6.5: Applications of Exponential and Logarithmic Functions

Compound and Continuous Interest Formulas

Recall that compound interest occurs when interest accumulated for one period is added to the principal investment before calculating interest for the next period. The amount \(A\) accrued in this manner over time \(t\) is modeled by the compound interest formula:

\(A(t)=P\left(1+\frac{r}{n}\right)^{n t}\)

Here the initial principal \(P\) is accumulating compound interest at an annual rate \(r\) where the value \(n\) represents the number of times the interest is compounded in a year.

Example \(\PageIndex{1}\):

Susan invested $\(500\) in an account earning \(4 \frac{1}{2}\)% annual interest that is compounded monthly.

a. How much will be in the account after \(3\) years?

b. How long will it take for the amount to grow to $\(750\)?

Solution

In this example, the principal \(P =\) $\(500\), the interest rate \(r = 4 \frac{1}{2}\)% \(= 0.045\), and because the interest is compounded monthly, \(n = 12\). The investment can be modeled by the following function:

\(A(t)=500\left(1+\frac{0.045}{12}\right)^{12 t}\)
\(A(t)=500(1.00375)^{12 t}\)

a. Use this model to calculate the amount in the account after \(t=3\) years.

\(\begin{aligned} A(\color{Cerulean}{3}\color{black}{)} &=500(1.00375)^{12(\color{Cerulean}{3}\color{black}{)}} \\ &=500(1.00375)^{36} \\ & \approx 572.12 \end{aligned}\)

Rounded off to the nearest cent, after \(3\) years, the amount accumulated will be $\(572.12\).

b. To calculate the time it takes to accumulate $\(750\), set \(A (t) = 750\) and solve for \(t\).

\(\begin{array}{l}{A(t)=500(1.00375)^{12 t}} \\ {\color{Cerulean}{750}\color{black}{=}500(1.00375)^{12 t}}\end{array}\)

This results in an exponential equation that can be solved by first isolating the exponential expression.

\(\begin{aligned} 750 &=500(1.00375)^{12 t} \\ \frac{750}{500} &=(1.00375)^{12 t} \\ 1.5 &=(1.00375)^{12 t} \end{aligned}\)

At this point take the common logarithm of both sides, apply the power rule for logarithms, and then solve for \(t\).

\(\begin{aligned} \log (1.5) &=\log (1.00375)^{12 t} \\ \log (1.5) &=12 t \log (1.00375)\\\frac{\log (1.5)}{\color{Cerulean}{12\log (1.00375)}}&\color{black}{=} \frac{\cancel{12}t\cancel{\log (1.00375)}}{\cancel{\color{Cerulean}{12\: \log (1.00375)}}} \\\frac{\log(1.5)}{12\:\log(1.00375)}&=t\end{aligned}\)

Using a calculator we can approximate the time it takes.

\(t=\log (1.5) /(12 * \log (1.00375)) \approx 9\) years

Answer:

a. $\(572.12\)

b. Approximately \(9\) years

The period of time it takes a quantity to double is called the doubling time²⁰. We next outline a technique for calculating the time it takes to double an initial investment earning compound interest.

Example \(\PageIndex{2}\):

Mario invested $\(1000\) in an account earning \(6.3\)% annual interest, that is compounded semi-annually. How long will it take the investment to double?

Solution

Here the principal \(P =\) $\(1,000\), the interest rate \(r = 6.3\)% \(= 0.063\), and because the interest is compounded semi-annually \(n = 2\). This investment can be modeled as follows:

\(A(t)=1,000\left(1+\frac{0.063}{2}\right)^{2 t}\)
\(A(t)=1,000(1.0315)^{2 t}\)

Since we are looking for the time it takes to double $\(1,000\), substitute $\(2,000\) for the resulting amount \(A (t)\) and then solve for \(t\).

\(\begin{aligned} \color{Cerulean}{2,000} &\color{black}{=}1,000(1.0315)^{2 t} \\ \frac{2,000}{1,000} &=(1.0315)^{2 t} \\ 2 &=(1.0315)^{2 t} \end{aligned}\)

At this point we take the common logarithm of both sides.

\(\begin{aligned} 2 &=(1.0315)^{2 t} \\ \log 2 &=\log (1.0315)^{2 t} \\ \log 2 &=2 t \log (1.0315) \\ \frac{\log 2}{2 \log (1.0315)} &=t \end{aligned}\)

Using a calculator we can approximate the time it takes:

\(t=\log (2) /(2 * \log (1.0315)) \approx 11.17\) years

Answer:

Approximately \(11.17\) years to double at \(6.3\)%.

If the investment in the previous example was one million dollars, how long would it take to double? To answer this we would use \(P =\) $\(1,000,000\) and \(A (t) =\) $\(2,000,000\):

\(\begin{aligned} A(t) &=1,000(1.0315)^{2 t} \\ \color{Cerulean}{2,000,000} &\color{black}{=}1,000,000(1.0315)^{2 t} \end{aligned}\)

Dividing both sides by \(1,000,000\) we obtain the same exponential function as before.

\(2=(1.0315)^{2 t}\)

Hence, the result will be the same, about \(11.17\) years. In fact, doubling time is independent of the initial investment \(P\).

Interest is typically compounded semi-annually \((n = 2)\), quarterly \((n = 4)\), monthly \((n = 12)\), or daily \((n = 365)\). However if interest is compounded every instant we obtain a formula for continuously compounding interest:

\(A(t)=P e^{r t}\)

Here \(P\) represents the initial principal amount invested, \(r\) represents the annual interest rate, and \(t\) represents the time in years the investment is allowed to accrue continuously compounded interest.

Example \(\PageIndex{3}\):

Mary invested $\(200\) in an account earning \(5 \frac{3}{4}\)% annual interest that is compounded continuously. How long will it take the investment to grow to $\(350\)?

Solution

Here the principal \(P =\) $\(200\) and the interest rate \(r = 5 \frac{3}{4}\)% \(= 5.75\)% \(= 0.0575\). Since the interest is compounded continuously, use the formula \(A (t) = Pe^{rt}\). Hence, the investment can be modeled by the following,

\(A(t)=200 e^{0.0575 t}\)

To calculate the time it takes to accumulate to $\(350\), set \(A (t) = 350\) and solve for \(t\).

\(\begin{array}{r}{A(t)=200 e^{0.0575 t}} \\ {\color{Cerulean}{350}\color{black}{=}200 e^{0.0575 t}}\end{array}\)

Begin by isolating the exponential expression.

\(\begin{aligned} \frac{350}{200} &=e^{0.0575 t} \\ \frac{7}{4} &=e^{0.0575 t} \\ 1.75 &=e^{0.0575 t} \end{aligned}\)

Because this exponential has base \(e\), we choose to take the natural logarithm of both sides and then solve for \(t\).

\(\begin{array}{l}{\ln (1.75)=\ln e^{0.0575 t}}\quad\quad\color{Cerulean}{Apply\:the\:power\:rule\:for\:logarithms.} \\ {\ln (1.75)=0.0575 t \ln e} \quad\color{Cerulean}{Recall\:that\: \ln e=1.} \\ {\ln (1.75)=0.0575 t \cdot 1} \\ {\frac{\ln (1.75)}{0.0575}=t}\end{array}\)

Using a calculator we can approximate the time it takes:

\(t=\ln (1.75) / 0.0575 \approx 9.73 \quad years\)

Answer:

It will aprroximately \(9.73\) years.

When solving applications involving compound interest, look for the keyword “continuous,” or the keywords that indicate the number of annual compoundings. It is these keywords that determine which formula to choose.

Exercise \(\PageIndex{1}\)

Mario invested $\(1,000\) in an account earning \(6.3\)% annual interest that is compounded continuously. How long will it take the investment to double?

Answer

Approximately \(11\) years.

www.youtube.com/v/Z_ilJHIQvWQ

Modeling Exponential Growth and Decay

In the sciences, when a quantity is said to grow or decay exponentially, it is specifically meant to be modeled using the exponential growth/decay formula²¹:

\(P(t)=P_{0} e^{k t}\)

Here \(P_{0}\), read “\(P\) naught,” or “\(P\) zero,” represents the initial amount, k represents the growth rate, and \(t\) represents the time the initial amount grows or decays exponentially. If \(k\) is negative, then the function models exponential decay. Notice that the function looks very similar to that of continuously compounding interest formula. We can use this formula to model population growth when conditions are optimal.

Example \(\PageIndex{4}\):

It is estimated that the population of a certain small town is \(93,000\) people with an annual growth rate of \(2.6\)%. If the population continues to increase exponentially at this rate:

1. Estimate the population in \(7\) years' time.
2. Estimate the time it will take for the population to reach 120,000 people.

Solution

We begin by constructing a mathematical model based on the given information. Here the initial population \(P_{0} = 93,000\) people and the growth rate \(r = 2.6\)% \(= 0.026\). The following model gives population in terms of time measured in years:

\(P(t)=93,000 e^{0.026 t}\)

a. Use this function to estimate the population in \(t = 7\) years.

\(\begin{aligned} P(t) &=93,000 e^{0006(\color{Cerulean}{7}\color{black}{)}} \\ &=93,000 e^{0.182} \\ & \approx 111,564 \quad people \end{aligned}\)

b. Use the model to determine the time it takes to reach \(P (t) = 120,000\) people.

\(\begin{aligned} P(t) &=93,000 e^{0.026 t} \\ \color{Cerulean}{120,000} &\color{black}{=}93,000 e^{0.026 t} \\ \frac{120,000}{93,000} &=e^{0.026 t} \\ \frac{40}{31} &=e^{0.026 t} \end{aligned}\)

Take the natural logarithm of both sides and then solve for \(t\).

\(\ln \left(\frac{40}{31}\right)=\ln e^{0.026 t}\)
\(\ln \left(\frac{40}{31}\right)=0.026 t \ln e\)
\(\ln \left(\frac{40}{31}\right)=0.026 t \cdot 1\)
\(\frac{\ln \left(\frac{40}{31}\right)}{0.026}=t\)

Using a calculator,

\(t=\ln (40 / 31) / 0.026 \approx 9.8\quad years\)

Answer:

1. \(111,564\) people
2. \(9.8\) years

Often the growth rate \(k\) is not given. In this case, we look for some other information so that we can determine it and then construct a mathematical model. The general steps are outlined in the following example.

Example \(\PageIndex{5}\):

Under optimal conditions Escherichia coli (E. coli) bacteria will grow exponentially with a doubling time of \(20\) minutes. If \(1,000\) E. coli cells are placed in a Petri dish and maintained under optimal conditions, how many E. coli cells will be present in \(2\) hours?

Figure \(\PageIndex{1}\): Escherichia coli (E. coli)

Solution

The goal is to use the given information to construct a mathematical model based on the formula \(P (t) = P_{0} e^{kt}\).

Step 1: Find the growth rate \(k\). Use the fact that the initial amount, \(P_{0} = 1,000\) cells, doubles in \(20\) minutes. That is, \(P (t) = 2,000\) cells when \(t = 20\) minutes.

\(\begin{aligned} P(t) &=P_{0} e^{k t} \\ \color{Cerulean}{2,000} &\color{black}{=}1,000 e^{k \color{Cerulean}{20}} \end{aligned}\)

Solve for the only variable \(k\).

\(\begin{aligned} 2,000 &=1,000 e^{k 20} \\ \frac{2,000}{1,000} &=e^{k 20} \\ 2 &=e^{k 20} \\ \ln (2) &=\ln e^{k 20} \\ \ln (2) &=k 20 \ln e \\ \ln (2) &=k 20 \cdot 1 \\ \frac{\ln (2)}{20} &=k \end{aligned}\)

Step 2: Write a mathematical model based on the given information. Here \(k ≈ 0.0347\), which is about \(3.5\)% growth rate per minute. However, we will use the exact value for \(k\) in our model. This will allow us to avoid round-off error in the final result. Use \(P_{0} = 1,000\) and \(k=\ln (2) / 20\):

\(P(t)=1,000 e^{(\ln (2) / 20) t}\)

This equation models the number of E. coli cells in terms of time in minutes.

Step 3: Use the function to answer the questions. In this case, we are asked to find the number of cells present in \(2\) hours. Because time is measured in minutes, use \(t = 120\) minutes to calculate the number of E. coli cells.

\(\begin{aligned} P(\color{Cerulean}{120}\color{black}{)} &=1,000 e^{(\ln (2) / 20)(\color{Cerulean}{120}\color{black}{)}} \\ &=1,000 e^{\ln (2) \cdot 6} \\ &=1,000 e^{\ln 2^{6}} \\ &=1,000 \cdot 2^{6} \\ &=64,000 \text { cells } \end{aligned}\)

Answer:

In two hours \(64,000\) cells will be present.

When the growth rate is negative the function models exponential decay. We can describe decreasing quantities using a half-life²², or the time it takes to decay to one-half of a given quantity.

Example \(\PageIndex{6}\):

Due to radioactive decay, caesium-137 has a half-life of \(30\) years. How long will it take a \(50\)-milligram sample to decay to \(10\) milligrams?

Solution

Use the half-life information to determine the rate of decay \(k\). In \(t = 30\) years the initial amount \(P_{0} = 50\) milligrams will decay to half \(P (30) = 25\) milligrams.

\(\begin{aligned} P(t) &=P_{0} e^{k t} \\ 25 &=50 e^{k 30} \end{aligned}\)

Solve for the only variable, \(k\).

\(\begin{aligned}25&=50 e^{130} \\ \frac{25}{50}&=e^{30 k} \\ \ln \left(\frac{1}{2}\right)&=\ln e^{30 k} \\ \ln \left(\frac{1}{2}\right)&=30 k \ln e \\ \frac{\ln 1-\ln 2}{30}&=k\quad\quad\quad\color{Cerulean}{Recall\:that\:\ln1=0.} \\ -\frac{\ln 2}{30}&=k\end{aligned}\)

Note that \(k=-\ln \frac{2}{30} \approx-0.0231\) is negative. However, we will use the exact value to construct a model that gives the amount of cesium-137 with respect to time in years.

\(P(t)=50 e^{(-\ln 2 / 30) t}\)

Use this model to find \(t\) when \(P (t) = 10\) milligrams.

\(\begin{aligned}10&=50 e^{(-\ln 2 / 30) t} \\ \frac{10}{50}&=e^{(-\ln 2 / 30) t} \\ \ln \left(\frac{1}{5}\right)&=\ln e^{(-\ln 2 / 30) t} \\ \ln 1-\ln 5&=\left(-\frac{\ln 2}{30}\right) t \ln e \quad\color{Cerulean} { Recall\: that\: \ln e=1. }\\ -30\frac{(\ln 1-\ln5)}{\ln 2}&=t\\-\frac{30(0-\ln 5)}{\ln 2}&=t \\ \frac{30\ln 5}{\ln 2}&=t\end{aligned}\)

Answer:

Using a calculator, it will take \(t ≈ 69.66\) years to decay to \(10\) milligrams.

Radiocarbon dating is a method used to estimate the age of artifacts based on the relative amount of carbon-14 present in it. When an organism dies, it stops absorbing this naturally occurring radioactive isotope, and the carbon-14 begins to decay at a known rate. Therefore, the amount of carbon-14 present in an artifact can be used to estimate the age of the artifact.

Example \(\PageIndex{7}\):

An ancient bone tool is found to contain \(25\)% of the carbon-14 normally found in bone. Given that carbon-14 has a half-life of \(5,730\) years, estimate the age of the tool.

Solution

Begin by using the half-life information to find \(k\). Here the initial amount \(P_{0}\) of carbon-14 is not given, however, we know that in \(t = 5,730\) years, this amount decays to half, \(\frac{1}{2} P_{0}\).

\(P(t)=P_{0} e^{k t}\)
\(\frac{1}{2} P_{0}=P_{0} e^{k 5,730}\)

Dividing both sides by \(P_{0}\) leaves us with an exponential equation in terms of \(k\). This shows that half-life is independent of the initial amount.

\(\frac{1}{2}=e^{k5,730}\)

Solve for \(k\).

\(\begin{aligned}\ln \left(\frac{1}{2}\right)&=\ln e^{k 5,730}\\ \ln1-\ln2 &= 5,730k\ln e \\ \frac{0-\ln 2}{5,730}&=k \\ -\frac{\ln 2}{5,730}&=k\end{aligned}\)

Therefore we have the model,

\(P(t)=P_{0} e^{(-\ln 2 / 5,730) t}\)

Next we wish to the find time it takes the carbon-14 to decay to \(25\)% of the initial amount, or \(P (t) = 0.25P_{0}\)

.\(0.25 P_{0}=P_{0} e^{(-\ln 2 / 5,730) t}\)

Divide both sides by \(P_{0}\) and solve for \(t\).

\(\begin{aligned} 0.25 &=e^{(-\ln 2 / 5,730) t} \\ \ln (0.25) &=\ln e^{(-\ln 2 / 5,730) t} \\ \ln (0.25) &=\left(-\frac{\ln 2}{5,730}\right)_{t \ln e} \\-\frac{5,730 \ln (0.25)}{\ln 2} &=t\\11,460&\approx t \end{aligned}\)

Answer:

The tool is approximately \(11,460\) years old.

Exercise \(\PageIndex{2}\)

The half-life of strontium-90 is about \(28\) years. How long will it take a \(36\) milligram sample of strontium-90 to decay to \(30\) milligrams?

Answer

\(7.4\) years

www.youtube.com/v/Oto0ihiyvBc

Key Takeaways

• When interest is compounded a given number of times per year use the formula \(A(t)=P\left(1+\frac{r}{n}\right)^{n t}\).
• When interest is to be compounded continuously use the formula \(A(t)=P e^{r t}\).
• Doubling time is the period of time it takes a given amount to double. Doubling time is independent of the principal.
• When amounts are said to be increasing or decaying exponentially, use the formula \(P(t)=P_{0} e^{k t}\).
• Half-life is the period of time it takes a given amount to decrease to one-half. Half-life is independent of the initial amount.
• To model data using the exponential growth/decay formula, use the given information to determine the growth/decay rate \(k\). Once \(k\) is determined, a formula can be written to model the problem. Use the formula to answer the questions.

Footnotes

²⁰ The period of time it takes a quantity to double.

²¹A formula that models exponential growth or decay: \(P(t)=P_{0} e^{k t}\).

²²The period of time it takes a quantity to decay to one-half of the initial amount.