Exponential and Logarithmic Models

Question

The count in a bacteria culture was 200 after 20 minutes and 1300 after 35 minutes. Assuming the count grows exponentially,

What was the initial size of the culture?   

Find the doubling period.   

Find the population after 110 minutes.   

When will the population reach 13000.

Answer 1

I do things different and I think more useful and accurate for you to use as model going forward.

Exponential growth does not have to involve the constant e.

y=2^t is an exponential function with the nice property of doubling exactly once ever increase of t by 1.

since the base of exponentiation 2=1+1 the exponential rate of growth =1x100=100%

y=1.2^t is an exponential function with a 20% growth rate, ,,,,,base =1.2

y=e^t is an exponentiial function with a 178....% growth rate ,,,,,,base = 2.78...

The general exponential; equation is y=A*b^k*t where A is quantity at time zero, b is base and k effects time axis (for growth k>0, and importantly b>1).

For this problem y=16.5*1.866^(t/5) meaning the growth rate is 86.6% every 5 minutes and at t=0 number of bacteria is 16.5. To find growth rate per minute, b₁^t = 1.866^(t/5) or b₁=1.866^(1/5)=1.133 or 13.3% growth per minute. equation can be written y=16.5*1.133^t. Check, 1300=16.5*1.133³⁵. Note e^.122=1.13, but actual growth rate per minute is 13.3% ((1.133-1)x100) = (b-1)x100) , not 12.2%.

check, when t=20 y=16.5*1.866⁴=200, when t=35 y=16.5*1.866⁷=1300

answering other queries, when t=110, y=number of bacteria=1.505x10⁷

if y=13000, then 13000=16.5*1.866^(t/5) or (t/5)=ln(13000/16.5)/ln(1.866), or t=53.5 minutes.

for doubling time, divide 2y₁=16.5*1.866^(t2/5) by y₁=16.5*1.866^(t1/5) or (t2-t1)/5=ln2/ln1.866

t2-t1 is douibling time =5.5 minutes.

check, is 400=16.5*1.866^(25.5/5), yes

there are your correct answers. Now, how does one find b, the base. If told function is exponential then quantity found by division from one interval to the next (kt going up one integer) will be a constant and equal b, the base (growth rate in percent is (b-1)x100). Since in this problem (I chose k=1/5 for convenience) kt increases friom 4 to 7,

Ab⁷/Ab⁴ is given as 1300/200, b= cubed root of 1300/200= 1.866 (see if kt went up just 1 then the division=b)

Once b is found with division , to find A see 200=A*1.866⁴, A=16.5, or 1300=A*1.866⁷, and A=16.5

Answer 2

200 = Ae^20r where A = initial amount, r = growth rate

divide by A

200/A = e^20r

take natural logs

ln(200/A) = 20r (ln(a/b) = lna -lnb)

ln200 - lnA = 20r (call this equation I)

1300 =Ae^35r

ln1300 -lnA = 35r (call this equation II)

ln200 -lnA = 20r subtract I from II to get

ln1300 -ln200 = 15r

15r = ln(1300/200) = ln(13/2)

r = ln(13/2)/15 = .1248 = 12.48% growth rate

A=200/e^20(.1248) = 200/e^2.496 = 16.483 = original amount of bacteria

2=e^.1248t where t= doubling period

ln2 = .1248t

t = ln2/.1248 = 5.554 minutes= doubling period

B = 16.483e^.1248(110) = 15,101,692.3 after 110 minutes

13,000= 16.483e^.1248t

ln(13,000/16.483) = .1248t

t = ln(13,000/16.483)/.1248 = 53.45 minutes to reach 13,000