# If the bacteria population doubles in 5 days. When was it 1/32 of its present population?

*print*Print*list*Cite

### 2 Answers

You need to come up with the following notation for the initial bacteria population such that:

p - initial bacteria population

The problem provides the information that the initial bacteria population doubles in 5 days such that:

`p_5 = 2p`

You may find the bacteria population after `10` days, such that:

`p_10 = 2p_5 => p_10 = 2*2p = 4p`

You may find the bacteria population after `15` days, such that:

`p_15 = 2*p_10 => p_15 = 2*4p => p_15 = 8p`

You may find the bacteria population after `20` days, such that:

`p_20 = 2*8p => p_20 = 16p`

You may find the bacteria population after `25` days, such that:

`p_25 = 2*16p => p_25 = 32p => p = (1/32)*p_25`

**Hence, ****`25`**** `` days before the actual date, the population of bacteria was `1/32` of the present population.**

The population of bacteria becomes double after 5 days. If the present population of bacteria is P, the population after T days is given by `P*2^(T/5)`

To determine the number of days before which the population was 1/32 the present population solve the equation `P/32 = P*2^(T/5)`

=> `1/32 = 2^(T/5)`

=> `2^-5 = 2^(T/5)`

=> `T/5 = -5`

=> `T = -25`

T is negative here as it indicates a date in the past.

**The population of bacteria was 1/32 the present population 25 days back.**