How many bacteria will be present after 5 hours if a culture of bacteria obeys the law of uninhibited growth where if 500 bacteria are present in the culture initially and there are 800 after 1 hour?

3 Answers
Jul 7, 2018

THe number of bacteria is =5243 " bacteria"

Explanation:

The growth of bacteria is according to the differential equation

(dN)/dt=kN

Integration of this equation yields

ln(N_2/N_1)=k(t_2-t_1)

or

(N_2/N_1)=e^(k(t_2-t_1))

The initial conditions are

N_1=500 at t_1=0

Also,

N_2=800 at t_2=1h

Therefore,

ln(800/500)=k(1-0)

Therefore,

k=ln(8/5)

Therefore,

After t_3=5h, there are

ln(N_3/N_1)=k(t_3-t_1)

ln(N_3/500)=ln(8/5)*5=2.35

N_3=500e^(2.35)=5243 " bacteria"

Jul 7, 2018

color(blue)(5242.88)

Explanation:

We need to find an equation of the form:

A(t)=A(0)e^(kt)

Where:

bb(A(t))= the amount after time t.

bb(A(0)= the amount at the start. i.e. t = 0.

bbk= the growth/decay factor.

bbe= Euler's number.

bbt= time, in this case hours.

We have:

A(0)=500

A(1)=800

Using these values:

800=500e^(k)

We now solve for bbk:

Divide by 500:

8/5=e^(k)

Taking natural logarithms of both sides:

ln(8/5)=kln(e)

ln(e)=1 ( The logarithm of the base is always 1 ):

Hence:

ln(8/5)=k

And:

A(t)=500e^(tln(8/5))

This simplifies to:

A(t)=500(8/5)^t

Amount of bacteria after 5 hours:

A(5)=500(8/5)^5=5242.88

Jul 7, 2018

If it obeys the law of uninhibited growth, growth occurs exponentially .

With B as the number of bacteria:

  • B(t) = B_o a^t qquad {(a in RR), (a gt 1):}

Use the given condition to find a:

B(1) = 800 = 500* a^1

implies a = 8/5

Then after 5 years:

B(5) = 500 * (8/5)^5 = 5242.88

Then round that to a whole number of bacteria :)