Answer to Question #209465 in Algebra for Maria

Question

It is observed that the number of bacteria present in the portion of the small intestine of an animal grows exponentially. Initially,there are 10,000 bacteria present. Three hours later,the bacteria grow to 500,000. Assuming no bacteria due in the progress,how many bacteria will there be after one day? After how many hours will the bacteria double it's number?

Answer 1

solution

10,000e^r(3) =500,000

10,000e^r(3)/10000 =500,000/10,000

introduce (ln) function

ln e^3r =50

3r=ln 50

3r/3=ln 50/3 use a calculator

r=1.304007668

exponential function becomes 10,000e^{1.304007668(t)}

Then the population after 1 day becomes

1day= 24 hours

Therefore

10000e^{1.304007668(24)}=(3.906249948 x 10) ¹⁷

b)how many hours the bacteria will take to double its number

10,000e^1.304007668t=20,000

10,000e^1.304007668t/10,000=20000/10000

ln e 1.304007668t=ln 2

1.304007668t/1.304007668=ln 2/1.304007668

t= ln2/1.304007668

t=0.5 hours or 30 minutes