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

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`.