Name: K. C.
If there are 120 bacteria at 0 hours, and the bacteria
grows at 5% per hour, how many bacteria will there be at 24 hours and how
many in one week and what is an algebraic formula that would tell me how
many bacteria equal t hours?
Let A be the number at 0 hours: A=120. Let r be the growth in one hour:
After 1 hour, you will have B=A(1+r).
After 2 hours, you will have C=B(1+r)=A(1+r)(1+r)=A(1+r)^2.
After 3 hours, you will have D=C(1+r)=A(1+r)^3.
Notice the pattern. After N hours, you will have A(1+r)^N bacteria. The
mathematics is the same as compound interest: a steady percent growth.
Dr. Ken Mellendorf
Illinois Central College
This is not an easy question to answer because it depends on the model you
want to use for the population. From your specification that the growth rate
is 0.05% per hour I will assume that you mean that the population increases
exponentially (i.e. the rate of growth is proportional to the population and
no bacteria die over the course of the "experiment"), that is not realistic,
but the results are simpler. If you let me change the value of 5% to 4.95%
the numbers come out neater. Assuming the latter value 0.0495 / hr, the half
life for a first order growth is
ln(2)/0.0495 = 14 hr = t1/2. So the population of bacteria will double every
14 hours. The formula is Ln(P/Po) = k*(t - to) where 'P' is the population
at time 't', and 'Po' is the population at time 'to', which in your case is
120 @ to = 0, and 'ln' is the natural logarithm which can be converted to
base 10 by the relation ln(x) =
2.303*log(x). This can also be cast in an equivalent form: P = Po*exp(k*(t -
to)), where 'exp' is the exponential function. Using your initial
conditions: P = 120* exp(0.0495*t). So at one week = 7*24 hrs = 168 hrs, the
population P = 21,183.
Real bacterial cultures are much more complicated because growth depends
on available nutrients, the "death" rate, and so on.
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Update: June 2012