Growth Rates

Bacteria 1 - population doubles every 60 minutes

  800, 1600,3200,  etc      $$FV=800*2^{n}$$

Bacteria 2 - population doubles every 20 minutes

 10,20,40,80......                   $$FV=10*2^{3n}$$

a) If Bacteria 1 starts at a polulation of 800, and Bacteria 2 starts at a population of 10, how long will it take for Bacteria 2 to out grow Bacteria 1?

I am going  to let the base unit of time be 1 hour ie n is in hours

$$\\10*2^{3n}>800*2^n}\\
2^{3n}>80*2^n\\
2^{3n}\div 2^n>80\\
2^{3n-n}>80\\
2^{2n}>80\\
4^n>80\\
log(4^n)>log80\\
nlog4>log80\\
n>log80/log4\\
n>3.16 hours$$

 

b) What population will the two Bacteria have when they are both equal? How long will that take?

$${\mathtt{800}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{{\mathtt{3.16}}} = {\mathtt{7\,150.637\: \!683\: \!662\: \!207\: \!786\: \!4}}$$ =   7151 bacteria after 3.16hours

check

$${\mathtt{10}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{3.16}}\right)} = {\mathtt{7\,141.087\: \!571\: \!714\: \!075\: \!752}}$$  7141 bactera after 3.16 hours

 

The answer are slightly different because n is rounded to 2 decimal places.

Let t be number of hours. then:

 

B1 = 800*2^t

B2 = 10*2^3t (as there are 3 lots of 20 minutes in an hour)

 

Set B1 = B2 and solve for t (Divide both sides by 10 first)

 

Take logs, so log(80) +t*log(2) = 3t*log(2)

2t*log(2) = log(80)

t = log(80)/(2*log(2))

$${\mathtt{t}} = {\frac{{log}_{10}\left({\mathtt{80}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{2}}\right)\right)}} \Rightarrow {\mathtt{t}} = {\mathtt{3.160\: \!964\: \!047\: \!443\: \!681}}$$

 

so t ≈ 3.16 hours

.