Math question?

Question

given that current world population is 6.45 * 10^9 and increasing at the rate of 1.14% per year, calculate the number of years it would take for the world population to double at the current rat

Answer 1

This can be calculated easily using the formula:
e^(rt) = end_population/init_population

e is Euler's Number, which is the base for natural logarithm (ln)
r is the rate of increase (or decrease if negative)
t is time (we're finding this) (in years, actually it can be anything, but since the increase rate is in years it could only be in years)
the rest explains itself

e^(rt) = end_population/init_population
e^(0.0114 * t) = 2
ln 2 = 0.0114 * t
t = (ln 2)/0.0114
t = 60.8023843 years

Notice that in your question, the number of initial population doesn't really matters, you just need to know the ratio of the end to initial population.

Doug's answer is not fully correct, it is flawed in a way that it assumed that population increases only at every end-of-year (while stays the same throughout the year). He uses a compound interest formula, while a population doesn't increase every end-of-year only, it increases throughout the year every day, every hour, every second. So we'll need to use continuous compound interest: http://en.wikipedia.org/wiki/Compound_interest#Continuous_compounding

Answer 2

If you set up an equation such as

P2=P1* e^(rate * time)

P1 is given 6.45 * 10^9 (initial population)
P2is double that 2*(6.45 * 10^9 ) or 12.9*10^9
rate is 1.14% - convert to decimal .0114
let t=time
now all you are looking for is the time - just plug all these values in to the above formula (this is a little long but it should help you know what you are doing and why)

12.9*10^9=6.45*10^9*e^(.0114*t)
Divide both sides by P1
2=e^(.0114*t)
use LN to get rid of e
LN2=(.0114*t)
Divide by .0114 to isolate t
(LN2)/.0114=t
60.8024=t

Answer 3

If n is the number of years, then (1+r)^n = (1.014)^n and we want that to be equal to 2 so
(1.014)^n=2 take teh log of both sides
n ln(1.014) = ln(2) and
n = ln(2)/ln(1.014) = 49.856 years.

Doug