Question

In a certain culture of bacteria, the rate of increase is proportional to the number present. If it is found that the number doubles in 4 hours, find the number of times the bacteria are increased in 12 hours.

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

Then the rate of increase of x is `"dx"/"dt"` which is proportional to x.

∴ `"dx"/"dt" ∝  "x"`

∴ `"dx"/"dt"` = kx, where k is a constant

∴ `square`

On integrating, we get

`int "dx"/"x" = "k" int "dt"`

∴ log x = kt + c

Initially, i.e. when t = 0, let x = x₀

∴ log x₀ = k × 0 + c

∴ c = `square`

∴ log x = kt + log x₀ 

∴ log x - log x₀ = kt

∴ `log ("x"/"x"_0)`= kt    ......(1)

Since the number doubles in 4 hours, i.e. when t = 4,

x = 2x₀ 

∴ `log ((2"x"_0)/"x"_0)` = 4k

∴ k = `square`

∴ equation (1) becomes, `log ("x"/"x"_0) = "t"/4` log 2

When t = 12, we get

`log ("x"/"x"_0) = 12/4` log 2 = 3 log 2

∴ `log ("x"/"x"_0)` = log 2³

∴ `"x"/"x"_0 = 8`

∴ x = `square`

∴ number of bacteria will be 8 times the original number in 12 hours.

Answer 1

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

Then the rate of increase of x is `"dx"/"dt"` which is proportional to x.

∴ `"dx"/"dt" ∝  "x"`

∴ `"dx"/"dt"` = kx, where k is a constant

∴ `"dx"/"x"` = k dt

On integrating, we get

`int "dx"/"x" = "k" int "dt"`

∴ log x = kt + c

Initially, i.e. when t = 0, let x = x₀

∴ log x₀ = k × 0 + c

∴ c = log x₀ 

∴ log x = kt + log x₀ 

∴ log x - log x₀ = kt

∴ `log ("x"/"x"_0)`= kt    .....(1)

Since the number doubles in 4 hours, i.e. when t = 4,

x = 2x₀ 

∴ `log ((2"x"_0)/"x"_0)` = 4k

∴ log 2 = 4k

∴ k = `1/4 log 2`

∴ equation (1) becomes, `log ("x"/"x"_0) = "t"/4` log 2

When t = 12, we get

`log ("x"/"x"_0) = 12/4` log 2 = 3 log 2

∴ `log ("x"/"x"_0)` = log 2³

∴ `"x"/"x"_0 = 8`

∴ x = 8x₀ 

∴ number of bacteria will be 8 times the original number in 12 hours.