multiplication. In particular,
\begin{equation*}
\log(10(1.75)^t)=\log(10)+\log((1.75)^t).
\end{equation*}
Often this property of logarithms is forgotten, and thus leads to mistakes. Even if we remember to apply this property, this leads to much more work than is necessary.
There are a few particular input values which are solved for quite frequently. In fact, they are used often enough to have their own name.
Doubling Time.
The Doubling Time is the amount of time it takes for a quantity to double in value or size. If we are finding the doubling time of the exponential function
\begin{equation*}
A(t)=a(b)^t
\end{equation*}
where \(1\lt b \text{,}\) then we are solving the equation
\begin{equation*}
2a=a(b)^t
\end{equation*}
for \(t \text{.}\)
Half Life.
The Half Life is the amount of time it takes for a quantity to shrink to half its value or size. If we are finding the half life of the exponential function
\begin{equation*}
A(t)=a(b)^t
\end{equation*}
where \(b\lt 1 \text{,}\) then we are solving the equation
\begin{equation*}
\frac{1}{2}a=a(b)^t
\end{equation*}
for \(t \text{.}\)
Notice that the requirement for \(b \) to be greater than \(1 \) in doubling time and less than \(1 \) in half life is simply to have the statement make sense. For instance, we can only solve for the time it takes for an investment to double if the investment is increasing in value, and it is only increasing in value when \(1\lt b \text{.}\)
Another important thing to notice in these definitions is that we could solve for the doubling time or half life without knowing the value of the principal (or initial) value \(a \text{,}\) we need only know the rate or growth factor. For example, in the formula for doubling time we are solving for the value of \(t \) in the formula
\begin{equation*}
2a=a(b)^t.
\end{equation*}
If we were to divide each side by \(a \) we would achieve the formula
\begin{equation*}
2=(b)^t.
\end{equation*}
Solving this equation for \(t \) is the same as solving the original equation for \(t \) and yet this new equation does not involve \(a \) at all.
Exercises 4.3.3 Exercises
1. Solving Using Logarithms.
Using the natural logarithm, solve for \(x\) in the equation \(4^x = 8\text{.}\)
\(x =\)
Hint.
For this problem, we are told to use the natural logarithm when solving. So, to do that, we take the natural logarithm of both sides.
Recall the following properties of logarithms:
\begin{equation*}
log_a(x\cdot y)=\log_a(x)+\log_a(y)
\end{equation*}
\begin{equation*}
\log_a(x/ y)=\log_a(x)-\log_a(y)
\end{equation*}
\begin{equation*}
\log_a(x^y)=y\log_a(x)
\end{equation*}
\begin{equation*}
\log_a(a^x)=x
\end{equation*}
\begin{equation*}
a^{\log_a(x)}=x.
\end{equation*}
2. Half Life.
Let \(P(t)=1000 (0.4)^t\) be the amount of bacteria in an experiment \(t\) hours after the experiment started. What is the half-life for the bacteria? (Round to two decimal places.)
hours
Hint.
We want to find out at what \(t\) value half of the original amount is left. Half of \(1000\) is \(500\text{,}\) so we want to solve for \(t\) in
\begin{equation*}
500 = 1000 (0.4)^t
\end{equation*}
Note that you will need to use logarithms in order to do this. We suggest you use natural or common logarithm.
3. Doubling Time.
The population of Middle-earth (in billions) is given by \(P(t)= 7 e^{0.011 t}\text{,}\) where \(t\) is the time in years since 2000. What is the doubling time for the population of Middle-earth? (Round to two decimal places).
years
Hint.
We want to know when the population will be double of \(7\) billion, which is \(14\) billion. That is, we want to solve the equation
\begin{equation*}
14 = 7 e^{0.011 t}
\end{equation*}
for \(t\text{.}\)
Note that you will need to use logarithms in order to do this. We suggest you use natural or common logarithm.
4. Doubling Time.
You deposit 180 dollars into an account that pays an interest rate of 6 \(\%\) per year.
a) Find an equation for the amount you have after \(t\) years.
\(A(t) =\)
b) Use the equation in part (a) to find the doubling time for the amount in your account. Round to two decimal places.
years
Hint.
a) Recall the annual interest formula:
\begin{equation*}
y=P(1+r)^t,
\end{equation*}
where \(P\) is the principal investment and \(r\) is the nominal growth rate.
b) We want to know at what value of \(t\) will the amount in the account be doubled. That is, we want to solve
\begin{equation*}
2P= P (1+r)^t
\end{equation*}
for \(t\) (after plugging in the appropriate values of \(P\) and \(r\)).
Note that you will need to use logarithms to do this, and we recommend using common or natural logarithm.
5. Rate of Growth.
a) Let \(f(t) = 45 e^{- 0.016 t}\text{.}\) Write \(f(t)\) in the form \(y=ab^t\text{.}\) (Round to three decimal places if needed.)
\(y =\)
What is the percent annual decay rate?
percent per year
What is the continuous percent decay rate?
percent
b) Let \(g(t) =330 (1.24)^t\text{.}\) Write \(g(t)\) in the form \(y=ae^{kt}\text{.}\) (Round to three decimal places when needed.)
\(y =\)
What is the percent annual growth rate?
percent per year
What is the continuous percent growth rate?
percent
Hint.
a) We want to write \(f\) as a function of form \(f(t) =ab^t\text{.}\) Consider what you need to do to find the values of \(a\) and \(b\) with the following equality:
\begin{equation*}
45 e^{- 0.016 t}=a(b)^t.
\end{equation*}
It may be useful to recall the following property of exponents: \((b^m)^n=b^{mn}.\)
For this part, we will use the equation we found in part (a). The percent annual decay rate (since the base is less than 1) is given by the growth factor minus 1. Remember to convert into a percentage.
In the equation \(y=ae^{kt}\text{,}\) \(k\) gives the growth/decay rate. Since the problem specifies decay, your answer should be positive. Remember to convert into a percentage.
b) We want to write \(g\) as a function of form \(g(t) = ae^{kt}\text{.}\) Consider what you need to do to find the values of \(a\) and \(k\) with the following equality:
\begin{equation*}
330 (1.24)^t=ae^{kt}.
\end{equation*}
As a hint, you will need to use a logarithm to find the value of \(k\text{.}\)
For the percent annual growth rate and the continuous growth rate, use the same process as in part a).
6. Comparing Rates.
Suppose a population doubles in size every 8 years.
a) Find the percent annual growth rate of the population. (Round to two decimal places.)
percent
b) Find the continuous percent growth rate of the population. (Round to two decimal places.)
percent
Hint.
a) For annual growth, we want to find \(r\) in the equation \(2a = a(1+r)^{8}\text{,}\) where \(a\) is the initial amount.
b) For continuous growth, we want to find \(r\) in the equation \(2a = a(e)^{8 r}\text{,}\) where \(a\) is the initial amount.
For both of these, be sure to convert to a percentage.
7. Calculating Decay.
The half-life of carbon-14, a radioactive isotope of carbon, is 5,730 years.
a) Write an equation for \(P(t)\text{,}\) the percentage of carbon-14 remaining after \(t\) years of radioactive decay.
\(P(t) =\)
b) What is the percent annual decay rate of carbon-14? Round to 4 decimal places.
percent
c) What is the continuous decay rate (as a percentage) of carbon-14? Round to 4 decimal places.
percent
d) In 2014, a tomb containing the alleged remains of Attila the Hun were discovered under the Danube river in Budapest, Hungary. At the time of discovery, tests showed that approximately 93\(\%\) of the carbon-14 originally present in the body remained. If Attila died in 453 A.D., could these be the remains of Attila the Hun?
-
Yes, the remains might have been left in Hungary around 600 AD
-
No, the remains can only have been decaying for about 600 years if they still have \(93 \%\) of the original carbon-14.
-
Yes, the remains were probably preserved well.
-
No, the remains can only have been decaying for about 1400 years if they still have \(93 \%\) of the original carbon-14.
Hint.
a) We want to find an equation to model the percentage of carbon-14 remaining after \(t\) years. We are given that the half-life of carbon-14 is 5730 years. Using this, we can write the equation
\begin{equation*}
\frac{1}{2}P = Pe^{k (5730)}.
\end{equation*}
Note that we are using base \(e\) here, which is fairly standard for radioactive materials, as the decay is continuous. We can use this equation to get the decay rate, which we will use for our model. Note that you will need to use logarithms to solve for \(k\text{.}\)
It may be useful to also consider what the initial value is, since we are dealing with percents. What percent of carbon-14 would be present initially?
b) Recall that the percent annual decay rate will be given by the growth factor minus 1. Consider what the growth factor is here. Remember to convert to a percentage.
c) The value of \(k\) gives the continuous growth rate. Remember to convert to a percentage.
d) We want to find the value of \(t\) at which \(93 \%\) of the carbon-14 would remain. Using the equation from part (a), we set
\begin{equation*}
93 = P(t)
\end{equation*}
and solve for \(t\text{.}\)
The value of \(t\) here will give the number of years the carbon-14 present would have been decaying. In other words, it will give the number of years the person has been dead.
