Posted by Z32 on Tuesday, October 6, 2009 at 11:58pm.
Since one of the questions deals with "doubling time", I will use 2 as the base.
(Of course we could use any base, many mathematicians would automatically use e as the base.)
P(t) = a(2)^(kt) , where P(t) is the population in millions, a is the beginning population, and t is the time in years.
clearly , a = 40
P(t) = 40(2)^(kt)
when t=10, (1990), N = 50
50 = 40(2)^(10k)
1.25 = 2^(10k)
take the ln of both sides, hope you remember your log rules
10k = ln 1.25/ln 2
10k = .32193
k = .032193
so P(t) = 40(2)^(.032193t)
in 2000, t = 20
P(20) = 40(2)^(.032193(20))
= 62.5 million
for the formula
P(t) = a(2)^(t/d), d = the doubling time
so changing .032193t to t/d
= .032193t
= t/31.06
so the doubling time is 31.06
another way would be to set
80 = 40(2)^(.032193t)
2 = (2)^(.032193t)
.032193t = ln 2/ln 2 = 1
t = 31.06
Curiously, this is very close to the current population doubling time for India.
Related Questions
calculus - The population of a country was 8.4 million people in 2000. Three ...
Math AP Calc - In 1951, the population of India was 357 million people. By 1981 ...
Math - I have no idea how to even start this problem. In 1951, the population of...
Math - In 1951, the population of India was 357 million people. By 1981 it had ...
Math-Reiny Help! Previous Problem - In 1951, the population of India was 357 ...
6th grade Math - I need some help figuring out how to work the word problem ...
Calculus - Year 1960 1970 1980 1990 2000 Rate 39 59 64 89 106 of change (million...
College Mathematics - Guys, could you please help me with this problem? I'm ...
MATH - A table of the U.S. population in millions indicated in 1960 the ...
Math - Imagine a country has a population of 210 million. Within the country ...
For Further Reading