Bacterial Growth ```Name: K. C. Status: student Age: N/A Location: N/A Country: N/A Date: N/A ``` Question: 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? Replies: KC, Let A be the number at 0 hours: A=120. Let r be the growth in one hour: r=0.05. 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 Physics Instructor 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. Vince Calder Click here to return to the Mathematics Archives

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