[y = et, then dy/dt = y. The one such solution for this latter equation. This property of e makes it very useful for problems of exponential growth and decay -problems of a fixed rate of expansion or contraction.]
Some things to know:
ln(e) = ln(e1) = 1
ln(ex) = x
ln(ert) = rt
log(ab) = log(a) + log(b) - for any log (including ln)
log(ab) = b[log(a)]
I want you to accustom yourself to this formula.
N(t) = N(0)ert, where t is time and r is a constant of inverse time representing the rate of change, i.e. r = 10% per year (0.1/yr). N(0) is the initial number of your stuff (be it humans on the planet or radioactive particles) at t = 0; and N(t) is the number at any time t.
If N(0) = $500 and r = 4%/year. What is N at t = 25 years?
What are we solving for? We have three of four variables given.
Thus N(25) = $500 x er4%/year x 25 years = $500 x e1 = $1429. This is continuously compounded interest.
Remember that r has a unit measure it is a time-1. You must keep track of the time (reference) being considered. If growth is 4%/month and I ask the same question as above - what is N at t = 25 years (25 years = 300 months).
N(25 years) = $500 x e.04/month x 300 months = $500 x e12 = $81.4 x 106. A lot more money.
The take home is that r is a rate - in this case a percent per time, where percent is just a fraction (unitless).
Why are things plotted on log plots - because ln(N(t)) = ln(N(0)rt) = ln(N(0) + rt, which is just y = ax + b, a linear graph. Important: If you wish to know whether a population (a substance, anything) is growing or decaying exponentially take the natural log (or log ten) of the population and plot this versus time. If it is exponential, or at least in part, the data, or part of the data, will plot as a straight line. You need many points for this, and it is easy to do on a spread sheet.
| Time | Population | Natural Log | Log 10 |
|---|---|---|---|
| 1 | 10 | 2.3 | 1 |
| 2 | 101 | 4.6 | 2 |
| 3 | 1003 | 6.9 | 3 |
Taking the example of world population growth that Dr. Simpson used:
| Year | Population (millions) |
|---|---|
| 1950 | 2516 |
| 1995 | 5750 |
What is the rate of growth over this time--assuming it to be exponential?
N(t) = N(0) ert.
Ln (N(t)) = Ln(N(0)) + rtHere t = 1995-1950 = 45 years, N(t) = 5.750 x 109, N(0) = 2.516 x 109.
Thus, Ln(5.750 x 109) = Ln(2.516 x 109) + r x 45 years
22.47 = 21.65 + r x 45
0.82 = 45r
r = 0.01822, or 1.822% per year.
The question then can be asked: if this growth rate remains, when will the population of the planet reach 10 billion people?
| Year | Population (millions) |
|---|---|
| 1995 | 5750 |
| ?? | 10000 |
Ln(1.000 x 1010) = Ln(5.750 x 109) + 0.01822 x t
23.03 = 22.47 + 0.01822t
0.56 = 0.01822t
t = 30.77 years, or by the end of 2026.
This is one way to create worst case and best case prediction scenarios. Researchers study demographic trends, the factors that affect into population growth, such as China's restrictions on large families, or overcrowding in Sao Paolo. Decisions are made as to whether the rate of growth is going up or down, etc.; margins of error are established based on what are considered reasonable limits (high and low) for the change of this rate, then the calculations are made. Suppose the growth rate drops to 0.01, then
0.56 = 0.01t
t = 56 years, or 10 billion people by 2051
Suppose the rate increases to 0.025, then
0.56 = 0.025t
t = 22.4 years, or 10 billion people by the end of 2018.
The same thing works for decay. Later we will work with radioactive materials. That value of r instead of being a positive rate is negative. If they want an example give it to them ($500, losing 5%/year, when will have lost half your money).
Point is with these problems - if you know three of the four variables - r, t, N(0) and N(t), you can solve for the fourth.
This example was originally part of the Tutorial section of U4735 Environmental Science for Policy Makers (9/21/99).