Question

A bacteria doubles its population in 8 hours. At this rate how many hours would it take the population of the bacteria to triple?

Answer 1

Time taken for 3 times the population to have grown is:

12 hours 40 minutes and say 47 seconds

Explanation:

Let the rate of growth be constant and of value `y%`
Let time in hours be `t`
Let count of bacteria at any time `t` be `b_t`
So initial count of bacteria would be `b_o`

Given that count of bacteria after 8 hours (`b_8`) is such that `b_8 =2b_o`

Required to determine unknown time `x" for "3b_o->t_x`
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`color(blue)("Building the model for initial condition")`

The increase in bacteria after 1 hour is

`b_o(1+y/100)^1`

The increase in bacteria after 2 hour is

`b_o(1+y/100)^2`

The increase in bacteria after 2 hour is

`b_o(1+y/100)^3`

So after 8 hours we have:

`color(blue)(b_o(1+y/100)^8 = 2b_o)`....................Equation(1)

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`color(blue)("Determine the value of "y)`

Using equation(1) divide both sides by `b_o`

`(1+y/100)^8=2`

Taking roots

`1+y/100 = root(8)(2)`

`color(blue)(y=100(root(8)(2)-1))`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the time it takes for "3b_o)`

Using Equation(1) we have:

`b_o(1+y/100)^x = 3b_o`

`=> (1+y/100)^x=3`

But `y= 100(root(8)(2)-1)` giving

`(1+root(8)(2)-1)^x=3`

`(root(8)(2))^x=3`

Taking logs ( I elect to use log to base 10 )

`xlog(root(8)(2))=log(3)`

But `root(8)(2) = 2^(1/8)` so we have `log(2^(1/8)) = 1/8log(2)`

`x/8log(2)=log(3)`

`x=(8log(3))/(log(2))`

My calculator gives:

`x=12.6797000058`

Is there could be a degree of error in the calculation lets say:

`x=12.6797` hours exactly

`x=`12 hours 40 minutes and say 47 seconds