Bacteria Growth!?

Question

I'm really stumped on this one; can anyone help me please?

The number N of bacteria in a culture is given by the model

N = 100e^kt

where t is the time (in hours). If N = 300 when t = 5, estimate the time required for the population to double in size.

Don't just give the answer; that is of no use to me. Step by step explanation please.

Thank you sahsjing, gudspeling, and Northstar for your excellent answers.

(gudspeling you rounded one too many times)

Smart Ambitious, I find your answer confusing, but thanks for trying to help!

Answer 1

First solve for k.

N = 100e^kt
300 = 100e^5k
300/100 = e^5k
3 = e^5k
ln 3 = 5k
k = (ln 3)/5

Now solve for the time needed to double.

N = 100e^kt
200 = 100e^kt
2 = e^kt
ln 2 = kt
t = (ln 2)/k
t = (ln 2) / [(ln 3)/5] = 5(ln 2) / (ln 3) ≈ 3.1546488 hours

Answer 2

Ok, let's take the equation:

N = 100e^kt

N = 300 when t = 5

Plug the values of N and t in the equation:

300 = 100e^5k

e^5k = 300/100
e^5k = 3

convert e^5k = 3 into from exponential into logarithmic function:

log (base e) 3 = 5k

Now, log (base e) is same as ln.

So, ln 3 = 5k

5k = 1.09861

k = 1.09861/5

k = 0.219722

So, now we plug the value of k in the original equation:

N = 100e^0.219722t

Test:

If t = 5, N = 100e^0.219722*5 = 300.

The population double in size means N = 100*2 = 200

200 = 100e^0.219722t

e^0.219722t = 200/100 = 2

log (base e) 2 = 0.219722t

t = ln 2/0.219722

t = 3.15466

So, the times required for the population to double in size is 3.15466 hours.



Good Luck!

Answer 3

N = 100e^kt

N = 300 when t=5
300 = 100e^5k
e^5k = 3
5k = ln(3)
k = (1/5)ln(3)

N = 100e^[(1/5)ln(3)t]
= 100*3^(t/5)

100 is a contant - can be ignored for the rest of the solution.
3^(t/5) must double

3^(ta/5) = 2 * 3^(tb/5)
(ta/5) log 3 = log 2 + (tb/5)log 3
(ta-tb)/5 = log(2) / log(3)
ta - tb = 5 log (2) / log (3) = 3.155 hours

The population will double every 3.15 hours (approx.)

Answer 4

N = 100e^kt
300 = 100e^5k
200 = 100e^(k*t)
Divide by 100 and take ln,
ln3 = 5k......(1)
ln2 = k*t......(2)
(2)/(1): t = 5ln2/ln3 = 3.1546

Answer 5

Ask the teacher.