Arithmetic, Population, Energy

Dr. Albert A. Bartlett
Professor Emeritus
Department of Physics
University of Colorado at Boulder
http://en.wikipedia.org/wiki/Albert_Bartlett

The greatest shortcoming of the human race is our inability to understand the Exponential Function.

The Exponential function is used to describe the size of anything that growing steadily,

for example, 5% per year.

The exponential function

http://en.wikipedia.org/wiki/Exponential_function

We are talking about a situation where the time that is required for the growing quantity

to increase by a fixed fraction is constant.

If it takes a fixed length of time to grow 5%, then it follows that it takes a longer fixed length of time to grow by 100%.

This longer time is called the doubling time.

http://en.wikipedia.org/wiki/Doubling_time

Then, we need to calculate the doubling time. Watch the video.

You take 70 divided by the % growth per unit time

So, if your growth rate is 5% per year, then the doubling time is

70/5 = 14 years

the growing size will double every 14 years.

Doubling Time Formula

http://www.controlgrowth.org/double.htm

How do you get 70?

In the video, the professor says, "it's approximately the natural logarithm of 2"

ln(2) = 0.6931471805599

70 = 100 * ln (2)

If tripled instead of double, you use the natural logarithm of 3

getting,

ln(3) = 1.0986122886681

The natural logarithm

http://en.wikipedia.org/wiki/Natural_logarithm

Read more: the rule of 72, the rule of 70 and the rule of 69

http://en.wikipedia.org/wiki/Rule_of_72

Related topic: Solving for the period needed to double money

http://en.wikipedia.org/wiki/Time_value_of_money#Example_3:_Solving_for_the_period_needed_to_double_money

The video and those links do not show clearly how this is done.

In the case of doubling the amount of money, we use simple compound interests.

Since the initial principal is to be doubled we have the equation (formula for compound interests):

2 = (1 + r/100)^n

Taking the log of both sides:

log 2 = n * log(1 + r/100)

or

n = log 2 / [log(100 + r) - 2]

Then we look for a number, x, such that when divided by r, the result is approximately n.
Hence,
x / r = log 2 / [log(100 + r) - 2]

x = 0.301 * r / [log(100 + r) - 2]

Suppose r = 5%

0.301 * 5 = 1.50500
log(100 + r) = log(105)
log(105) - 2 = 0.0211892991

So, x = 1.50500 / 0.0211892991 = 71.0264173 (approximately 70)