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

Answer:

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

Answer:

#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 :)