Question

Bacteria increases at the rate proportional to the number of bacteria present. If the original number N doubles in 4 hours, find in how many hours the number of bacteria will be 16N.

Solution: Let x be the number of bacteria in the culture at time t.

Then the rate of increase of x is `("d"x)/"dt"` which is proportional to x.

∴ `("d"x)/"dt" ∝ x`

∴ `("d"x)/"dt"` = kx, where k is a constant

∴ `("d"x)/x` = kdt

On integrating, we get

`int ("d"x)/x = "k" int "dt"`

∴ log x = kt + c    .....(1)

∴ x = ae^kt where a = e^c 

Initially, i.e.,when t = 0, let x = N

∴ N = ae^k(0)

∴ a = `square`

∴ a = N, x = Ne^kt    ......(2)

When t = 4, x = 2N

From equation (2), 2N = Ne^4k

∴ e^4k = 2

∴  e^k = `square`

Now we have to find out t, when x = 16N

From equation (2),

16N = Ne^kt 

∴ 16 = e^kt 

∴ `"t"/4 = square` hours

Hence, number of bacteria will be 16N in `square` hours

Answer 1

Let x be the number of bacteria in the culture at time t.

Then the rate of increase of x is `("d"x)/"dt"` which is proportional to x.

∴ `("d"x)/"dt" ∝ x`

∴ `("d"x)/"dt"` = kx, where k is a constant

∴ `("d"x)/x` = kdt

On integrating, we get

`int ("d"x)/x = "k" int "dt"`

∴ log x = kt + c    .....(1)

∴ x = ae^kt where a = e^c 

Initially, i.e.,when t = 0, let x = N

∴ N = ae^k(0)

∴ a = N

∴ a = N, x = Ne^kt    ......(2)

When t = 4, x = 2N

From equation (2), 2N = Ne^4k

∴ e^4k = 2

∴  e^k = `bb(2^(1/4))`

Now we have to find out t, when x = 16N

From equation (2),

16N = Ne^kt 

∴ 16 = e^kt

∴ 2⁴ = `2^("t"/4)`

∴ `"t"/4` = 4 hours

∴ t = 16 hours

Hence, number of bacteria will be 16N in 16 hours