Question

The table shows the estimated number of E. coli bacteria in a lab dish t minutes after the start of an experiment.

Time (min)	0	10	20	30	40	50	60
Bacteria	300	423	596	842	1188	1686	2354

A. Using t as the independent variable, find the model that best fits the data. Round values to the nearest thousandths.

B. How long does it take the population of E. coli to triple?

Answer 1

A. A scatterplot of the given data is attached, which resembles the graph of an exponential function. Let y = f(t) = ab^t be the exponential function representing the given data set, where t is the time in minutes, y = f(t) is the number of bacteria, and a,b are arbitrary real numbers.

On substituting t = 0, y = 300 in the above equation, we get a*b⁰ = 300 or, a = 300.

On substituting t = 60, y = 2354 in the above equation, we get 300b⁶⁰ = 2354 or, b⁶⁰= 2354/300. Now, on taking log of both the sides, we get 60logb=log2354-log300(as logm/n=log m–log n and logmⁿ= nlogm). Thus, log b =(log2354-log300)/60 = (3.371806459-2.477121255)/60 = 0.894685203/60 = 0.01491142. Hence b = 10^0.01491142 = 1.034931058. Hence y = f(t) = 300*(1.034931058)^t = 300*(1.035)^t ( on rounding off to the nearest thousandth).

B. Let it take t minutes for the population of E. coli to triple. Then 3*300 = 300*(1.034931058)^t or, (1.034931058)^t = 3. Now, on taking log of both the sides, we get t log 1.034931058 = log 3 or, t = log 3/ log 1.034931058 = 0.477121254/0.01491142 = 31.997 minutes( on rounding off to the earest thousandth) or, 32 minutes (on rounding off to the nearest whole number).