Question

The E. coli bacteria has a volume of 6 µm3 . In optimal conditions, an E. coli bacteria will double about every 30 minutes. Under these conditions, how long will it take for a single bacterium to grow to fill a thimble with volume 1 cm3 ? How long will it take for the volume to fill the entire earth 1.08 × 108 km3 ?

Answer 1

It is going to take around 19 hours and 15 minutes for a single bacteria to grow to fill a thimble with volume 1 cm3.

It will take 57.6 hours to for the volume to fill the entire earth

Explanation:

We can say that the volume of the bacteria is a geometric sequence, and each time moment is an arithmetic sequence.

Each geometric sequence has the following format:

`{a, ar, ar^(2), ar^(3),...}`

In which:

a is the first term

r is the common ratio.

We can find any term of the sequence by the following equation:

`x_(n) = ar^((n-1))`

Each arithmetic sequence has the following format:

`{a, a+d, a+2d,...}`

In which:

a is the first term

d is the difference between the terms.

We can find any term of the sequence by the following equation:

`x_(n) = a + d(n)`

How i am going to solve this problem.

We have the sequence that is the volume of the bacteria:

Obs:
`1(um)^(3) = 10^(-18)m^(3)`

`V_(n) = {6*10^(-18), 12*10^(-18),...}`

`a = 6*10^(-18)`

`r = 2`

And the following arithmetic sequence that are the time(in hours).

`T_(n) = {0,0.5,1,...}`

`a = 0`

`d = 0.5`

`T_(n) = 0.5n`

How long will it take for a single bacterium to grow to fill a thimble with volume 1 cm3 ?

I am going to find the value of n for which
`V_(n) = 1cm^(3)`, then i find the value at this position in the arithmetic sequence. So

`1cm^(3) = 10^(-6)m^(3)`

`V_(n) = 6*10^(-18)*(2^((n-1)))`

`10^(-6) = 6*10^(-18)*(2^((n-1)))`

`(10^(-6))/(6*10^(-18)) = 2^((n-1)))`

Obs:
`a^(b-c) = (a^(b))/(a^(c))`

`(10^(12))/(6) = (2^(n))/(2)`

`2^(n) = (10^(12))/(3)`

Now, we have to apply these following logarithim proprierties to find the value of n:

`log a^(n) = n log a`

`log((a)/(b)) = log a - log b`

`log 10^(n) = n`

log 2 = 0.30

log 3 = 0.48

`log 2^(n) = log((10^(12))/(3))`

`n log 2 = log(10^(12)) - log 3`

`0.30n = 12 - 0.48`

`0.30n = 11.52`

`n = 38.4`

Lets find
`T_(38)` and
`T_(39)` in the arithmetic sequence.

`T_(n) = 0.5n`

`T_(38) = 0.5*38 = 19`

`T_(39) = 0.5*39 = 19.5`

It is going to take around 19 hours and 15 minutes for a single bacteria to grow to fill a thimble with volume 1 cm3.

How long will it take for the volume to fill the entire earth 1.08 × 108 km3 ?

`1 (km)^(3) = 10^(9)m^(3)`

So

`1.08*10^(8) km^(3) = 1.08*10^(17) m^(3) = 108*10^(15)m^(3)`

Lets solve the same way as the first question.

`V_(n) = 6*10^(-18)*(2^((n-1)))`

`108*10^(15) = 6*10^(-18)*(2^((n-1)))`

`(108*10^(15))/(6*10^(-18)) = 2^((n-1)))`

`18*10^(33) = (2^(n))/(2)`

`2^(n) = 36 * 10^(33)`

Aside from the proprierties seen in the first exercise, we also have that

log a*b = log a + log b

`log (2^(n)) = log(36 * 10^(33))`

`n log 2 = log 36 + log 10^(33)`

`0.30n = 1.56 + 33`

`0.30n = 34.56`

`n = (34.56)/(0.30)`

`n = 115.2`

`T_(115) = 0.5*115.2 = 57.6`

It will take 57.6 hours to for the volume to fill the entire earth