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Question
If the population of a country doubles in 60 years, in how many years will it be triple under
the assumption that the rate of increase in proportional to the number of inhabitants?
[Given : log 2 = 0.6912 and log 3 = 1.0986.]
Solution
Let P be the population of the country at time t.
Given `"dP"/dt prop P`
`therefore "dP"/dt =kP` (where k is a constant)
`therefore 1/PdP=kdt`
Integrating both the side w.r.t x
`int 1/Pdp=kint 1 dt+c`
`logP=kt+c`
`P=e^(kt+c)=e^(kt).e^c`
Let `e^c=alpha`
`therefore P=alpha.e^(kt)`
Let initial population at t = 0
`therefore N=alpha.e^0 thereforeN=alpha`
`P=N.e^(kt)`
Given P = 2N when t = 60 years,
`therefore 2N=Ne^(60k)`
`therefore 2=e^(60k)=>k=1/60 log 2`
`therefore P=N.e^(60k)`
Required t when P = 3N
`3=e^(kt)=>log3=kt`
`log3=(1/60log2).t`
`t=(60log3)/log2`
`=(60xx1.0986)/0.6912`
`=95.4 years(approx.)`
The population of the countr will triple approximately in 95.4 years.
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