[In RHWN #197, we introduced some tools that allow anyone to understand things that grow exponentially, which is to say, things that grow by a constant percentage of the whole in a constant time period, such as a bank account that grows at the rate of 6% each year. Most things that affect the environment are growing exponentially. Tools for understanding exponential growth are very useful because they allow us to see why we're in the situation we're in today.]

RULE 5: If you know some original amount, called N_sub_o, and some amount that it has grown to, N_sub_t, at some later time, t, you can calculate the value of k (the annual rate of change, expressed as a decimal fraction) by the following equation:

k = (ln(N_sub_t/N_sub_o))/t [Rule 5]

where:

ln means "natural logarithm of" (your $15 Radio Shack scientific calculator will give you the natural log of any number instantly);

N_sub_o is some original amount;

N_sub_t is the amount that it has grown to at some later time, t;

k = the annual percentage increase expressed as a decimal fraction (in other words, it's the value we've been calling p, divided by 100) [for example, if something is growing at 6% per year, p = 6 and k = 0.06];

t = time (in any units you care to choose).

In other words, take the amount N_sub_t at time t, divide it by N_sub_o (the original amount), take the natural log of the result, then divide that natural log by t. That will give you k; if you then multiply k by 100, you have p. Then you can use k and p with Rules 1 through 4 (given in RHWN #197) to learn important characteristics of the thing that's growing.

For example, the consumer price index grew from 133.1 in 1973 to 246.8 in 1980. What was the annual percentage rate of increase, p? To find p, first we calculate k. The annual increase rate (expressed as a decimal fraction), k, was:

(ln(246.8/133.1))/t

Since t = 1980-1973 = 7 years, and 246.8/133.1 = 1.85 and the natural log of 1.85 = 0.62, the value of k = 0.62/7, or 0.088; thus p = 0.088 x 100 or 8.8% per year increase. With this number in hand, we can use rules 1 through 4 to learn its other growth characteristics.

RULE 6: If a quantity is growing exponentially and you want to know how rapidly it is increasing by a factor of x (i.e., how fast it's increasing by a factor of, say, 5, or, as we might say, how long it takes to increase 5-fold), use this equation:

x folding time = ln(x)/k [Rule 6]

Example: If a bank account is growing at 6% per year, how long will it take for the amount in the bank to increase five-fold? The natural log of 5 = 1.61; 1.61/0.06 = 26.8. Therefore, a bank account growing at 6% per year will increase 5-fold in a period of 26.8 years.

RULE 7: If an exponentially-growing quantity has increased by a factor of x during some period of time, t, then the fractional increase, k, per unit time can be found by the equation

k = ln(x)/x folding time [Rule 7]

Example: The annual production of a very useful chemical increased 6-fold from 1949 to 1958. What was the annual percentage increase?

My scientific calculator tells me that the natural log of 6 = 1.79. The x folding time = 1958-1949 = 9. Thus k = 1.79/9 = 0.198 or 19.8% per year, an impressive annual rate of increase, which, by Rule 1, would lead to a doubling of that chemical's production every 70/19.8 = 3.5 years, and, by Rule 4, would lead to an increase by a factor of a million during one human lifetime. Impressive growth, indeed. (See Table 1 in RHWN #197.)

RULE 8: If you want to know the total amount of an exponentially-growing quantity produced during time-period t, use this equation:

C = (N_sub_o/k)*(e**((k*t)-1)) [Rule 8]

where:

** means "raise to the power of"

* means "multiply by"

C = total amount produced during time-period t;

N_sub_o = the original amount;

k = the increase (expressed as a decimal fraction) per unit of time (for example, per year);

e = 2.718, the base of natural logarithms;

t = time (in any units).

For example, if 40,000 automobiles were produced in 1920 and car production increased 10% per year for the next 15 years, how many cars were produced, total, during the 15 years? N_sub_o = 40,000; k = 0.10; t = 15. Therefore, C = (40,000/0.10) (e**((0.10*15)-1)) = 1,392,676.

In writing the equation above, we've used an asterisk to indicate multiplication; in other words, 2*2 = 4, and we've used ** to indicate exponentiation, meaning "raised to the power of;" i.e., 2**3 = 8.

RULE 9: During one doubling time, the growth in an exponentially-growing quantity equals all the growth that has occurred in all previous time. Thus if production of chemical XYZ is growing at 10% per year (doubling in 7 years), during the next 7 years we will produce an amount of chemical XYZ equal to all of the chemical XYZ that has been produced up to today.

These 9 rules can help us all understand why environmental problems have sneaked up on us, and why the growth of some things must be curbed.

Let's take an example. Let's say we learn from a magazine article that the synthetic organic chemical industry produced 38 billion pounds of chemicals in 1945 and since then has grown steadily at 6.5% per year. From that small amount of information, we can learn a lot. For example, from Rule 3 we can learn that annual synthetic organic chemical production in 1990 was about 708 billion pounds. By Rule 1 we can learn that total production is doubling every 10.8 years (we'll round it off and call it 11 years). By Rule 9, we can learn that during the most recent doubling-time (from 1979 through 1990), which happens to be the time since Love Canal was discovered, the chemical industry produced an amount of chemicals equal to the amount it produced during all time prior to the discovery of Love Canal. By Rule 8, we can learn that since 1945, the chemical industry has produced a total of 10.3 million million (or 1.03 x 10**13) pounds of chemicals, all of which has gone somewhere, and Rule 9 tells us that during the next 11 years they'll produce another 10.3 million million pounds. From Rule 4, we can learn that if the chemical industry keeps growing at 6.5% per year, during one human lifetime it will increase its size 90-fold; in other words, if the chemical industry keeps growing at its current rate, for every chemical factory in existence today, there will be 90 chemical factories just 70 years from now.

These facts help us understand why more and more people are complaining about chemicals affecting their lives: it isn't because of growing "chemophobia," it is because people are being exposed to more and more chemicals that are being released into the air and water that we all breathe and drink. A massive change--the "chemicalization of the environment"--has occurred during the past 50 years, and the problem is doubling in size every 11 years.

These facts can lead us to one more conclusion: since the chemical industry in all its trade publications constantly stresses the need for continuous growth at rates that match or exceed the historical rate of growth, we can conclude that executives in the chemical industry do not understand the implications of exponential growth. No one who has seen the heavily industrialized sections of chemical-producing states like New Jersey, Ohio, Louisiana, Texas and California can believe that we could survive 90 chemical factories where each chemical factory stands today. Growth of this industry has got to be curbed. By what means? By whatever non-violent means are necessary. Whatever it takes.

Further reading: Ralph Lapp, THE LOGARITHMIC CENTURY (Englewood Cliffs, NJ: Prentice-Hall, 1973). And: chapters 1 and 2 of Donella H. Meadows and others, THE LIMITS TO GROWTH (NY: Universe Books, 1972). And: Albert A. Bartlett, "The Exponential Function," THE PHYSICS TEACHER (October, 1976), pgs. 393-401.
--Peter Montague, Ph.D.

Descriptor terms: mathematics; chemical industry;