Question

The population of city doubles in 80 years, in how many years will it be triple when the rate of increase is proportional to the number of inhabitants. `("Given" log3/log2 = 1.5894)`

Solution: Let p be the population at time t.

Then the rate of increase of p is `"dp"/"dt"` which is proportional to p.

∴ `"dp"/"dt"  ∝  "p"`

∴ `"dp"/"dt"` = kp, where k is a constant

∴ `"dp"/"p"` = kdt

On integrating, we get

`int "dp"/"p" = "k" int "dt"`

∴ log p = kt + c

Initially, i.e., when t = 0, let p = N

∴ log N = k × 0 + c

∴ c = `square`

When t = 80, p = 2N

∴ log 2N = 80k + log N

∴ log 2N – log N = 80k

∴ `log ((2"N")/"N")` = 80k

∴ log (2) = 80k

∴ k = `square`

∴ p = 3N, then t = ?

∴ log p = `log2/80  "t" + log "N"`

∴ log 3N – log N = `square`

∴ t = `square` = `square` years

Answer 1

Let p be the population at time t.

Then the rate of increase of p is `"dp"/"dt"` which is proportional to p.

∴ `"dp"/"dt"  ∝  "p"`

∴ `"dp"/"dt"` = kp, where k is a constant

∴ `"dp"/"p"` = kdt

On integrating, we get

`int "dp"/"p" = "k" int "dt"`

∴ log p = kt + c

Initially, i.e., when t = 0, let p = N

∴ log N = k × 0 + c

∴ c = log N

When t = 80, p = 2N

∴ log 2N = 80k + log N

∴ log 2N – log N = 80k

∴ `log ((2"N")/"N")` = 80k

∴ log (2) = 80k

∴ k = `1/80 log 2`

∴ p = 3N, then t = ?

∴ log p = `log2/80  "t" + log "N"`

∴ log 3N – log N = `"t"/80 log 2` 

∴ `log ((3"N")/"N") = "t"/80 log 2`

∴ log 3 = `"t"/80 log 2`

∴ t = `80 (log3)/(log 2)`

= 80(1.5894)

= 127.15 years