The rate of growth of bacteria is proportional to the number present. If initially, there were 1000 bacteria and the number doubles in 1 hour, find the number of bacteria after 5/2 hours.

(Given: `sqrt2 = 1.414`)

#### Solution

Let ‘x’ be the number of bacteria present at time ‘t’.

∴ `dx/dt ∝ x`

∴ `dx/dt = kx`, where k is the constant of proportionality.

∴ `dx/x = k dt`

Integrating on both sides, we get

`int dx/x = k int dt`

∴ log x = kt + c …(i)

When t = 0, x = 1000

∴ log (1000) = k(0) + c

∴ c = log (1000)

∴ log x = kt + log (1000) …(ii)[From (i)]

When t = 1, x = 2000

∴ log (2000) = k(1) + log (1000)

∴ log (2000) - log (1000) = k

∴ `k = log (2000/1000) = log 2` ...(iii)

When t = `5 /2` , we get

log x = `5/2 k + log(1000)` …[From (ii)]

∴ log x = `(5/ 2) log 2 + log (1000)` ...[From (iii)

= log `(2^(5/2)) + log (1000)`

= `log (4sqrt2)+ log (1000)`

= `log (4000sqrt2)`

= log (4000 x1.414)

∴ log x = log (5656)

∴ x = 5656

Thus, there will be 5656 bacteria after `5/2` hours.