Posted by **Rani** on Wednesday, October 7, 2009 at 3:06pm.

If a population consists of 10,000 individuals at time t=0 years (P0), and the annual growth rate (excess of births over deaths) is 3% (GR), what will the population be after 1, 15 and 100 years (n)? Calculate the "doubling time" for this growth rate. Given this growth rate, how long would it take for this population of 10,0000 individuals to reach 1.92 million?

One equation that may be useful is:

Pt = Po * (1 + {GR/100})n

Science!! Please HELP - MathMate, Wednesday, October 7, 2009 at 12:08am

Apply the given formula,

P(t) = Po(1+r)n

where

r=0.03 is the annual growth rate in percent divided by a hundred.

n=number of years

Po=initial population=10000

For 1 year,

P(1) = 10000*(1.03)1

=10300

P(15)=10000*(1.03)15

= ______

P(100)=10000*(1.03)100

=_______

The doubling time is rougely 24 years using the rule of 72 (72 divided by the annual rate of interest).

The exact doubling time is log(2)/log(1.03)

=23.45 years.

Time required to grow from 10000 to 1920000 is

log(1920000/10000)/log(1.03)

=______ years

OK I understand the 1og2/log1.03 but when i calculate it i get

0.01280....

and why do you need to divide by 72? what is the prupose and where do you divide it i dont understand that part.

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