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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.
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