The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 20, 000 in 1999 and 25000 in the year 2004, what will be the population of the

Explanation

Let the population at time `t` be `y`, then `(dy)/(dt)prop y`
`implies (dy)/(dt)=ky`, where `k` is a constant
`implies (dy)/(y)=k dt`
On integration, `logy=kt+C`……..`(1)`
In year `1999`, `t=0`, `y=20000`
from equation `(1)`, `log20000=k(0)+C`
`implies log20000=C`.......`(2)`
In year `2004`, `t=5`, `y=25000`
Therefore, from equation `(1)`,
`log25000=k*5+C`
`implies log25000=5k+log20000` [from equation `(2)` ]
`implies 5k=log((25000)/(20000))=log((5)/(4))`
`implies k=(1)/(5)log(5)/(4)`
For the year `2009`, `t=10`years
Now, put the values of `t`, `k` and `C` in equation `(1)`,
`logy=10xx(1)/(5)log((5)/(4))+log(20000)`
`implies logy=log[20000xx(5)/(4))^(2)]`
`implies y=20000xx(5)/(4)xx(5)/(4)impliesy=31250`
Therefore, the population of the village in the year `2009` will be `31250`.