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

Answer» 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`.


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