Q:

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 ?

Accepted Solution

A:
Answer: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 earthStep-by-step 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:[tex]{a, ar, ar^{2}, ar^{3},...}[/tex]In which:a is the first termr is the common ratio.We can find any term of the sequence by the following equation:[tex]x_{n} = ar^{(n-1)}[/tex]Each arithmetic sequence has the following format:[tex]{a, a+d, a+2d,...}[/tex]In which:a is the first termd is the difference between the terms.We can find any term of the sequence by the following equation:[tex]x_{n} = a + d(n)[/tex]How i am going to solve this problem.We have the sequence that is the volume of the bacteria:Obs: [tex]1(um)^{3} = 10^{-18}m^{3}[/tex][tex]V_{n} = {6*10^{-18}, 12*10^{-18},...}[/tex][tex]a = 6*10^{-18}[/tex][tex]r = 2[/tex]And the following arithmetic sequence that are the time(in hours).[tex]T_{n} = {0,0.5,1,...}[/tex][tex]a = 0[/tex][tex]d = 0.5[/tex][tex]T_{n} = 0.5n[/tex]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 [tex]V_{n} = 1cm^{3}[/tex], then i find the value at this position in the arithmetic sequence. So[tex]1cm^{3} = 10^{-6}m^{3}[/tex][tex]V_{n} = 6*10^{-18}*(2^{(n-1)})[/tex][tex]10^{-6} = 6*10^{-18}*(2^{(n-1)})[/tex][tex]\frac{10^{-6}}{6*10^{-18}} = 2^{(n-1)})[/tex]Obs: [tex]a^{b-c} = \frac{a^{b}}{a^{c}}[/tex][tex]\frac{10^{12}}{6} = \frac{2^{n}}{2}[/tex][tex]2^{n} = \frac{10^{12}}{3}[/tex]Now, we have to apply these following logarithim proprierties to find the value of n:[tex]log a^{n} = n log a[/tex][tex]log(\frac{a}{b}) = log a - log b[/tex][tex]log 10^{n} = n[/tex]log 2 = 0.30log 3 = 0.48[tex]log 2^{n} = log(\frac{10^{12}}{3})[/tex][tex]n log 2 = log(10^{12}) - log 3[/tex][tex]0.30n = 12 - 0.48[/tex][tex]0.30n = 11.52[/tex][tex]n = 38.4[/tex]Lets find [tex]T_{38}[/tex] and [tex]T_{39}[/tex] in the arithmetic sequence.[tex]T_{n} = 0.5n[/tex][tex]T_{38} = 0.5*38 = 19[/tex][tex]T_{39} = 0.5*39 = 19.5[/tex]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 ?[tex]1 (km)^{3} = 10^{9}m^{3}[/tex]So[tex] 1.08*10^{8} km^{3} = 1.08*10^{17} m^{3} = 108*10^{15}m^{3}[/tex]Lets solve the same way as the first question.[tex]V_{n} = 6*10^{-18}*(2^{(n-1)})[/tex][tex]108*10^{15} = 6*10^{-18}*(2^{(n-1)})[/tex][tex]\frac{108*10^{15}}{6*10^{-18}} = 2^{(n-1)})[/tex][tex]18*10^{33} = \frac{2^{n}}{2}[/tex][tex]2^{n} = 36 * 10^{33}[/tex]Aside from the proprierties seen in the first exercise, we also have thatlog a*b = log a + log b[tex]log (2^{n}) = log(36 * 10^{33})[/tex][tex]n log 2 = log 36 + log 10^{33}[/tex][tex]0.30n = 1.56 + 33[/tex][tex]0.30n = 34.56[/tex][tex]n = \frac{34.56}{0.30}[/tex][tex]n = 115.2[/tex][tex]T_{115} = 0.5*115.2 = 57.6[/tex]It will take 57.6 hours to for the volume to fill the entire earth