The rate o growth of bacteria is proportional to the number present . IT intially, there were 1000 bacteria and the number doubles in 1 hours. Find the number of bacteria after ` 2""1/2` hours . [ take ` sqrt 2 = 1.414]`

Explanation

Let x be number of bacteria present at time 't' `:'` The rate of growth of bacteria is proportional to the number present, we get
`(dx)/(dt)=propx`
`:.(dx)/(dt)=kx`
`:.(dx)/(x)=k.dt`
Integrating both sides,
`int(dx)/(x)=kintdt`
`:.logx=kt+C" ".......(i)`
when t=0, x=1000
`:.` we get,`log1000=kxx0+C`
`:.C=log1000`
`:.` Equation (i) can be written as
`logx=kt+log100" "......(ii)`
when t=1, x=2000
`:.` Equation (ii) can be written as,
`log2000=k.1+log1000`
`impliesk=log2000-log1000`
`impliesk=log((2000)/(1000))=log2`
`impliesk=log2`
`:.` Equation (ii) can be weitten as
`logx=t.log2+log1000" ".......(iii)`
when `t=2(1)/(2)=(5)/(2),` we get
`implieslogx=(5)/(2).log2+log1000`
`=log(2^(5//2))+log1000`
`=log(4sqrt2)+log1000`
`=log(4sqrt2xx1000)`
`=log(4000xxsqrt2)`
`=log(4000xx1.414)`
`implieslogx=log(5656)`
`impliesx=5656`
Thus, after `2(1)/(2)` hrs, number of bacteria =5656.