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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:
$$A\left(t\right)=P{\left(1+\frac{r}{n}\right)}^{nt}$$
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.
Susan invested $500 in an account earning $4\frac{1}{2}$% annual interest that is compounded monthly.
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:
$$\begin{array}{ccc}\hfill A\left(t\right)& =& 500{\left(1+\frac{0.045}{12}\right)}^{12t}\hfill \\ \hfill A\left(t\right)& =& 500{\left(1.00375\right)}^{12t}\hfill \end{array}$$
Use this model to calculate the amount in the account after $t=3$ years.
$$\begin{array}{ccc}A({3})& =& 500{(1.00375)}^{12({3})}\\ & =& 500{(1.00375)}^{36}\hfill \\ & \approx & 572.12\hfill \end{array}$$
Rounded off to the nearest cent, after 3 years, the amount accumulated will be $572.12.
To calculate the time it takes to accumulate $750, set $A\left(t\right)=750$ and solve for t.
$$\begin{array}{c}A(t)=500{(1.00375)}^{12t}\hfill \\ {750}=500{(1.00375)}^{12t}\hfill \end{array}$$
This results in an exponential equation that can be solved by first isolating the exponential expression.
$$\begin{array}{ccc}\hfill 750& =& 500{\left(1.00375\right)}^{12t}\hfill \\ \hfill \frac{750}{500}& =& {\left(1.00375\right)}^{12t}\hfill \\ \hfill 1.5& =& {\left(1.00375\right)}^{12t}\hfill \end{array}$$
At this point take the common logarithm of both sides, apply the power rule for logarithms, and then solve for t.
$$\begin{array}{ccc}\hfill \mathrm{log}\text{\hspace{0.17em}}\left(1.5\right)& =& \mathrm{log}\text{\hspace{0.17em}}{\left(1.00375\right)}^{12t}\hfill \\ \hfill \mathrm{log}\text{\hspace{0.17em}}\left(1.5\right)& =& 12t\mathrm{log}\text{\hspace{0.17em}}\left(1.00375\right)\hfill \\ \hfill \frac{\mathrm{log}\text{\hspace{0.17em}}\left(1.5\right)}{{12\mathrm{log}\text{\hspace{0.17em}}\left(1.00375\right)}}& =& \frac{\overline{)12}t\overline{)\mathrm{log}\text{\hspace{0.17em}}\left(1.00375\right)}}{{\overline{)12}\overline{)\mathrm{log}\text{\hspace{0.17em}}\left(1.00375\right)}}}\hfill \\ \hfill \frac{\mathrm{log}\text{\hspace{0.17em}}\left(1.5\right)}{12\mathrm{log}\text{\hspace{0.17em}}\left(1.00375\right)}& =& t\hfill \end{array}$$
Using a calculator we can approximate the time it takes.
$$t=\mathrm{log}\text{\hspace{0.17em}}\left(1.5\right)/\left(12*\mathrm{log}\text{\hspace{0.17em}}\left(1.00375\right)\right)\approx 9\text{\hspace{0.17em}\hspace{0.17em}}years$$
Answer:
The period of time it takes a quantity to double is called the doubling timeThe period of time it takes a quantity to double.. We next outline a technique for calculating the time it takes to double an initial investment earning compound interest.
Mario invested $1,000 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 $P=\$\mathrm{1,000}$, 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:
$$\begin{array}{ccc}\hfill A\left(t\right)& =& \mathrm{1,000}{\left(1+\frac{0.063}{2}\right)}^{2t}\hfill \\ \hfill A\left(t\right)& =& \mathrm{1,000}{\left(1.0315\right)}^{2t}\hfill \end{array}$$
Since we are looking for the time it takes to double $1,000, substitute $2,000 for the resulting amount $A\left(t\right)$ and then solve for t.
$$\begin{array}{ccc}\hfill {\mathrm{2,000}}& =& \mathrm{1,000}{\left(1.0315\right)}^{2t}\hfill \\ \hfill \frac{\mathrm{2,000}}{\mathrm{1,000}}& =& {\left(1.0315\right)}^{2t}\hfill \\ \hfill 2& =& {\left(1.0315\right)}^{2t}\hfill \end{array}$$
At this point we take the common logarithm of both sides.
$$\begin{array}{ccc}\hfill 2& =& {\left(1.0315\right)}^{2t}\hfill \\ \hfill \mathrm{log}\text{\hspace{0.17em}}2& =& \mathrm{log}\text{\hspace{0.17em}}{\left(1.0315\right)}^{2t}\hfill \\ \hfill \mathrm{log}\text{\hspace{0.17em}}2& =& 2t\mathrm{log}\text{\hspace{0.17em}}\left(1.0315\right)\hfill \\ \hfill \frac{\mathrm{log}\text{\hspace{0.17em}}2}{2\mathrm{log}\text{\hspace{0.17em}}\left(1.0315\right)}& =& t\hfill \end{array}$$
Using a calculator we can approximate the time it takes:
$$t=\mathrm{log}\text{\hspace{0.17em}}\left(2\right)/\left(2*\mathrm{log}\text{\hspace{0.17em}}\left(1.0315\right)\right)\approx 11.17\text{\hspace{0.17em}\hspace{0.17em}}\text{years}$$
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 $P=\$\mathrm{1,000,000}$ and $A\left(t\right)=\$\mathrm{2,000,000}$:
$$\begin{array}{ccc}\hfill A\left(t\right)& =& \mathrm{1,000}{\left(1.0315\right)}^{2t}\hfill \\ \hfill {\mathrm{2,000,000}}& =& \mathrm{1,000,000}{\left(1.0315\right)}^{2t}\hfill \end{array}$$
Dividing both sides by 1,000,000 we obtain the same exponential function as before.
$$2={\left(1.0315\right)}^{2t}$$
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:
$$A\left(t\right)=P{e}^{rt}$$
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.
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 $A\left(t\right)=P{e}^{rt}.$ Hence, the investment can be modeled by the following,
$$A\left(t\right)=200{e}^{0.0575t}$$
To calculate the time it takes to accumulate to $350, set $A\left(t\right)=350$ and solve for t.
$$\begin{array}{ccc}\hfill A\left(t\right)& =& 200{e}^{0.0575t}\hfill \\ \hfill {350}& =& 200{e}^{0.0575t}\hfill \end{array}$$
Begin by isolating the exponential expression.
$$\begin{array}{ccc}\hfill \frac{350}{200}& =& {e}^{0.0575t}\hfill \\ \hfill \frac{7}{4}& =& {e}^{0.0575t}\hfill \\ \hfill 1.75& =& {e}^{0.0575t}\hfill \end{array}$$
Because this exponential has base e, we choose to take the natural logarithm of both sides and then solve for t.
$$\begin{array}{cccc}\hfill \mathrm{ln}\text{\hspace{0.17em}}\left(1.75\right)& =& \mathrm{ln}\text{\hspace{0.17em}}{e}^{0.0575t}\hfill & \text{\hspace{0.17em}}\text{\hspace{0.17em}}{Apply\text{\hspace{0.17em}}the\text{\hspace{0.17em}}power\text{\hspace{0.17em}}rule\text{\hspace{0.17em}}for\text{\hspace{0.17em}}logarithms.}\hfill \\ \hfill \mathrm{ln}\text{\hspace{0.17em}}\left(1.75\right)& =& 0.0575t\mathrm{ln}\text{\hspace{0.17em}}e\hfill & \text{\hspace{0.17em}}\text{\hspace{0.17em}}{Recall\text{\hspace{0.17em}}that\text{\hspace{0.17em}}\mathrm{ln}\text{\hspace{0.17em}}e=1.}\hfill \\ \hfill \mathrm{ln}\text{\hspace{0.17em}}\left(1.75\right)& =& 0.0575t\cdot 1\hfill \\ \hfill \frac{\mathrm{ln}\text{\hspace{0.17em}}\left(1.75\right)}{0.0575}& =& t\hfill \end{array}$$
Using a calculator we can approximate the time it takes:
$$t=\mathrm{ln}\text{\hspace{0.17em}}\left(1.75\right)/0.0575\approx 9.73\text{\hspace{0.17em}\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}years$$
Answer: It will take approximately 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.
Try this! 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.
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 formulaA formula that models exponential growth or decay: $P\left(t\right)={P}_{0}{e}^{kt}.$:
$$P\left(t\right)={P}_{0}{e}^{kt}$$
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.
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:
Solution:
We begin by constructing a mathematical model based on the given information. Here the initial population ${P}_{0}=\mathrm{93,000}$ people and the growth rate $r=2.6\%=0.026.$ The following model gives population in terms of time measured in years:
$$P\left(t\right)=\mathrm{93,000}{e}^{0.026t}$$
Use this function to estimate the population in $t=7$ years.
$$\begin{array}{ccc}P(t)& =& \mathrm{93,000}{e}^{0.026({7})}\hfill \\ & =& \mathrm{93,000}{e}^{0.182}\hfill \\ & \approx & \mathrm{111,564}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}people\hfill \end{array}$$
Use the model to determine the time it takes to reach $P\left(t\right)=\mathrm{120,000}$ people.
$$\begin{array}{ccc}\hfill P\left(t\right)& =& \mathrm{93,000}{e}^{0.026t}\hfill \\ \hfill {\mathrm{120,000}}& =& \mathrm{93,000}{e}^{0.026t}\hfill \\ \hfill \frac{\mathrm{120,000}}{\mathrm{93,000}}& =& {e}^{0.026t}\hfill \\ \hfill \frac{40}{31}& =& {e}^{0.026t}\hfill \end{array}$$
Take the natural logarithm of both sides and then solve for t.
$$\begin{array}{ccc}\hfill \mathrm{ln}\text{\hspace{0.17em}}\left(\frac{40}{31}\right)& =& \mathrm{ln}\text{\hspace{0.17em}}{e}^{0.026t}\hfill \\ \hfill \mathrm{ln}\text{\hspace{0.17em}}\left(\frac{40}{31}\right)& =& 0.026t\mathrm{ln}\text{\hspace{0.17em}}e\hfill \\ \hfill \mathrm{ln}\text{\hspace{0.17em}}\left(\frac{40}{31}\right)& =& 0.026t\cdot 1\hfill \\ \hfill \frac{\mathrm{ln}\text{\hspace{0.17em}}\left(\frac{40}{31}\right)}{0.026}& =& t\hfill \end{array}$$
Using a calculator,
$$t=\mathrm{ln}\text{\hspace{0.17em}}\left(40/31\right)/0.026\approx 9.8\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}years$$
Answer:
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.
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 7.3
Escherichia coli (E. coli) (Wikipedia)
Solution:
The goal is to use the given information to construct a mathematical model based on the formula $P\left(t\right)={P}_{0}{e}^{kt}.$
Step 1: Find the growth rate k. Use the fact that the initial amount, ${P}_{0}=\mathrm{1,000}$ cells, doubles in 20 minutes. That is, $P\left(t\right)=\mathrm{2,000}$ cells when $t=20$ minutes.
$$\begin{array}{ccc}\hfill P\left(t\right)& =& {P}_{0}{e}^{kt}\hfill \\ \hfill {\mathrm{2,000}}& =& \mathrm{1,000}{e}^{k{20}}\hfill \end{array}$$
Solve for the only variable k.
$$\begin{array}{ccc}\hfill \mathrm{2,000}& =& \mathrm{1,000}{e}^{k20}\hfill \\ \hfill \frac{\mathrm{2,000}}{\mathrm{1,000}}& =& {e}^{k20}\hfill \\ \hfill 2& =& {e}^{k20}\hfill \\ \hfill \mathrm{ln}\text{\hspace{0.17em}}\left(2\right)& =& \mathrm{ln}\text{\hspace{0.17em}}{e}^{k20}\hfill \\ \hfill \mathrm{ln}\text{\hspace{0.17em}}\left(2\right)& =& k20\mathrm{ln}\text{\hspace{0.17em}}e\hfill \\ \hfill \mathrm{ln}\text{\hspace{0.17em}}\left(2\right)& =& k20\cdot 1\hfill \\ \hfill \frac{\mathrm{ln}\text{\hspace{0.17em}}\left(2\right)}{20}& =& k\hfill \end{array}$$
Step 2: Write a mathematical model based on the given information. Here $k\approx 0.0347$, 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 ${P}_{0}=\mathrm{1,000}$ and $k=\mathrm{ln}\text{\hspace{0.17em}}\left(2\right)/20$:
$$P\left(t\right)=\mathrm{1,000}{e}^{\left(\mathrm{ln}\text{\hspace{0.17em}}\left(2\right)/20\right)t}$$
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.
$$\begin{array}{ccc}\hfill P\left({120}\right)& =& \mathrm{1,000}{e}^{\left(\mathrm{ln}\text{\hspace{0.17em}}\left(2\right)/20\right)\left({120}\right)}\hfill \\ & =& \mathrm{1,000}{e}^{\mathrm{ln}\text{\hspace{0.17em}}\left(2\right)\cdot 6}\hfill \\ & =& \mathrm{1,000}{e}^{\mathrm{ln}\text{\hspace{0.17em}}{2}^{6}}\hfill \\ & =& \mathrm{1,000}\cdot {2}^{6}\hfill \\ & =& \mathrm{64,000}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}cells\hfill \end{array}$$
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-lifeThe period of time it takes a quantity to decay to one-half of the initial amount., or the time it takes to decay to one-half of a given quantity.
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 $P\left(30\right)=25$ milligrams.
$$\begin{array}{ccc}\hfill P\left(t\right)& =& {P}_{0}{e}^{kt}\hfill \\ \hfill 25& =& 50{e}^{k30}\hfill \end{array}$$
Solve for the only variable, k.
$$\begin{array}{ccc}\hfill 25& =& 50{e}^{k30}\hfill \\ \hfill \frac{25}{50}& =& {e}^{30k}\hfill \\ \hfill \mathrm{ln}\text{\hspace{0.17em}}\left(\frac{1}{2}\right)& =& \mathrm{ln}\text{\hspace{0.17em}}{e}^{30k}\hfill \\ \hfill \mathrm{ln}\text{\hspace{0.17em}}\left(\frac{1}{2}\right)& =& 30k\mathrm{ln}\text{\hspace{0.17em}}e\hfill \\ \hfill \frac{\mathrm{ln}\text{\hspace{0.17em}}1-\mathrm{ln}\text{\hspace{0.17em}}2}{30}& =& k\hfill & \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{Recall\text{\hspace{0.17em}}that\text{\hspace{0.17em}}\mathrm{ln}\text{\hspace{0.17em}}1=0.}\\ \hfill -\frac{\mathrm{ln}\text{\hspace{0.17em}}2}{30}& =& k\hfill \end{array}$$
Note that $k=-\mathrm{ln}\text{\hspace{0.17em}}\frac{2}{30}\approx -0.0231$ 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.
$$P\left(t\right)=50{e}^{\left(-\mathrm{ln}\text{\hspace{0.17em}}2/30\right)t}$$
Use this model to find t when $P\left(t\right)=10$ milligrams.
$$\begin{array}{ccc}\hfill 10& =& 50{e}^{\left(-\mathrm{ln}\text{\hspace{0.17em}}2/30\right)t}\hfill \\ \hfill \frac{10}{50}& =& {e}^{\left(-\mathrm{ln}\text{\hspace{0.17em}}2/30\right)t}\hfill \\ \hfill \mathrm{ln}\text{\hspace{0.17em}}\left(\frac{1}{5}\right)& =& \mathrm{ln}\text{\hspace{0.17em}}{e}^{\left(-\mathrm{ln}\text{\hspace{0.17em}}2/30\right)t}\hfill \\ \hfill \mathrm{ln}\text{\hspace{0.17em}}1-\text{\hspace{0.17em}}\mathrm{ln}\text{\hspace{0.17em}}5& =& \left(-\frac{\mathrm{ln}\text{\hspace{0.17em}}2}{30}\right)t\text{\hspace{0.17em}}\mathrm{ln}\text{\hspace{0.17em}}e\hfill & \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{Recall\text{\hspace{0.17em}}that\text{\hspace{0.17em}}\mathrm{ln}\text{\hspace{0.17em}}e=1.}\\ \hfill -\frac{30\left(\mathrm{ln}\text{\hspace{0.17em}}1-\mathrm{ln}\text{\hspace{0.17em}}5\right)}{\mathrm{ln}\text{\hspace{0.17em}}2}& =& t\hfill \\ \hfill \frac{-30\left(0-\mathrm{ln}\text{\hspace{0.17em}}5\right)}{\mathrm{ln}\text{\hspace{0.17em}}2}& =& t\hfill \\ \hfill \frac{30\mathrm{ln}\text{\hspace{0.17em}}5}{\mathrm{ln}\text{\hspace{0.17em}}2}& =& t\hfill \end{array}$$
Answer: Using a calculator, it will take $t\approx 69.66$ 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.
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 $t=\mathrm{5,730}$ years, this amount decays to half, $\frac{1}{2}{P}_{0}.$
$$\begin{array}{ccc}\hfill P\left(t\right)& =& {P}_{0}{e}^{kt}\hfill \\ \hfill \frac{1}{2}{P}_{0}& =& {P}_{0}{e}^{k\mathrm{5,730}}\hfill \end{array}$$
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.
$$\frac{1}{2}={e}^{k\mathrm{5,730}}$$
Solve for k.
$$\begin{array}{ccc}\hfill \mathrm{ln}\text{\hspace{0.17em}}\left(\frac{1}{2}\right)& =& \mathrm{ln}\text{\hspace{0.17em}}{e}^{k\mathrm{5,730}}\hfill \\ \hfill \mathrm{ln}\text{\hspace{0.17em}}1-\mathrm{ln}\text{\hspace{0.17em}}2& =& \mathrm{5,730}k\text{\hspace{0.17em}}\mathrm{ln}\text{\hspace{0.17em}}e\hfill \\ \hfill \frac{0-\mathrm{ln}\text{\hspace{0.17em}}2}{\mathrm{5,730}}& =& k\hfill \\ \hfill -\frac{\mathrm{ln}\text{\hspace{0.17em}}2}{\mathrm{5,730}}& =& k\hfill \end{array}$$
Therefore we have the model,
$$P\left(t\right)={P}_{0}{e}^{\left(-\mathrm{ln}\text{\hspace{0.17em}}2/\mathrm{5,730}\right)t}$$
Next we wish to the find time it takes the carbon-14 to decay to 25% of the initial amount, or $P\left(t\right)=0.25{P}_{0}.$
$$0.25{P}_{0}={P}_{0}{e}^{\left(-\mathrm{ln}\text{\hspace{0.17em}}2/\mathrm{5,730}\right)t}$$
Divide both sides by ${P}_{0}$ and solve for t.
$$\begin{array}{ccc}\hfill 0.25& =& {e}^{\left(-\mathrm{ln}\text{\hspace{0.17em}}2/\mathrm{5,730}\right)t}\hfill \\ \hfill \mathrm{ln}\text{\hspace{0.17em}}\left(0.25\right)& =& \mathrm{ln}\text{\hspace{0.17em}}{e}^{\left(-\mathrm{ln}\text{\hspace{0.17em}}2/\mathrm{5,730}\right)t}\hfill \\ \hfill \mathrm{ln}\text{\hspace{0.17em}}\left(0.25\right)& =& \left(-\frac{\mathrm{ln}\text{\hspace{0.17em}}2}{\mathrm{5,730}}\right)t\mathrm{ln}\text{\hspace{0.17em}}e\hfill \\ \hfill -\frac{\mathrm{5,730}\text{\hspace{0.17em}}\mathrm{ln}\text{\hspace{0.17em}}\left(0.25\right)}{\mathrm{ln}\text{\hspace{0.17em}}2}& =& t\hfill \\ \hfill \mathrm{11,460}& \approx & t\hfill \end{array}$$
Answer: The tool is approximately 11,460 years old.
Try this! 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
Jill invested $1,450 in an account earning $4\frac{5}{8}$% annual interest that is compounded monthly.
James invested $825 in an account earning $5\frac{2}{5}$% annual interest that is compounded monthly.
Raul invested $8,500 in an online money market fund earning 4.8% annual interest that is compounded continuously.
Ian deposited $500 in an account earning 3.9% annual interest that is compounded continuously.
Bill wants to grow his $75,000 inheritance to $100,000 before spending any of it. How long will this take if the bank is offering 5.2% annual interest compounded quarterly?
Mary needs $25,000 for a down payment on a new home. If she invests her savings of $21,350 in an account earning 4.6% annual interest that is compounded semi-annually, how long will it take to grow to the amount that she needs?
Joe invested his $8,700 savings in an account earning $6\frac{3}{4}$% annual interest that is compounded continuously. How long will it take to earn $300 in interest?
Miriam invested $12,800 in an account earning $5\frac{1}{4}$% annual interest that is compounded monthly. How long will it take to earn $1,200 in interest?
Given that the bank is offering 4.2% annual interest compounded monthly, what principal is needed to earn $25,000 in interest for one year?
Given that the bank is offering 3.5% annual interest compounded continuously, what principal is needed to earn $12,000 in interest for one year?
Jose invested his $3,500 bonus in an account earning $5\frac{1}{2}$% annual interest that is compounded quarterly. How long will it take to double his investment?
Maria invested her $4,200 savings in an account earning $6\frac{3}{4}$% annual interest that is compounded semi-annually. How long will it take to double her savings?
If money is invested in an account earning 3.85% annual interest that is compounded continuously, how long will it take the amount to double?
If money is invested in an account earning 6.82% annual interest that is compounded continuously, how long will it take the amount to double?
Find the annual interest rate at which an account earning continuously compounding interest has a doubling time of 9 years.
Find the annual interest rate at which an account earning interest that is compounded monthly has a doubling time of 10 years.
Alice invested her savings of $7,000 in an account earning 4.5% annual interest that is compounded monthly. How long will it take the account to triple in value?
Mary invested her $42,000 bonus in an account earning 7.2% annual interest that is compounded continuously. How long will it take the account to triple in value?
Calculate the doubling time of an investment made at 7% annual interest that is compounded:
Calculate the doubling time of an investment that is earning continuously compounding interest at an annual interest rate of:
Billy’s grandfather invested in a savings bond that earned 5.5% annual interest that was compounded annually. Currently, 30 years later, the savings bond is valued at $10,000. Determine what the initial investment was.
In 1935 Frank opened an account earning 3.8% annual interest that was compounded quarterly. He rediscovered this account while cleaning out his garage in 2005. If the account is now worth $11,294.30, how much was his initial deposit in 1935?
The population of a small town of 24,000 people is expected grow exponentially at a rate of 1.6% per year. Construct an exponential growth model and use it to:
During the exponential growth phase, certain bacteria can grow at a rate of 4.1% per hour. If 10,000 cells are initially present in a sample, construct an exponential growth model and use it to:
In 2000, the world population was estimated to be 6.115 billion people and in 2010 the estimate was 6.909 billion people. If the world population continues to grow exponentially, estimate the total world population in 2020.
In 2000, the population of the United States was estimated to be 282 million people and in 2010 the estimate was 309 million people. If the population of the United States grows exponentially, estimate the population in 2020.
An automobile was purchased new for $42,500 and 2 years later it was valued at $33,400. Estimate the value of the automobile in 5 years if it continues to decrease exponentially.
A new PC was purchased for $1,200 and in 1.5 years it was worth $520. Assume the value is decreasing exponentially and estimate the value of the PC four years after it is purchased.
The population of the downtown area of a certain city decreased from 12,500 people to 10,200 people in two years. If the population continues to decrease exponentially at this rate, what would we expect the population to be in two more years?
A new MP3 player was purchased for $320 and in 1 year it was selling used online for $210. If the value continues to decrease exponentially at this rate, determine the value of the MP3 player 3 years after it was purchased.
The half-life of radium-226 is about 1,600 years. How long will a 5-milligram sample of radium-226 take to decay to 1 milligram?
The half-life of plutonium-239 is about 24,000 years. How long will a 5-milligram sample of plutonium-239 take to decay to 1 milligram?
The half-life of radioactive iodine-131 is about 8 days. How long will it take a 28-gram initial sample of iodine-131 to decay to 12 grams?
The half-life of caesium-137 is about 30 years. How long will it take a 15-milligram sample of caesium-137 to decay to 5 milligrams?
The Rhind Mathematical Papyrus is considered to be the best example of Egyptian mathematics found to date. This ancient papyrus was found to contain 64% of the carbon-14 normally found in papyrus. Given that carbon-14 has a half-life of 5,730 years, estimate the age of the papyrus.
A wooden bowl artifact carved from oak was found to contain 55% of the carbon-14 normally found in oak. Given that carbon-14 has a half-life of 5,730 years, estimate the age of the bowl.
The half-life of radioactive iodine-131 is about 8 days. How long will it take a sample of iodine-131 to decay to 10% of the original amount?
The half-life of caesium-137 is about 30 years. How long will it take a sample of caesium-137 to decay to 25% of the original amount?
The half-life of caesium-137 is about 30 years. What percent of an initial sample will remain in 100 years?
The half-life of radioactive iodine-131 is about 8 days. What percent of an initial sample will remain in 30 days?
If a bone is 100 years old, what percent of its original amount of carbon-14 do we expect to find in it?
The half-life of plutonium-239 is about 24,000 years. What percent of an initial sample will remain in 1,000 years?
Find the amount of time it will take for 10% of an initial sample of plutonium-239 to decay. (Hint: If 10% decays, then 90% will remain.)
Find the amount of time it will take for 10% of an initial sample of carbon-14 to decay.
Solve for the given variable:
Solve for t: $A=P{e}^{rt}$
Solve for t: $A=P{\left(1+r\right)}^{t}$
Solve for I: $M=\mathrm{log}\text{\hspace{0.17em}}\left(\frac{I}{{I}_{0}}\right)$
Solve for ${H}^{+}$: $pH=-\mathrm{log}\text{\hspace{0.17em}}\left({H}^{+}\right)$
Solve for t: $P=\frac{1}{1+{e}^{-t}}$
Solve for I: $L=10\mathrm{log}\text{\hspace{0.17em}}\left(I/{10}^{-12}\right)$
The number of cells in a certain bacteria sample is approximated by the logistic growth model $N\left(t\right)=\frac{1.2\times {10}^{5}}{1+9{e}^{-0.32t}}$, where t represents time in hours. Determine the time it takes the sample to grow to 24,000 cells.
The market share of a product, as a percentage, is approximated by the formula $P\left(t\right)=\frac{100}{3+{e}^{-0.44t}}$ where t represents the number of months after an aggressive advertising campaign is launched.
In chemistry, pH is a measure of acidity and is given by the formula $\mathrm{pH}=-\mathrm{log}\text{\hspace{0.17em}}\left({H}^{+}\right)$, where ${H}^{+}$ is the hydrogen ion concentration (measured in moles of hydrogen per liter of solution.) What is the hydrogen ion concentration of seawater with a pH of 8?
Determine the hydrogen ion concentration of milk with a pH of 6.6.
The volume of sound, L in decibels (dB), is given by the formula $L=10\mathrm{log}\text{\hspace{0.17em}}\left(I/{10}^{-12}\right)$ where I represents the intensity of the sound in watts per square meter. Determine the sound intensity of a hair dryer that emits 70 dB of sound.
The volume of a chainsaw measures 110 dB. Determine the intensity of this sound.
Which factor affects the doubling time the most, the annual compounding n or the interest rate r? Explain.
Research and discuss radiocarbon dating. Post something interesting you have learned as well as a link to more information.
Is exponential growth sustainable over an indefinite amount of time? Explain.
Research and discuss the half-life of radioactive materials.
5.6 years
$\frac{1}{2}$ year
$583,867
12.7 years
18 years
7.7%
24.5 years
$2,006.44
About 7.806 billion people
About $23,269.27
8,323 people
3,715 years
9.8 days
About 3,689 years old
26.6 days
9.9%
98.8%
3,648 years
$I={I}_{0}\cdot {10}^{M}$
Approximately 2.5 hours
${10}^{-8}$ moles per liter
${10}^{-5}$ watts per square meter
Answer may vary
Answer may vary