If the population of a country doubles in 60 years; in how many years will it be triple (treble) under the assumption that the rate of increase is proportional to the number of inhabitants?

(Given log = 20.6912, log 3 = 1.0986)

#### Solution

Let P be the population at time t years. Then `"dP"/"dt"`, the rate of increase of population is proportional to P.

∴ `"dP"/"dt" ∝ "P"`

∴ `"dP"/"dt"` = kP, where k is a constant

∴ `"dP"/"P"` = k dt

On integrating, we get

`int "dP"/"P" = "k" int "dt" + "c"`

∴ log P = kt + c

Initially, i.e. when t = 0, let P = P_{0}

∴ log P_{0} = k × 0 + c

∴ c = log P_{0}

∴ log P = kt + log P_{0}

∴ log P - log P_{0} = kt

∴ `log ("P"/"P"_0)`= kt ...(1)

Since the population doubles in 60 hours, i.e. when t = 60, P = 2P_{0}

∴ `log ((2"P"_0)/"P"_0)` = 60k

∴ k = `1/60` log 2

∴ (1) becomes, `log ("P"/"P"_0) = "t"/60` log 2

When population becomes triple, i.e. when P = 3P_{0} , we get

`log ((3"P"_0)/"P"_0) = "t"/60` log 2

∴ `log 3 = ("t"/60)` log 2

∴ t = `60 ((log 3)/(log 2)) = 60 (1.0986/0.6912)`

= 60 × 1.5894 = 95.364 ≈ 95.4 years

∴ the population becomes triple in 95.4 years (approximately).