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Just Enough Algebra

Dr. Suzanne Dorée, Dr. Spencer Bagley

Section 5.3 Growth factors

Obesity among children ages 6-11 continues to increase. From 1994 to 2010, the proportion of children classified as obese rose from an average of 1.1 out of every ten children in 1994 to around 2 out of every ten children in 2010.

Aside

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Assuming that the prevalence of childhood obesity increases exponentially, what is the annual percent increase and what does the equation project for the year 2020? Well, unless we are able to make drastic improvements in how children eat and how much they exercise.
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Because we are told obesity is increasing exponentially we can use the template for an exponential equation.
\begin{equation*} \text{dep }=\text{ start } \ast \text{growth factor}^{\text{indep}} \end{equation*}
The variables are
\begin{align*} C \amp= \text{ obese children (out of every ten) } \sim \text{ dep} \\ Y \amp= \text{ time (years since 1994) } \sim \text{ indep} \end{align*}
The starting amount is 1.1 children out of every ten in 1994 so our equation is of the form
\begin{equation*} C = 1.1 \ast g^Y \end{equation*}
Trouble is we don’t actually know what the growth factor \(g\) is. Yet.
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We do know that in 2010 we have \(Y = 2010 - 1994 = 16\) years and \(C = 2\text{.}\) We can put those values into our equation to get
\begin{equation*} 1.1 \ast g^{16}=2 \end{equation*}
No particular reason for switching sides, just wanted to have the variable on the left. That’s supposed to be true but we don’t know what number \(g\) is so we can’t check. Argh.
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Oh, wait a minute. The only unknown in that equation is the growth factor \(g\text{.}\) What if we solve for \(g\text{?}\) First, divide each side by 1.1 to get
\begin{equation*} \frac{\cancel{1.1}\ast g^{16}}{\cancel{1.1}} = \frac{2}{1.1} \end{equation*}
which simplifies to
\begin{equation*} g^{16} = \frac{2}{1.1} = 2 \div 1.1= 1.818181818\ldots \end{equation*}
Since we want to solve for the base (not the exponent), we have a power equation. We use the Root Formula with power \(n=16\) and value \(v=1.818181818\) to get
\begin{equation*} g = \sqrt[n]{v} = \sqrt[16]{1.818181818} = 16 \sqrt[x]{~\text{ }} 1.818181818 = 1.038071653 \approx 1.0381 \end{equation*}
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Want a quicker way to find the growth factor? Forget the entire calculation we just did. It all boils down to two steps:
\begin{equation*} \frac{2}{1.1} = 2 \div 1.1 = 1.818181818\ldots \end{equation*}
and then
\begin{equation*} g = \sqrt[16]{1.818181818} = 16 \sqrt[x]{~\text{ }} 1.818181818 = 1.038071653 \approx 1.0381 \end{equation*}
We can even do this calculation all at once as
\begin{equation*} g = \sqrt[16]{\frac{2}{1.1}} = 16 \sqrt[x]{~\text{ }} (2 \div 1.1)= 1.038071653 \approx 1.0381 \end{equation*}
Notice we added parentheses because the normal order of operations would do the root first and division second. We wanted the division calculated before the root.
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Here’s the easy version in a formula.
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Growth Factor Formula

If a quantity is growing (or decaying) exponentially, then the growth (or decay) factor is
\begin{equation*} \displaystyle g = \sqrt[t]{\frac{a}{s}} \end{equation*}
where \(s\) is the starting amount and \(a\) is the amount after \(t\) time periods.
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We knew from the beginning that our equation was in the form \(C = 1.1 \ast g^Y\text{.}\) Now that we found the growth factor \(g \approx 1.0381\) we get our final equation
\begin{equation*} C = 1.1 \ast 1.0381^Y \end{equation*}
For example, we can check that in 2010, we have \(Y=16\) still and so
\begin{equation*} C = 1.1 \ast 1.0381^{16} = 1.1 \times 1.0381 \wedge \underline{16} = 2.000874004 \approx 2 \quad \checkmark \end{equation*}
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You might wonder why we didn’t just round off and use the equation
\begin{equation*} C = 1.1 \ast 1.04^Y \end{equation*}
Look what happens when we evaluate at \(Y=16\) then. We would get
\begin{equation*} C = 1.1 \ast1.04^{16} = 1.1 \times 1.04 \wedge \underline{16} = 2.06027937 \approx 2.1 \end{equation*}
Not a big difference (2.1 vs. 2.0) but enough to encourage us to keep extra digits in the growth factor in our equation. Lesson here is: don’t round off the growth factor too much.
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Back to the more reliable equation
\begin{equation*} C = 1.1 \ast 1.0381^Y \end{equation*}
We can now answer the two questions. First, in 2020 we have \(Y = 2020-1994 = 26\) and so
\begin{equation*} C = 1.1 \ast 1.0381^{26} = 1.1 \times 1.0381 \wedge \underline{26} = 2.908115507 \approx 2.9 \end{equation*}
According to our equation, by 2020 there would be approximately 2.9 obese children for every ten children.
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The other question was what the annual percent increase is. Think back to an earlier example. Remember that Jocelyn was analyzing health care costs in Section 2.2? They began at $2.26 million and grew 6.7% per year. She had the equation
\begin{equation*} H=2.26\ast1.067^Y \end{equation*}
So the growth factor \(g=1.067\) in the equation came from the growth rate \(r=6.7\%=0.067\text{.}\) Our equation modeling childhood obesity is
\begin{equation*} C = 1.1 \ast 1.0381^Y \end{equation*}
The growth factor of \(g=1.0381\) in our equation must come from the growth rate \(r= 0.0381=3.81\%\text{.}\) Think of it as converting to percent \(1.0381 = 103.81 \%\) and then ignoring the 100% to see the 3.81% increase. Childhood obesity has increased around 3.81% each year. Well, on average.
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Here’s the general formula relating the growth rate and growth factor.
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Percent Change Formula

  • If a quantity changes by a percentage corresponding to growth rate \(r\text{,}\) then the growth factor is
    \begin{equation*} \displaystyle g=1+r \end{equation*}
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  • If the growth factor is \(g\text{,}\) then the growth rate is
    \begin{equation*} r = g-1 \end{equation*}
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Let’s check. We have \(g=1.0381\) and so the growth rate is
\begin{equation*} r=g-1 = 1.0381-1 = 0.0381= 3.81\% \end{equation*}
Not sure we really need these formulas, but there you have it.
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By the way, formula works just fine if a quantity decreases by a fixed percent. One example we saw was Joe, who drank too much coffee. The growth (or should I say decay) factor was \(g=0.87\text{.}\) That corresponds to a growth (decay) rate of
\begin{equation*} r=g-1=0.87-1=-0.83=-13\% \end{equation*}
Again, the negative means that we have a percent decrease.
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Do you know …

  1. How to find the growth/decay factor given the starting amount and another point of information?
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  2. How to find the growth/decay factor given the doubling time or half-life?
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  3. When we use the Percent Change Formula, and when we use the Growth Factor Formula instead? Ask your instructor if you need to remember the Percent Change Formula and Growth Factor Formula or if they will be provided during the exam.
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  4. How to evaluate the Percent Change Formula and Growth Factor Formula using your calcuator?
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  5. How to read the starting amount and percent increase/decrease from the equation?
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If you’re not sure, work the rest of exercises and then return to these questions. Or, ask your instructor or a classmate for help.
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Exercises Exercises

Exercises 1-4 are available in a separate workbook format.
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1.

In 1962, my grandfather had savings bonds that matured to $200. He gave those to my mother to keep for me. These bonds have continued to earn interest at a fixed, guaranteed rate so I have yet to cash them in. The table lists the value at various times since then.
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year 1962 1970 1980 1990 2000 2010
years since 1962 0 8 18 28 38 48
value 200.00 318.77 570.87 1,022.34 1,830.85 3,278.77
(Story also appears in 1.2.1 and 4.1.3.a)
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(c)
What do you notice? What in the story told you that would happen?
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(d)
What is the corresponding interest rate?
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(e)
Write an equation for the value of bonds over time.
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(f)
Use your equation to check the information for 1990, 2000, and 2010.
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(g)
In what year will the bond be worth over $5,000? Set up and solve an equation to decide.
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(h)
Draw a graph using the data in the table, but not your answer to part (g). Include another year that is later than your answer to part (g).
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(i)
Does your answer to part (g) agree with your graph? If not, fix your work.
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2.

Have you read news stories about archaeological digs where a specimen (like a bone) is found that dates back thousands of years? How do scientists know how old something is? One method uses the radioactive decay of carbon.
Aside
After an animal dies the carbon-14 in its body very slowly decays. By comparing how much carbon-14 remains in the bone to how much carbon-14 should have been in the bone when the animal was alive, scientist can estimate how long the animal has been dead. Clever, huh? Actually, it is so clever that Willard Libby won the Nobel Prize in Chemistry for it. The key information to know is that the half-life of carbon-14 (the amount of time it takes for half of the original amount of carbon-14 to decay) is about 5,730 years. For this problem, suppose a bone were found that should have contained 300 milligrams of carbon-14 when the animal was alive.
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(a)
Find the annual “growth” factor. Keep at least six digits after the decimal place for your calculations.
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Hint.
If the bone started off with 300 mg of carbon-14, how much carbon-14 would be left after 5,730 years?
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(b)
Name the variables and write an equation describing the dependence.
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(c)
How many milligrams of carbon-14 should remain in this bone after 1,000 years? After 10,000 years? After 100,000 years?
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(d)
How many milligrams of carbon-14 should remain in this bone after 1 million years? Explain the “scientific notation” answer your calculator gives you.
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(e)
Draw a graph that shows up to 10,000 years.
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(f)
If the bone is determined to have 100 milligrams of carbon-14, approximately how long ago did it die? Start by estimating the answer from your graph. Then revise your estimate using successive approximation. Display your guesses in a table.
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(g)
Solve the equation exactly.
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3.

For each story, find the annual growth factor \(g\) and annual growth rate \(r\) as a percent.
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Hint: First decide if you can use the Percent Change Formula or if you will need to use the Growth Factor Formula. Don’t forget to include the negative sign for decay rates.
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(a)
Donations to the food shelf have increased 35% per year for the past few years.
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\begin{equation*} g= \end{equation*}
\begin{equation*} r = \end{equation*}
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(b)
People picking up food at the food shelf has increased exponentially too, from 120 per week in 2005 to 630 per week in 2011.
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\begin{equation*} g= \end{equation*}
\begin{equation*} r = \end{equation*}
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(c)
The crime rate has dropped 3% each year recently.
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\begin{equation*} g= \end{equation*}
\begin{equation*} r = \end{equation*}
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(d)
The new stop sign has decreased accidents exponentially, from 40 in 2008 to 17 in 2013.
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\begin{equation*} g= \end{equation*}
\begin{equation*} r = \end{equation*}
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(e)
The creeping vine taking over Fiona’s lawn will double in area each year.
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\begin{equation*} g= \end{equation*}
\begin{equation*} r = \end{equation*}
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(f)
Attendance at parent volunteer night has doubled every 3 years.
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\begin{equation*} g= \end{equation*}
\begin{equation*} r = \end{equation*}
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(g)
The number of people addicted to prescription drugs was estimated to have tripled in the past 5 years. Assume the number is increasing exponentially.
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\begin{equation*} g= \end{equation*}
\begin{equation*} r = \end{equation*}
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(h)
The number of high school students arrested for driving under the influence is half what it was 5 years ago. Assume the number is falling exponentially.
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\begin{equation*} g= \end{equation*}
\begin{equation*} r = \end{equation*}
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4.

For each equation, find the growth rate and state its units. For example, something might “grow 2% per year” while something else might “drop 7% per hour”.
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(a)
The number of households watching reality television \(R\) (in millions) was estimated by the equation
\begin{equation*} R=2.5 \ast 1.072^T \end{equation*}
where \(T\) is the time in years since 1990. (Story also appears in 5.1.10)
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(b)
Chlorine is often used to disinfect water in swimming pools, but the concentration of chlorine \(C\) (in ppm) drops as the swimming pool is used for \(T\) hours according to the equation
\begin{equation*} C = 2.5 \ast 0.975^T \end{equation*}
(Story also appears in 3.4.2)
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(c)
The number of players of a wildly popular mobile app drawing game has been growing exponentially according to the equation
\begin{equation*} N = 2 \ast 1.57^W \end{equation*}
where \(N\) is the number of players (in millions) and \(T\) is the time in weeks since people started playing the game. (Story also appears in 5.1.7)
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5.

Estimates for childhood obesity for 2010 were revised to 2.1 out of every ten children. (The 1994 figure of 1.1 out of every ten children remains accurate.)
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(a)
Calculate the revised growth factor. What is the revised percent increase?
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(c)
Use your new equation to project childhood obesity rates for 2020.
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(d)
Graph both the original and revised estimates on the same set of axes.
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6.

For each equation, find the growth rate (percent increase or percent decrease) and state the units. (For example, something might “grow 2% per year” while something else might “drop 7% per hour”)
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(a)
The light \(L\%\) that passes through panes of glass \(W\) inches thick is given by the equation
\begin{equation*} L = 100\ast 0.75^W \end{equation*}
(Story also appears in 2.4.7 and 3.4.7)
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(b)
The population of bacteria (\(B\)) in a culture dish after \(D\) days is given by the equation
\begin{equation*} B=2{,}000\ast 3^D \end{equation*}
(Story also appears in 5.2.6)
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(c)
The remaining contaminants (\(C\) grams) in a waste water sample after \(M\) months of treatment is given by
\begin{equation*} C=8 \ast 0.25^M \end{equation*}
(Story also appears in 5.2.8)
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7.

Years ago, Whitney bought an antique mahogany table worth $560. Now, 30 years later, she had the table appraised for $3,700.
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(a)
Calculate the annual growth factor, assuming the value of Whitney’s table has increased exponentially.
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(b)
What should she expect the set to be worth in another 10 years? As part of your work, name the variables and write an equation relating them.
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(c)
Find the corresponding annual percent increase.
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8.

The opiate drug morphine leaves the body quickly. After 72 hours about 10% remains. A patient receives 100 mg of morphine.
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(a)
How much morphine will remain in the patient’s body after 72 hours?
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(d)
What is the corresponding percent decrease?
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(e)
Name the variables and write an equation relating them. Check that 72 hours gives you the same answer as in part (a).
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(f)
What is the half-life of morphine? Set up and solve an appropriate equation.
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(g)
Draw a graph showing this patient’s morphine levels for 10 days following the injection.
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9.

Unemployment figures were just released. At last report there were 20,517 unemployed adults and now, 10 months later, we have 39,061 unemployed adults.
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(a)
Calculate the monthly growth factor, assuming unemployment increases exponentially.
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(b)
Write an equation relating the variables.
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(c)
According to your equation, what is the expected number of unemployed adults 6 months from now. Notice: the report was issued 10 months ago.
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(d)
Make a table of values and draw a graph showing the number of unemployed adults for the past 10 months and the next 2 years.
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10.

Wetlands help support fish populations, various plant and animal populations, control floods and erosion from nearby lakes and streams, filter water, and help preserve our supply of ground water. Minnesota wetlands acreage in 1850 was 18.6 million acres. By 2003, that number had dropped to 9.3 million acres.
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Aside
(a)
Assuming the acreage decreased exponentially, name the variables, find the annual decay factor and write an exponential equation showing how Minnesota wetlands have decreased.
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(b)
With some effective management, many wetlands have been restored. By 2012, it’s up to about 10.6 million acres. Assuming acreage has increased exponentially from 2003, name the variables (you may now want to start the years in 2003), find the growth factor and write an exponential equation showing how Minnesota wetlands have been restored.
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