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
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.
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