Applications of Exponential and Logarithmic Functions Lesson

Let's begin our lesson by looking at a model for exponential growth. Earlier in our course, we studied continuously compounded interest. $$A=Pe^{rt}$$ Where A is the account balance after t years, P is the principal or amount invested, e is the special number (Euler's number), r is the interest rate as a decimal, and t is the time in years.
For a given principal, interest rate, and time period in years, this formula will give us the account balance when the interest is continuously compounded. This same formula can be used to describe the "exponential law" or the "law of uninhibited growth". $$n(t)=n_0e^{rt}, r > 0$$ Where n(t) is the population at time t, n₀ is the initial size of the population, r is the relative growth rate, which is expressed as a proportion of the population, and t is the time. When we think about this model for exponential growth, our quantities grow at a rate directly proportional to their size. The r in the formula or relative growth rate is the rate of population growth expressed as a proportion of the population at any time. Let's suppose that in a problem, we are given r = 0.03, then at any time t, the growth rate is 3% of the population at time t. Let's look at an example.
Example #1: Solve each word problem.
For a period of time, an island's population grows exponentially. If the relative growth rate is 3% per year and the current population is 1,442, what will the population be 6 years from now?
To start the problem, let's first create a function that models the island's population after t years. $$n(t)=n_0e^{rt}$$ We know the current population is 1,442. We will use this as the n₀ or initial size of the population. $$n_0=1{,}442$$ $$n(t)=1{,}442e^{0.03t}$$ Since we want to know the island's population 6 years from now, we can just plug in a 6 for t. $$n(t)=1{,}442e^{0.03(6)}≈ 1{,}726$$ Six years from now, the island's population will be about 1,726.

Time to Double

In many problems, we will be asked to find the doubling time or time to double. The formula is very easy to derive. $$n(t)=n_0e^{rt}$$ Since we are doubling the initial amount, the n(t) will be 2n₀: $$2n_0=n_0e^{rt}$$ Divide both sides by n₀: $$2=e^{rt}$$ Take the natural log of each side: $$\ln(2)=\ln(e^{rt})$$ $$rt=\ln(2)$$ Solve for t, divide both sides by r: $$t=\frac{\ln(2)}{r}$$ Let's look at an example.
Example #2: Solve each word problem.
For a period of time, a colony of bacteria grows exponentially. If the relative growth rate is 4.5% per day and the initial population is 100, how long will it take for the population to double in size?
Here we can use our formula from above and just plug in for r, the relative growth rate. $$t=\frac{\ln(2)}{r}$$ $$r=0.045$$ $$t=\frac{\ln(2)}{0.045}≈ 15.4$$ The population of the bacteria colony will take about 15.4 days to double in size.

Exponential Growth (Doubling Time)

In some problems, we will be given the doubling time or maybe even the tripling time. When this happens, it's usually better to write a model with a base of 2 (doubling) or 3 (tripling). You can still use e as the base if you would like.
If the initial size of a population is n₀ and the doubling time is a, then the size of the population at time t is given by: $$\large{n(t)=n_{0}\cdot 2^{\frac{t}{a}}}$$ where a and t are measured in the same units of time (seconds, minutes, hours, days, years, and so on). Let's look at an example.
Example #3: Solve each word problem.
Under ideal conditions, a certain bacteria population doubles every six hours. Initially, there are 25 bacteria in a colony.
a) Find a model for the bacteria population after t hours.
The initial size of the population is 25, this will be our n₀. Additionally, we are given that the population doubles every six hours. This will be our value for a. $$\large{n(t)=n_{0}\cdot 2^{\frac{t}{a}}}$$ $$\large{n_0=25, a=6}$$ $$\large{n(t)=25 \cdot 2^{\frac{t}{6}}}$$ b) How many bacteria are in the colony after 12 hours?
We just need to plug in a 12 for our variable t. $$\large{n(12)=25 \cdot 2^{\frac{12}{6}}}$$ $$\large{n(12)=25 \cdot 2^{2}=100}$$ After 12 hours, the number of bacteria is 100.
c) After how many hours will the bacteria count reach 800?
We will replace n(t) with 800 and solve for t. $$\large{800=25 \cdot 2^{\frac{t}{6}}}$$ Divide both sides by 25: $$\large{32=2^{\frac{t}{6}}}$$ Take the base 2 log of each side: $$\large{32=2^{\frac{t}{6}}}$$ $$\large{\log_2(32)=\log_2(2^{\frac{t}{6}})}$$ Simplify: $$\frac{t}{6}=5$$ Multiply both sides by 6: $$t=30$$ The bacteria level reaches 800 after 30 hours.
d) How could we use a base of e instead of a base of 2 to obtain a model?
In our lesson on special properties of exponential and logarithmic expressions, we saw the following:
For a > 0 and a ≠ 1: $$a^{\log_a(x)}=x, x > 0$$ Applying this to our current situation. $$e^{\ln(b)}=b, b > 0$$ $$y=ab^{x}$$ Can be converted into: $$y=ae^{\ln(b) \cdot x}$$ Let's return to the problem and think about our model from part a: $$\large{n(t)=25 \cdot 2^{\frac{t}{6}}}$$ To use the base of e instead of 2. $$e^{\ln(2)}=2$$ $$\large{n(t)=25 \cdot e^{\frac{\ln(2) \cdot t}{6}}}$$

Exponential Decay

What happens when we use our formula that describes the exponential law and the r-value is negative? $$n(t)=n_0e^{rt}, r < 0$$ Since t represents time and must be non-negative, for positive values of t, the exponent rt will now be negative. From the rules of exponents: $$n(t)=n_0 \cdot \left(\frac{1}{e}\right)^{rt}, r > 0$$ Here we can see the base is between 0 and 1, which means we have a function that is decreasing over its entire domain.

Radioactive Decay Model

We can derive a simple formula for dealing with radioactive decay. For these problems, we are given a half-life h. $$m(t)=m_{0}e^{rt}, r < 0$$ Given the half-life of a substance (h) or the amount of time for half of a given quantity of that substance to decay away, we can set up the following: $$\frac{1}{2}m_0=m_0e^{rh}$$ The h is plugged in for t. After the time period of h, the initial amount will be half. $$\frac{1}{2}m_0=m_0e^{rh}$$ Divide both sides by m₀: $$\frac{1}{2}=e^{rh}$$ Take the natural log of each side: $$\ln\left(\frac{1}{2}\right)=\ln(e^{rh})$$ $$rh=\ln\left(\frac{1}{2}\right)$$ Divide both sides by h: $$r=\frac{\ln\left(\frac{1}{2}\right)}{h}$$ Convert ln(1/2) into -ln(2): $$r=-\frac{\ln(2)}{h}$$ If m₀ is the initial mass of a radioactive substance with half-life h, then the mass remaining at time t is given by: $$m(t)=m_{0}e^{-rt}$$ $$r=\frac{\ln(2)}{h}$$ r is known as the relative decay rate.
Use the above formula carefully. Notice there was a slight change with r versus what we derived above. We applied the negative out in front of the rt part and then made the formula for r positive. Keep this in mind as you work on these problems. Since you are dealing with exponential decay, the part multiplying t is going to be negative. Let's look at an example.
Example #4: Solve each word problem.
Berkelium-247 has a half-life of about 1,400 years. If a sample of this substance has a current mass of 755.9 ounces, what will the mass be 8,000 years from now?
Let's begin with our model. $$m(t)=m_{0}e^{-rt}$$ We are given the initial amount of 755.9, this is our m₀. $$m_0=755.9$$ $$m(t)=755.9e^{-rt}$$ We can calculate r using our formula. $$r=\frac{\ln(2)}{h}$$ The half-life h is 1,400. $$r=\frac{\ln(2)}{1{,}400}$$ Plug in for r (don't forget the negative in front): $$m(t)=755.9e^{-\frac{\ln(2)}{1{,}400}t}$$ Plug in 8,000 for t: $$m(8{,}000)=755.9e^{-\frac{\ln(2)}{1{,}400}\cdot(8,000)}$$ $$m(8{,}000) ≈ 14.4 \hspace{.1em}\text{ounces}$$ 8,000 years from now, the mass will be about 14.4 ounces.
It's worth noting that many people prefer to solve this type of problem using a base of 1/2. $$m(t)=m_0 \cdot \left(\frac{1}{2}\right)^{\frac{t}{a}}$$ where m(t) is the mass remaining at time t, m₀ is the initial amount, and a is the half-life. We will also state that a and t are measured in the same time units (second, minutes, hours, days, years, and so on). Let's try the problem again with this approach. $$m(t)=m_0 \cdot \left(\frac{1}{2}\right)^{\frac{t}{a}}$$ Plug in a 755.9 for m, 8,000 for t, and 1,400 for a. $$m(8{,}000)=755.9 \cdot \left(\frac{1}{2}\right)^{\frac{8{,}000}{1{,}400}}$$ $$m(8{,}000) ≈ 14.4\hspace{.1em}\text{ounces}$$ Notice that we get the same result as before. From what we discussed earlier, this should be expected: $$e^{-\ln(2)}=\frac{1}{2}$$

Carbon Dating

Another common example of exponential decay is carbon dating. This is sometimes used to determine the age of certain fossils or artifacts. The method is based on the percentage of carbon-14 that remains in the fossil or artifact since carbon-14 decays according to the exponential law with a half-life of about 5,730 years. Note this half-life for carbon-14 is an estimate and varies based on your textbook. Since the problem given may use a different half-life, always go with what's in the problem to obtain a matching answer. Let's look at an example.
Example #5: Solve each word problem.
Assuming the half-life of carbon-14 is 5,730 years, suppose the animal bones from an ancient burial site contains 65% of the carbon-14 that is present in the same living animals. How long ago were these animals alive?
Let's begin with our model from the previous problem. $$m(t)=m_{0}e^{-rt}$$ Let's find r first, remember that h is the half-life. $$r=\frac{\ln(2)}{h}$$ The half-life h is 5,730. $$r=\frac{\ln(2)}{5{,}730}$$ Plug into the formula, we want m(t) to be 65% of m₀: $$m(t)=m_{0}e^{-rt}$$ Again, notice the negative in front of r. $$0.65m_0=m_{0}e^{-\frac{\ln(2)}{5{,730}}t}$$ Divide both sides by m₀: $$0.65=e^{-\frac{\ln(2)}{5{,730}}t}$$ Take the natural log of each side: $$\ln(0.65)=\ln\left(e^{-\frac{\ln(2)}{5{,730}}t}\right)$$ $$-\frac{\ln(2)}{5{,}730}t=\ln(0.65)$$ Solve for t: $$t=-\frac{5{,}730 \cdot \ln(0.65)}{\ln(2)}≈ 3{,}561$$ These animals were alive about 3,561 years ago.

Logistic Growth Models

When we think about growth, nothing can realistically grow exponentially indefinitely. Growth is always limited. We can use the logistic growth model for this type of situation.
A model for limited logistic growth is given by: $$f(t)=\frac{c}{1 + ae^{-bt}}$$ Where a, b, and c are constants, with c > 0 and b > 0. As time → ∞, f(t) gets closer and closer to c. y = c is a horizontal asymptote for the graph of the function f(t). This means c represents the limiting size that f(t) can attain.
Let's look at an example.
Example #6: Solve each word problem.
In a remote village with a population of 10,000 plants, a rare plant disease outbreak occurs. The number of infected plants after t weeks is described by the following function. $$f(t)=\frac{10{,}000}{1 + 15e^{-0.8t}}$$ a) How many plants were infected by the end of the third week?
For this problem, we just need to plug in a 3 for t. $$f(3)=\frac{10{,}000}{1 + 15e^{-0.8(3)}}≈ 4{,}236$$ After three weeks about 4,236 plants were infected.
b) How many weeks does it take for 7,000 plants to become infected?
For this problem, we just need to replace f(t) with 7,000 and solve for t: $$7{,}000=\frac{10{,}000}{1 + 15e^{-0.8t}}$$ Cross multiply: $$7{,}000(1 + 15e^{-0.8t})=10{,}000$$ Divide both sides by 7,000 $$1 + 15e^{-0.8t}=\frac{10}{7}$$ Subtract 1 away from each side: $$15e^{-0.8t}=\frac{10}{7}- \frac{7}{7}$$ $$15e^{-0.8t}=\frac{3}{7}$$ Divide both sides by 15: $$e^{-0.8t}=\frac{1}{35}$$ Take the natural log of each side: $$\ln(e^{-0.8t})=\ln\left(\frac{1}{35}\right)$$ $$-0.8t=\ln\left(\frac{1}{35}\right)$$ Divide both sides by -0.8: $$t=\frac{\ln\left(\frac{1}{35}\right)}{-0.8}$$ $$t ≈ 4.4$$ It takes about 4.4 weeks for 7,000 plants to become infected.

Newton's Law of Cooling

If D₀ is the initial temperature difference between an object and its surroundings, and if its surroundings have temperature T_s, then the temperature of the object at time t is given by: $$T(t)=T_s + D_0 \cdot e^{-kt}$$ Note: in this formula, the capital letter "T" is used for temperature, while the lowercase "t" is used for time. Let's look at an example.
Example #7: Solve each word problem.
A freshly baked chocolate cake is taken out of the oven, initially having a temperature of 200°F. The baker places the cake in a refrigerator that has a temperature of 35°F to cool it down. After 15 minutes, the baker checks the cake's temperature and it has dropped to 160°F. The baker intends to ice the cake when its temperature reaches 50°F. After how many minutes will the baker be able to ice the cake?
Let's begin by revisiting our formula: $$T(t)=T_s + D_0 \cdot e^{-kt}$$ T_s is the temperature of the refrigerator (35) and D₀ is 165 (200 - 35). Let's plug those values in and see where we are: $$T(t)=35 + 165 \cdot e^{-kt}$$ We know that we want to figure out time, which is t. At this point, we don't know the value of k. How can we find that? In the problem, it tells us that after 15 minutes, the cake's temperature has dropped to 160°F.
Let's plug in 160 for T(t) and 15 for t: $$160=35 + 165 \cdot e^{-15k}$$ Subtract 35 away from each side: $$125=165 \cdot e^{-15k}$$ Divide both sides by 165: $$\frac{25}{33}=e^{-15k}$$ Take the natural log of each side: $$\ln\left(\frac{25}{33}\right)=\ln(e^{-15k})$$ $$-15k=\ln\left(\frac{25}{33}\right)$$ Divide both sides by -15: $$k=-\frac{\ln\left(\frac{25}{33}\right)}{15}≈ 0.0185$$ Now, let's return to the problem. The baker wants the cake to be 50°F in order to ice the cake. $$T(t)=35 + 165 \cdot e^{-kt}$$ Plug in a 50 for T(t) and a 0.0185 for k (notice the negative in front of k in the formula). $$50=35 + 165 \cdot e^{-0.0185t}$$ Subtract 35 away from each side: $$15=165 \cdot e^{-0.0185t}$$ Divide both sides by 165: $$\frac{1}{11}=e^{-0.0185t}$$ Take the natural log of each side: $$\ln\left(\frac{1}{11}\right)=\ln(e^{-0.0185t})$$ $$-0.0185t=\ln\left(\frac{1}{11}\right)$$ Divide both sides by t: $$t=\frac{\ln\left(\frac{1}{11}\right)}{-0.0185}≈ 130$$ The time required for the cake to drop to 50°F and allow the baker to ice the cake is about 130 minutes.

---

Skills Check: