4.3: Special Cases- Doubling Time and Half-Life
Let’s say that on April 1st I say I will give you a penny, on April 2nd two pennies, four pennies on April 3rd, and that I will double the amount each day until the end of the month. How much money would I have agreed to give you on April 30? With \(P_{0} = $0.01\), we get the following table:
| Day | Dollar Amount |
|---|---|
| April 1 = \(P_{0}\) | 0.01 |
| April 2 = \(P_{1}\) | 0.02 |
| April 3 = \(P_{2}\) | 0.04 |
| April 4 = \(P_{3}\) | 0.08 |
| April 5 = \(P_{4}\) | 0.16 |
| April 6 = \(P_{5}\) | 0.32 |
| …. | …. |
| April 30 = \(P_{29}\) | ? |
In this example, the money received each day is 100% more than the previous day. If we use the exponential growth model \(P(t) = P_{0}(1+r)^{t}\) with r = 1, we get the doubling time model.
\[P(t) = P_{0}(1+1)^{t} = P_{0}(2)^{t} \nonumber \]
We use it to find the dollar amount when \(t = 29\) which represents April 30
\[P(29) = 0.01(2)^{29} = $5,368,709.12 \nonumber \]
Surprised? That is a lot of pennies.
Doubling Time Model
A water tank up on the San Francisco Peaks is contaminated with a colony of 80,000 E. coli bacteria. The population doubles every five days. We want to find a model for the population of bacteria present after \(t\) days. The amount of time it takes the population to double is five days, so this is our time unit. After \(t\) days have passed, then \(t/5\) is the number of time units that have passed. Starting with the initial amount of 80,000 bacteria, our doubling model becomes:
\[P(t) = 80,000(2)^{\frac{t}{5}} \nonumber \]
Using this model, how large is the colony in two weeks’ time? We have to be careful that the units on the times are the same; 2 weeks = 14 days.
Solution
\[P(14) = 80,000(2)^{\frac{14}{5}} = 557,152 \nonumber \]
The colony is now 557,152 bacteria.
If \(D\) is the doubling time of a quantity (the amount of time it takes the quantity to double) and\(P_{0}\) is the initial amount of the quantity then the amount of the quantity present after \(t\) units of time is \(P(t) = P_{0}(2)^{\frac{t}{D}}\)
The doubling time of a population of flies is eight days. If there are initially 100 flies, how many flies will there be in 17 days?
Solution
To solve this problem, use the doubling time model with \(D=8\) and \(P_{0} = 100\) so the doubling time model for this problem is:
\[P(t) = 100(2)^{t/8} \nonumber \]
When \(t = 17\, days\),
\[P(17) = 100(2)^{\frac{17}{8}} = 436 \nonumber \]
There are 436 flies after 17 days.
Note: The population of flies follows an exponential growth model.
Sometimes we want to solve for the length of time it takes for a certain population to grow given their doubling time. To solve for the exponent, we use the log button on the calculator.
Suppose that a bacteria population doubles every six hours. If the initial population is 4000 individuals, how many hours would it take the population to increase to 25,000?
Solution
\(P_{0} = 4000\) and \(D = 6\), so the doubling time model for this problem is:
\[P(t) = 4000(2)^{\frac{t}{6}}\nonumber \]
Now, find t when \(P(t) = 25,000\)
\[25,000 = 4000(2)^{\frac{t}{6}}\nonumber \]
\[\dfrac{25,000}{4000} = \dfrac{\cancel{4000}(2)^{\frac{t}{6}}}{\cancel{4000}}\nonumber \]
\[6.25 = (2)^{\frac{t}{6}}\nonumber \]
Now, take the log of both sides of the equation.
\[\text{log}6.25 = \text{log}(2)^{\frac{t}{6}}\nonumber \]
The exponent comes down using rules of logarithms.
\[\text{log}6.25 = (\dfrac{t}{6}) \text{log}(2) \nonumber \]
Now, calculate log6.25 and log2 with your calculator.
\[\begin{align*} 0.7959 &= (\dfrac{t}{6}) \cdot 0.3010 \\ \dfrac{0.7959}{0.3010} &= \dfrac{t}{6} \\ 2.644 &= \dfrac{t}{6} \\ t &= 15.9 \end{align*} \nonumber \]
The population would increase to 25,000 bacteria in approximately 15.9 hours.
Rule of 70
There is a simple formula for approximating the doubling time of a population. It is called the rule of 70 and it is an approximation for growth rates less than 15%. Do not use this formula if the growth rate is 15% or greater.
For a quantity growing at a constant percentage rate (not written as a decimal), \(R\), per time period, the doubling time is approximately given by
\[D \approx \dfrac{70}{R} \nonumber \]
A bird population on a certain island has an annual growth rate of 2.5% per year. Approximate the number of years it will take the population to double. If the initial population is 20 birds, use it to find the bird population of the island in 17 years.
Solution
To solve this problem, first approximate the population doubling time.
Doubling time \(D \approx \dfrac{70}{2.5} = 28\) years.
With the bird population doubling in 28 years, we use the doubling time model to find the population is 17 years.
\[P(t) = 20(2)^{\frac{t}{28}} \nonumber \]
When \(t = 17\) years
\[P(16) = 20(2)^{\frac{17}{28}} = 30.46 \nonumber \]
There will be 30 birds on the island in 17 years.
A certain cancerous tumor doubles in size every six months. If the initial size of the tumor is four cells, how many cells will there in three years? In seven years?
Solution
To calculate the number of cells in the tumor, we use the doubling time model. Change the time units to be the same. The doubling time is six months = 0.5 years.
\[P(t) = 4(2)^{\frac{t}{0.5}} \nonumber \]
When \(t = 3\) years
\[P(3) = 4(2)^{\frac{3}{0.5}} = 256 \text{cells} \nonumber \]
When \(t = 7\) years
\[P(7) = 4(2)^{\frac{7}{0.5}} = 65,536 \text{cells} \nonumber \]
Suppose that a certain city’s population doubles every 12 years. What is the approximate annual growth rate of the city?
Solution
By solving the doubling time model for the growth rate, we can solve this problem.
\[\begin{align*} D &\approx \dfrac{70}{R} \\ R \cdot D &\approx \dfrac{70}{\cancel{R}} \cdot \cancel{R} \\ RD &\approx 70 \\ \dfrac{R\cancel{D}}{D} &\approx \dfrac{70}{D} \\ \text{Annual growth rate R} &\approx \dfrac{70}{D} \\ R &= \dfrac{70}{12} = 5.83 \% \end{align*} \nonumber \]
The annual growth rate of the city is approximately 5.83%
Exponential Decay and Half-Life Model
The half-life of a material is the time it takes for a quantity of material to be cut in half. This term is commonly used when describing radioactive metals like uranium or plutonium. For example, the half-life of carbon-14 is 5730 years.
If a substance has a half-life, this means that half of the substance will be gone in a unit of time. In other words, the amount decreases by 50% per unit of time. Using the exponential growth model with a decrease of 50%, we have
\[P(t) = P_{0}(1-0.5)^{t} = P_{0}(\dfrac{1}{2})^{t} \nonumber \]
Let’s say a substance has a half-life of eight days. If there are 40 grams present now, how much is left after three days?
Solution
We want to find a model for the quantity of the substance that remains after t days. The amount of time it takes the quantity to be reduced by half is eight days, so this is our time unit. After t days have passed, then t8 is the number of time units that have passed. Starting with the initial amount of 40, our half-life model becomes:
\[P(t) = 40(\dfrac{1}{2})^{\frac{t}{8}} \nonumber \]
With \(t=3\)
\[P(3) = 40(\dfrac{1}{2})^{\frac{3}{8}} = 30.8 \nonumber \]
There are 30.8 grams of the substance remaining after three days.
If \(H\) is the half-life of a quantity (the amount of time it takes the quantity be cut in half) and \(P_{0}\) is the initial amount of the quantity then the amount of the quantity present after t units of time is
\[P(t) = P_{0}(\dfrac{1}{2})^{\frac{t}{H}} \nonumber \]
Lead-209 is a radioactive isotope. It has a half-life of 3.3 hours. Suppose that 40 milligrams of this isotope is created in an experiment, how much is left after 14 hours?
Solution
Use the half-life model to solve this problem.
\(P_{0} = 40\) and \(H = 3.3\), so the half-life model for this problem is:
\[P(t) = 40(\dfrac{1}{2})^{\frac{t}{3.3}} \nonumber \]
With \(t=14\) hours,
\[P(14) = 40(\dfrac{1}{2})^{\frac{14}{3.3}} = 2.1 \nonumber \]
There are 2.1 milligrams of Lead-209 remaining after 14 hours.
Note: The milligrams of Lead-209 remaining follows a decreasing exponential growth model.
Nobelium-259 has a half-life of 58 minutes. If you have 1000 grams, how much will be left in two hours?
Solution
We solve this problem using the half-life model. Before we begin, it is important to note the time units. The half-life is given in minutes and we want to know how much is left in two hours. Convert hours to minutes when using the model: two hours = 120 minutes.
\(P_{0} = 1000\) and \(H = 58\) minutes, so the half-life model for this problem is:
\[P(t) = 1000(\dfrac{1}{2})^{\frac{t}{58}} \nonumber \]
With \(t=120\) minutes,
\[P(120) = 1000(\dfrac{1}{2})^{\frac{120}{58}} = 238.33 \nonumber \]
There are 238 grams of Nobelium-259 is remaining after two hours.
Radioactive carbon-14 is used to determine the age of artifacts because it concentrates in organisms only when they are alive. It has a half-life of 5730 years. In 1947, earthenware jars containing what are known as the Dead Sea Scrolls were found. Analysis indicated that the scroll wrappings contained 76% of their original carbon-14. Estimate the age of the Dead Sea Scrolls.
Solution
In this problem, we want to estimate the age of the scrolls. In 1947, 76% of the carbon-14 remained. This means that the amount remaining at time t divided by the original amount of carbon-14, \(P_{0}\), is equal to 76%. So, \(\dfrac{P(t)}{P_{0}} = 0.76\) we use this fact to solve for t .
\[\begin{align*} P(t) &= P_{0}(\dfrac{1}{2})^{\frac{t}{5730}} \\ \dfrac{P(t)}{P_{0}} &= (\dfrac{1}{2})^{\frac{t}{5730}} \\ 0.76 &= (\dfrac{1}{2})^{\frac{t}{5730}} \end{align*} \nonumber \]
Now, take the log of both sides of the equation.
\[\text{log} 0.76 = \text{log}(\dfrac{1}{2})^{\frac{t}{5730}} \nonumber \]
The exponent comes down using rules of logarithms.
\[\text{log} 0.76 = (\frac{t}{5730}) \text{log}\dfrac{1}{2}\nonumber \]
Now, calculate log0.76 and log\(\dfrac{1}{2}\)with your calculator.
\[\begin{align*} -0.1192 &= (\frac{t}{5730}) \cdot (-0.3010) \\ \dfrac{-0.1192}{-0.3010} &= \dfrac{t}{5730} \\ 0.3960 &= \dfrac{t}{5730} \\ t &= 2269.08 \end{align*} \nonumber \]
The Dead Sea Scrolls are well over 2000 years old.
Plutonium has a half-life of 24,000 years. Suppose that 50 pounds of it was dumped at a nuclear waste site. How long would it take for it to decay into 10 lbs?
Solution
\(P_{0} = 50\) and \(H = 24,000\) minutes, so the half-life model for this problem is:
\[P(t) = 50(\dfrac{1}{2})^{\frac{t}{24,000}} \nonumber \]
Now, find \(t\) when \(P_{t} = 10\).
\[\begin{align*} 10 &= 50(\dfrac{1}{2})^{\frac{t}{24,000}} \\ \dfrac{10}{50} &= (\dfrac{1}{2})^{\frac{t}{24,000}} \\ 0.2 &= (\dfrac{1}{2})^{\frac{t}{24,000}} \end{align*} \nonumber \]
Now, take the log of both sides of the equation.
\[\text{log} 0.2 = \text{log}(\dfrac{1}{2})^{\frac{t}{24,000}} \nonumber \]
The exponent comes down using rules of logarithms.
\[\text{log} 0.2 = (\frac{t}{24,000}) \text{log}\dfrac{1}{2}\nonumber \]
Now, calculate log0.2 and log\(\dfrac{1}{2}\)with your calculator.
\[\begin{align*} -0.6990 &= (\frac{t}{24,000}) \cdot (-0.3010) \\ \dfrac{-0.6990}{-0.3010} &= \dfrac{t}{24,000} \\ 2.322 &= \dfrac{t}{24,000} \\ t &= 55,728 \end{align*} \nonumber \]
The quantity of plutonium would decrease to 10 pounds in approximately 55,728 years.
There is simple formula for approximating the half-life of a population. It is called the rule of 70 and is an approximation for decay rates less than 15%. Do not use this formula if the decay rate is 15% or greater.
For a quantity decreasing at a constant percentage (not written as a decimal), R , per time period, the half-life is approximately given by:
\[\text{Half-life } H \approx \dfrac{70}{R} \nonumber \]
The population of wild elephants is decreasing by 7% per year. Approximate the half -life for this population. If there are currently 8000 elephants left in the wild, how many will remain in 25 years?
Solution
To solve this problem, use the half-life approximation formula.
\[\text{Half-Life } H \approx \dfrac{70}{7} = 10 \text{ years} \nonumber \]
\(P_{0} = 7000\) and \(H = 10\) years, so the half-life model for this problem is:
\[P(t) = 7000(\dfrac{1}{2})^{\frac{t}{10}} \nonumber \]
When \(t=25\),
\[P(25) = 7000(\dfrac{1}{2})^{\frac{25}{10}} = 1237.4 \nonumber \]
There will be approximately 1237 wild elephants left in 25 years.
Note: The population of elephants follows a decreasing exponential growth model.
| Rules of Exponents | Rules of Logarithm for the Common Logarithm (Base 10) |
|---|---|
|
Definition of an Exponent \(a^{n} = a\cdot a\cdot a\cdot a\cdot ..... \cdot a\) (n a's multiplied together) |
Definition of a Logarithm \(10^{y} = x \text{ if and only if } \text{log}x = y\) |
| Zero Rule \(a^{0} = 1\) | |
| Product Rule \(a^{m} \cdot a^{n}= a^{m+n}\) | Product Rule \(\text{log}(xy) =\text{log }(x) + \text{log }(y)\) |
| Quotient Rule \(\dfrac{a^{m}}{a^{n}}= a^{m-n}\) | Quotient Rule \(\text{log}(\dfrac{x}{y}) =\text{log }(x) - \text{log }(y)\) |
| Power Rule \((a^{n})^{m}= a^{n\cdot m}\) | Power Rule \(\text{log }x^{r} =r \text{log }(x) (x > 0)\) |
| Distributive Rules \((ab)^{n}= a^{n} \cdot a^{n}, (\dfrac{a}{b})^{n} = \dfrac{a^{n}}{b^{n}} \) | \(\text{log}10^{x} =x \text{log}(10) = x\) |
| Negative Exponent Rules \( a^{-n} = \dfrac{1}{a^{n}}, (\dfrac{a}{b})^{-n} = (\dfrac{b}{a})^{n} \) | \(10^{\text{log}x} = x (x > 0)\) |