In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?

Explanation

Let at time `t`, the number of bacteria be `y` then
`(dy)/(dt)propyimplies (dy)/(dt)=ky`,
where `k` is proportional constant.
`implies (dy)/(y)=k dt`
`impliesint(dy)/(y)=intk dt`
`implies log|y|=kt+C` .........`(1)`
Initially at `t=0`, `y=100000`
`:. log100000=C` ........`(2)`
when `t=2`, `y=110000`
`log110000=2k+C` ..........`(3)`
`:.` Subtracting equation `(2)` from equation `(3)`,
`log110000-log100000=2k`
`implies log"(110000)/(100000)=2k`
`implies k=(1)/(2)log((11)/(10))`
put the value of `k` and `C` in equation `(1)`,
`logy=(1)/(2)log((11)/(10))t+log100000`
when `y=200000` then
`log200000=(1)/(2)((11)/(10))t+log100000`
`implies log((200000)/(100000))=(1)/(2)log((11)/(10))t`
`implies 2log(2)=log((11)/(10))t`
`implies t=(2log2)/(log((11)/(10)))`
Therefore, the number of bacteria will increase from `100000` to `200000` in `(2log2)/(log((11)/(10)))` time.