7.4 Applications
- Page ID
- 155747
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- implement exponential and logarithmic functions and equations to solve common problems in science and business
In this section, I want to demonstrate common real-world application problems one-by-one, doing an example with you and then giving you a related exercise. There are just a bunch of different configurations these problems can take, and I think a learn-by-example approach will be the most efficient. In these problems, we will generally need to use calculators/technology. Let's start in the area of money...
Say you make an initial investment of \($P\) at an annual interest rate of 4%. That means that after one year, 4% of \(P\) will be calculated and added on to the amount in your account. Thus you will have \(P(1+0.04)\) dollars in the account. After another year, the same thing happens, but with your new total, so that you have \( (P(1.04) )(1+0.04) = P(1.04)^2 \). Each year, you pick up another factor of \( (1.04)\), so if you want to compute the amount of money after \(t\) years at interest rate \(i = 0.04\), you can use an exponential function
\[ A(t) = P(1+i)^t. \notag \]
In reality, interest can be calculated and paid to you more than once a year, so let's start with that concept.
When interest on an investment is compounded multiple times a year (quarterly, monthly, weekly, etc.), an exponential function is used to compute the amount of money in the account after a certain number of years. In the formula below, \(P\) (the principal) is the initial amount, \(r\) is the (nominal) interest rate per year (even though it's compounded \(n\) times per year) expressed as a decimal, and \( A(t)\) is amount after \(t\) years.
\[ A(t) = P \left( 1 + \frac{r}{n} \right)^{nt} \notag \]
- Explain why the formula makes sense for this scenario.
- If $1000 is invested at an interest rate of 8% per year, compounded semiannually, find the amount in the account after 5 years.
- For the same situation, find the interest rate that, compounded annually, would result in the same amount of money after 5 years.
- Now find the amount in the account after 5 years if interest is compounded quarterly. What about monthly? What do you notice about the amounts the more times you compound?
- How long will it take for the account to double in the situation of part (2.)?
Solution
1. Every time the interest is compounded, the percentage used to calculate the interest money is \( \frac{r}{n} \). To find the total amount including the principal and this extra interest money added on after one compounding period, we would multiply \( P \left( 1 + \frac{r}{n} \right) \). In a year, there are \(n\) periods, and every single time, we are multiplying by another copy of \( \left( 1 + \frac{r}{n} \right)\). Thus, after \(t\) years, aka \(nt\) periods, we will have \(A(t) = P \left( 1 + \frac{r}{n} \right)^{nt} \).
2. We identify all the information given to us: \(P = 1000\), \(r = 0.08\), \(n = 2\) (twice a year), and we want to plug in \(t = 5\) years. Putting all of this into the formula, we are computing \(A(5) = 1000\left( 1 + \frac{0.08}{2} \right)^{2(5)} \). Simplifying and then plugging into a calculator, we have
\[A(5) = 1000\left( 1.04 \right)^{10} = $1480.24 \notag \]
3. If interest is only being calculated once per year, we would have some rate \(i\) and calculate the amount after \(t\) years using \(P(1+i)^t\). What we want is to end up with the exact same amount of money after 5 years as our previous answer, $1480.24. We can set that amount equal to our hypothetical annual situation and solve for that new rate \(i\).
\[ $1480.24 = 1000(1+i)^5 \quad \rightarrow \quad (1+i)^5 = 1.48024 \quad \rightarrow \quad 1+i = \sqrt[5]{1.48024} \approx 1.081599... \notag \]
This tells us that \(i\) must be 8.16% to achieve the same amount, if interest is only compounded once per year.
4. All we need to do here is go back to our compound interest formula and adjust the \(n\) accordingly. First, for quarterly (there are four quarters in a whole year) we use \(n = 4\). We compute \(A(5) = 1000\left( 1 + \frac{0.08}{\textcolor{red}{4}} \right)^{\textcolor{red}{4}(5)} = $1485.95\) in this case. Next, for monthly, there are twelve months in a year so we use \(n = 12\). We compute \(A(5) = 1000\left( 1 + \frac{0.08}{\textcolor{red}{12}} \right)^{\textcolor{red}{12}(5)} = $1489.85\) in this case. We notice that every time we increase \(n\), the amount of money goes up. The more times you compound, the more interest you make in the end.
5. If we want the account to double and we started with \( P = 1000\), we want to have \(A(t) \) turn out to be 2000. We just plug that in where we see \(A(t)\) and solve for the amount of time \(t\) needed to make the equation true. I first plug in all my numbers and simplify a bit.
\[ 2000 = 1000\left( 1 + \frac{0.08}{2} \right)^{2t} \quad \rightarrow \quad (1.04)^{2t} = 2 \notag \]
Now I see my variable \(t\) trapped upstairs. I apply \( \ln\) to both sides so I can bring him down outta there. Then finish isolating \(t\).
\[ \ln ( (1.04)^{2t}) = \ln 2 \quad \rightarrow \quad 2t \ln (1.04) = \ln 2 \quad \rightarrow \quad t = \frac{ \ln 2}{2 \ln (1.04)} \approx 8.84 \text{ years} \notag \]
Ian invests $1,000,000 at a (nominal) rate of 15% per year, compounded quarterly.
- Write the function that will compute the amount of money in Ian's account after \(t\) years.
- Fill in the table of values for different \(t\) values below.
\(t\) \( A(t) \) \(t\) \(A(t)\) 1 5 2 10 - Find the interest rate that, compounded annually, would result in the same amount of money after one year as in the compounded quarterly situation. This is called the annual percentage yield.
- How long will it take for the account to reach $3,000,000?
- Answer
-
- \(A(t) = 1000000 \left( 1 + \dfrac{0.15}{4} \right)^{4t} = 1000000 (1.0375)^{4t} \).
-
\(t\) \( A(t) \) \(t\) \(A(t)\) 1 $1,158,650 5 $2,088,150 2 $1,342,470 10 $4,360,380 - We want the factor \( \left(1+\frac{.15}{4}\right)^4 \) to have the same effect as a factor \( (1+i)^1 \). Setting them equal and solving for \(i\) gives \(0.15865\), so about 15.87%.
- \(t \approx 7.46\) years.
If we increase the number \(n\) of compounding periods to every day...twice a day...every hour...every minute...every second...(!) until \(n \rightarrow \infty\), we are talking about continuously compounding interest.
The formula that calculates the amount of money \(A(t)\) in an account after \(t\) years, with principal \(P\) and interest rate \(r\) compounded continuously, is
\[ A(t) = P e^{rt} \notag \]
- If $5000 is invested at an interest rate of 5% per year, compounded continuously, write the function \(A(t)\).
- How much money will be in the account after 10 years?
- For any initial investment \(P\), how long will it take for the money to double?
- What is the equivalent percentage annual yield? (That is, the rate \(i\) for which \((1+i)^1\) is equal to \( e^{r(1)} \).)
Solution
1. We identify \(P = 5000\) and \(r = 0.05\), and write our function \(A(t) = 5000e^{0.05t} \).
2. We are looking for \(A(10)\), so we just plug in \(t = 10\) and compute \(A(10) = 5000e^{0.05(10)} = $8243.61\).
3. If we put in \(P\) and we want double that, we simply want \(A(t)\) to turn out to be \(2P\). Plug that in and solve for \(t\). It's a good idea to use \(\ln\) here because it has base \(e\).
\[ 2P = P e^{0.05t} \quad \rightarrow \quad e^{0.05t} = 2 \quad \rightarrow \quad 0.05t = \ln 2 \quad \rightarrow \quad t \approx 13.86 \text{ years} \notag \]
4. We want to solve for \(i\) in the equation \( 1+i = e^{0.05} \). Plugging in \(e^{0.05}\) to a calculator, we get 1.051271... so we know \(i \approx 0.05127\) or about 5.13%.
A sum of $45,000 is invested at a rate of 13% compounded continuously.
- Write the function \(A(t)\) for this scenario.
- Fill in the table of values for different times \(t\).
\(t\) \( A(t) \) \(t\) \(A(t)\) 1 5 2 10 - How long will it take for the investment to double in value?
- Answer
-
- \(A(t) = 45000e^{0.13t} \)
-
\(t\) \( A(t) \) \(t\) \(A(t)\) 1 $51,247 5 $86,199 2 $58,362 10 $165,118 - \(t \approx 5.33\) years.
Switching gears to biology: exponential functions show up all the time in population growth modeling...
Bacteria reproduce by splitting themselves in two every certain amount of time. Thus, after a length of time \(a\) (minutes, hours, days, years, whatever), their population will double. If you start with a population of size \(n_0\), then to find the size of the population \(n(t)\) at time \(t\) (minutes, hours, days, years, whatever, consistent with \(a\)'s units), we can use the formula
\[ n(t) = n_0 2^{t/a} \notag \]
- Explain why this formula makes sense.
- If we start with a Petri dish containing 200 bacteria, and they are doubling every 3 hours, how many bacteria will we have after 24 hours?
- How long will it take to reach 1000 bacteria?
Solution
1. Say the population is doubling every half hour, so that \(a = 0.5\)hr. Then after \(t \) hours, the population will have doubled (multiplied itself by 2) a total of \(2t\) times. So if we started with \(n_0\) and multiplied it by 2 over and over, we will end up with \(n(t) = n_0 2^{t/0.5} = n_0 2^{2t} \) after \(t\) hours.
2. We identify \(n_0 = 200\) and \(a = 3\)hrs. At \(t = 24\), we can just calculate \(n(24) = 200 \cdot 2^{24/3} = 200 \cdot 2^{8} = 51,200\) bacteria.
3. All we are saying here is that we want to have \(n(t)\) turn out to be 1000 for some \(t\). We just plug in what we want and solve for \(t\). I'll rescue the \(t\) trapped upstairs using \( \ln\) since he's my favorite log...
\[ 1000 = 200 \cdot 2^{t/3} \quad \rightarrow \quad 5 = 2^{t/3} \quad \rightarrow \quad \ln 5 = \frac{t}{3} \ln 2 \notag \]
Then just finish solving to get \(t = 3\frac{ \ln 5}{\ln 2} \approx 6.97\) hours.
Rabbits have discovered the strawberry patch :( and on this nutritious banquet provided free of charge, their population is booming. At this point, the rabbit population in the garden is about 10 and doubling every 28 days.
- Write the formula that gives the rabbit population after \(t\) weeks.
- There are 15 more weeks of summer. If this keeps up, what size will the population be at the end of the summer?
- How long will it take to reach my lucky number of 51 rabbits?
- Answer
-
- \(n(t) = 10\cdot 2^{t/4} \) (watch out for the differences in units)
- \(n(15) = 135\) rabbits (round to a whole number, you can't have a half a rabbit (alive at least))
- \(t \approx 9.4\) weeks.
In the population-doubling scenario, the base used in the exponential function is 2, but this is not required. You could talk about "tripling time" and use base 3... If we choose to use \(e\) as the base in a model, giving for example \(n(t) = n_0 e^{rt}\), then \(t\)'s coefficient \(r\) is called the relative growth rate, which is the rate of growth expressed as a proportion of the population at any time. This scenario is basically the same as compound interest, just with different letters, lol.
The civilian noninstitutionalized population (people 16yo and up, not living in institutions, not on active military duty) in the United States is projected to increase at a relative growth rate around 0.41% between 2023 and 2053 (down from a rate of 1.1% over the past 40 years). The mathematical model for this exponential growth is
\[ n(t) = n_0 e^{rt} \notag \]
where \(n(t)\) gives population in millions, \(t\) years after 2023, with initial population size \(n_0\), and relative growth rate \(r\).
- In 2023, the civilian noninstitutionalized population was 266 million people. Find the formula that will estimate the projected population \(t\) years after 2023 in this scenario.
- Estimate the population at the end of this projection period, in 2053.
- In what year do you expect the population to cross 275 million?
Solution
1. We identify \(n_0 = 266\) and \( r = 0.0041\). Then we just plug into the formula to get \(n(t) = 266e^{0.0041 t} \).
2. The year 2053 is 30 years after 2023, so \(t = 30\). We plug this into the formula we just found to get \(n(30) = 266 e^{0.0041(30)} \approx 301\) million people.
3. If we want to know when \(n(t) = 275\), we just set the formula equal to that and solve for the necessary \(t\), using \(\ln\) to take care of the \(e\) situation.
\[ 275 = 266e^{0.0041t} \quad \rightarrow \quad 1.033835 = e^{0.0041t} \quad \rightarrow \quad 0.0041t = \frac{\ln 1.033835}{0.0041} \approx 8.12 \notag \]
So we expect it to happen sometime in 2031.
A few weeks have passed, the strawberry patch has been picked clean, and the sudden lack of food is having a negative effect on the rabbit population in our yard. I've observed that the population is declining at a relative rate of 10% per week. (This means you should use a negative number as your relative growth rate.) Currently, I count about 17 rabbits in the yard.
- Write a model for the population \(n(t)\) we expect \(t\) weeks from now.
- Assuming trends stay the same, approximately when will we be down to just two rabbits in the yard?
- Answer
-
- \( n(t) = 17 e^{-0.1 t} \)
- \( t \approx 21\) weeks from now.
Radioactive substances decay as they emit radiation over time. The rate of decay is proportional to the mass of the substance, which is similar to the population growth situation, but decreasing, like the rabbit population we just saw. Physicists talk about this process in terms of half-life: the time it takes for a sample to decay to half its original mass. However, it's still the general move to express the mathematical model for a radioactive decay situation in terms of base \(e\).
A substance has initial mass \(m_0\) and half-life \(h\). We want to express an exponential model for its radioactive decay in the form
\[ m(t) = m_0 e^{-rt} \notag \]
where \(r\) is the relative decay rate (it'll be positive, but the effect will be decay because of the negative sign in the exponent).
- Find a way to express \(r\) in terms of the half-life time \(h\).
- If I have 10 grams of radium-226, which has a half-life of 1,600 years, then how long (in years) will it take to decay to 6 grams?
Solution
1. If the amount of time \(h\) is the half-life, that means that when \(t = h\), we have \(m(t) = \frac{m_0}{2} \). We can just plug this data point into the model and solve for \(r\).
\[ \frac{m_0}{2} = m_0 e^{-rh} \quad \rightarrow \quad \frac{1}{2} = e^{-rh} \quad \rightarrow \quad \ln (0.5) = -rh \notag \]
To finish isolating \(r\), we divide both sides by \(-h\) and find that, given \(h\), you can compute \(r = -\frac{\ln(0.5)}{h}\), or equivalently and prettier, \(r = \frac{\ln 2}{h} \).
2. We identify that \(m_0 = 10\) and \(h = 1600\), and that I want \(m(t) = 6\). We just figured out that we can compute \(r = \frac{ \ln 2}{1600} \approx 0.0004332 \). Now we can just plug everything into the general formula and solve for \(t\).
\[ 6 = 10 e^{-0.0004332 t} \quad \rightarrow \quad 0.6 = e^{-0.0004332 t} \quad \rightarrow \quad t = \frac{\ln 0.6}{-0.0004332} \approx 1179.19 \text{ years} \notag \]
Does this answer make sense? Always stop and check. Going from 10g to 6g is decaying by less than half, so I expect my answer to be less than the half-life, 1600 years. So yeah, 1180ish years sounds pretty reasonable.
The substance used for carbon dating of ancient objects, carbon-14, has half-life 5,730 years. If I have a wooden artifact containing 3mg of carbon-14 now, and I know that originally it should have contained 12mg of carbon-14 (based on the amount that is present in non-aged wood), about how old is it?
- First find \(r\) and the formula for \(m(t)\).
- Of the two numbers given, 3mg and 12mg, which one is \(m_0\) and which one is a value for \(m(t)\)?
- Plug in the information appropriately and solve for \(t\).
- Answer
-
- \(r = 0.00012097\) so \(m(t) = m_0 e^{-.00012197t} \)
- \(m_0 = 12\) and \(m(t) = 3\) now.
- \(t \approx 11,366\) years old.
As evidenced by the rabbit population situation, maybe plain old exponential growth isn't always the best way to model population, because in real life, a given environment can't actually support an unlimited population. Limited resources result in a set maximum population possible, called the carrying capacity. Populations start off growing exponentially, but eventually the limiting factors cause them to level off around the carrying capacity. The way to model these situations is using a logistic function.
A population that experiences logistic growth increases according to the model
\[ n(t) = \frac{M}{1+\frac{M-n_0}{n_0} e^{-rt}}\notag \]
where \(n(t)\) is the population at time \(t\) years, \(M\) is the carrying capacity (a plain ol' number), \(r\) is the initial relative growth rate, and \(n_0\) is the initial population.
- My koi pond can support a maximum population of 40 fish. I introduce 4 koi fish as my initial population, and the (highly technical) pet store told me their initial relative growth rate is \(0.15\). Find the logistic growth function that will model the population in my pond.
- When will I have 20 fish?
- Graph the function using technology to see what will happen "eventually" as time goes on forever.
Solution
1. We identify \(M = 40\), \(r = 0.15\), and \(n_0 = 4\). The model is thus
\[ n(t) = \frac{40}{1+\frac{40-4}{4} e^{-0.15t}} = \frac{40}{1+9e^{-0.15t}} \notag \]
2. I want \(n(t) = 20\), so I plug that in and start solving for \(t\).
\[ 20 = \frac{40}{1+9^{-0.15t}} \quad \rightarrow \quad (1+9e^{-0.15t})(20) = 40 \quad \rightarrow \quad 1+9e^{-0.15t} = 2 \notag \]
Now we just isolate the exponential term and solve as usual.
\[ \rightarrow \quad e^{-0.15t} = \frac{1}{9} \quad \rightarrow \quad -0.15 t = \ln \left( \frac{1}{9} \right) \quad \rightarrow \quad t \approx 14.6 \text{ years} \notag \]
3. This is the graph of the function:
We can see that the population first grows quickly and then slows down, leveling out around the carrying capacity as time goes on.
An infectious disease begins turning people into zombies in a small city in Florida (obviously). The city has a population of 12,000, and initially 4 downhome country boys are infected after swimming in a particular swamp hole. The initial relative growth rate of the epidemic is 0.67 per day, and the number of people infected after \(t\) days can be predicted by the model
\[ n(t) = \frac{8000}{1+1999e^{-.67t}} \notag \]
- Will the entire city's population become zombies?
- After 1 week, how many zombies will there be?
- Answer
-
- This model seems to fit the formula for logistic growth, so we can identify the carrying capacity as \(M = 8000\) by looking at the numerator. We confirm that \(\frac{8000-4}{4} = 1999 \) accounts for the correct coefficient in the denominator as well. So we expect the infection to spread initially but level out around 8000 zombies in the long run.
- At \(t = 7\) days, we compute \( n(7) \approx 413\) zombies.
Okay, last little topic here is Newton's Law of Cooling, which you'll learn a lot more about if you take Differential Equations.
If an object with internal temperature \(T_0\) is placed into an environment with surrounding temperature \(T_s\), then it will cool down to match the surrounding temperature as time goes on. The temperature \(T(t)\) of the object at time \(t\) is given by the formula
\[ T(t) = T_s + (T_0 - T_s)e^{-rt} \notag \]
where \(r\) is the coefficient of heat transfer, depending on the type of object.
- Hold \(T_s\) constant and look at graphs (using technology) trying various options for \(T_0\). What do you notice about the rate of cooling relative to the difference between \(T_0\) and \(T_s\)?
- I take a turkey out of the oven with an internal temperature of \(165^\circ\)F and place it in an ambient room temperature of \(65^\circ\)F. After about 20 minutes, it's cooled to a temperature of \(125^\circ\)F. Find \(r\).
- After 30 minutes, what will be the internal temperature of the turkey?
- I want to know how long it will take for the turkey to cool to various temperatures. First find a function that expresses time \(t\) in terms of a desired temperature \(T\) (aka the inverse function to \(T(t)\), which you could name \(t(T)\) to be clear). Then use that function and technology to fill in the table of values.
Temperature \(T\) time to reach \(T\) Temperature \(T\) time to reach \(T\) 150 100 130 70
Solution
1. Using desmos, I generated these graphs by choosing \(T_s = 20\) and \(r = 0.1 \) and then picking a few different options for \(T_0\).
We observe that when the difference between the object's temperature and the surrounding temperature is large, the object's temperature drops rapidly, initially. You can see that when \(T_0 = 60\), much higher than \(T_s = 20\), the blue graph falls steeply at the beginning of time. However, as the gap between the temperatures gets smaller, the rate of cooling slows down; i.e. each graph begins to level out, getting less steep, as time goes on.
2. We are given essentially two data points here: at \(t = 0\), we have \(T(0) = T_0 = 165\), and at \( t = 20\), we have \(T(20) = 125\). We want a function of the form \( T(t) = T_s + (T_0 - T_s)e^{-rt} \) that passes through both of these points. First off, we recognize that \(T_s = 65\) and \(T_0 = 165\). Let's plug in the values of the second data point:
\[ \textcolor{red}{125} = 65 + (165-65)e^{-r\textcolor{red}{(20)}} \quad \rightarrow \quad 125 = 65 + 100e^{-20r} \notag \]
This gives us an equation with only one unknown: the \(r\) that we are trying to figure out. Solve for \(r\).
\[ 60 = 100 e^{-20r} \quad \rightarrow \quad 0.6 = e^{-20r} \quad \rightarrow \quad \ln(0.6) = -20 r \quad \rightarrow \quad r \approx 0.02554 \notag \]
We can update our model for this situation to be \(T(t) = 65+ 100 e^{-0.02554t} \).
3. At \(t = 30\), we have \(T(30) = 65+100e^{-0.02554(30)} \approx 111^\circ\)F.
4. We have done this type of problem a few times thus far, but generally by taking specific numbers that we plug in for \(T(t)\) and then solving for \(t\). This time, we're going to solve for \(t\) in general first, and then use the result to do computations with technology. Aka, we want to see \(t\) as depending on some input temperature \(T\), so we'll write our function as \(t(T)\) in the end.
\[ T = 65+ 100 e^{-0.02554t} \quad \rightarrow \quad \frac{T-65}{100} = e^{-0.02554t} \quad \rightarrow \quad -0.02554t = \ln \left( \frac{T-65}{100} \right) \quad \rightarrow \quad t(T) = -\frac{1}{0.02554}\ln \left( \frac{T-65}{100} \right) \notag \]
So for any \(T\) we are interested in, we can just plug that value into the right hand side of the equation above and use a calculator to compute the time it would take to cool to that temperature. I'm going to use WolframAlpha to help me fill in the table efficiently by typing in "table of values -(1/0.02554) ln((T-65)/100) for T = 150, 130, 100, 70" and hitting Enter. I'll round my answers to one decimal place.
| Temperature \(T^\circ\)F | time to reach \(T\) | Temperature \(T^\circ\)F | time to reach \(T\) |
| 150 | 6.4 min | 100 | 41.1 min |
| 130 | 16.9 min | 70 | 117.3 min (almost 2 hours) |
The melting point of a certain metal is \(800^\circ\)C. I melt a small amount of metal in a crucible and pour it into a mold, leaving it in an ambient temperature of \(20^\circ\)C to cool off. After 60 minutes, it has cooled to \(49^\circ\)C. Find the appropriate \(r\) for the model
\[ T(t) = T_s + (T_0 - T_s)e^{-rt} \notag \]
and write a function that takes as input a particular temperature and gives as output the time it takes to cool to that temperature. Fill in the table below using this formula.
| Temperature \(T^\circ\)C | time to reach \(T\) | Temperature \(T^\circ\)C | time to reach \(T\) |
| 90 | 40 | ||
| 60 | 30 |
- Answer
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We find \(r = 0.055\) and \(T(t) = 20+(800-20)e^{-0.055t}\). Then \(t(T) = -\frac{1}{0.055}\ln \left( \frac{T-20}{780} \right) \). Using technology,
Temperature \(T^\circ\)C time to reach \(T\) Temperature \(T^\circ\)C time to reach \(T\) 90 43.8 min 40 65.7 min 60 54.0 min 30 79.2 min
Okay, that's your survey of application problems where exponential and logarithmic functions and equations show up! We're done with this chapter!