A type of bacteria has an exponential grow rate at 70% every hour, what is the number of bacteria after 10 hr, 1 day and 3 days.
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The bacteria have an exponential growth rate of 70%. The formula for the exponential growth rate can be given by x (t) = a*b^ (t/T), where a is the initial number of bacteria, b is the increase in time T.
Here we have the initial number as 5. The population of bacteria increases by 0.7 in an hour.
Therefore in
10 hours, the number of bacteria is 5*(1.7)^(10/1)
=> 5*(1.7)^10
=> 1007 bacteria.
In 1 day:
=> 5* (1.7) ^24
=> 1697243
In 3 days:
=> 5*(1.7)^72
=> 1.955* 10^17
Therefore the number of bacteria in 1 hr, 1 day and 3 days are:
1007, 1697243 and 1.955* 10^17 resp.
Initial population of bacteria = 5.
The rate of growth of population of bacteria = 70% per hour.
Therefore in 10 hours the population of 5 bacteria becomes 5*1.7^10 = 1008.
The population of bacteria in after one day (=24 hours) = 5*1.7^24 = 1697243.
The population of bacteria after 3 days (= 72 hours) = 1.9557*10^17 nearly.
Population of bacteria after 10 hours is 1008, after 24 hours 1697243 and after 3 days 1.9557*10^17(approx) respectively.
We'll write the relation that allows to determine the population of bacteria, at specific time.
Population = Initial Population*(1+growth percentage)^time period in hours.
In this case, we'll have:
Population = 5*(1+70/100)^n
Population = 5*[(100+70)/100]^n
Population = 5*(1.7)^n
We'll calculate the population in 10 hours:
Population = 5*(1.7)^10
Population = 5*201.5993900449
We'll calculate the population in 1 day = 24 hours:
Population = 5*(1.7)^24
We'll calculate the population in 3 days = 3*24 hours:
Population = 5*(1.7)^72
Population(10 hours) = 5*(1.7)^10
Population(1 day) = 5*(1.7)^24
Population (3days) = 5*(1.7)^72
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