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

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 don't understand that part.

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1. log 2/log 1.03 = 23.44977. The mistake you are making is log 1.03 is 0.0128. You must divide that number into log 2 which is 0.30103 and you will obtain 23.45 if you do that (23.44977 to be exact about it).

There is no divide by 72 in Mathmate'e response although there is a divide into 72. The rule of 72 is a quick short cut rule which is 72/annual rate of interest and this gives the approximate time for doubling. 72/3 = 24 years which is close to the actual time of 23.45 which Mathmate calculated for you.

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