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.

Answer 1

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

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

∴ `dx/dt prop x`

∴ `dx/dt = kx`, where k is a constant

∴ `dx/x = k dt`

On integrating, we get

`int dx/x = k int 1dt + c`

∴ 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 ((2x_0)/x_0) = 4k`

∴ `k = 1/4 log 2`

∴ (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 8`

∴ `x/x_0 = 8`

∴ x = 8x₀ 

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