I'm unsure of how to approach this problem because of the 'semi-log' part, would I find the line of best fit, then log both sides on that equation until it is in exponential form?
Thanks in advance.
E. coli has the potential to double every 20 minutes. A graduate student in biology collected the following data. A solution was inoculated with an indeterminate amount of bacteria at t = 0, and then the number of E. coli cells per milliliter was measured at 20-minute intervals. For the first two time periods the number of bacteria was so small that it could not be measured.
Time Periods (t): 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12
Billions of cells/ml (C): | 1.34 | 2.87 | 4.25 | 7.51 | 14.65 | 25.30 | 45.96 | 84.72 | 146.52 | 269.55
Use semi-log re-expression to find an exponential model for the given data using time as the independent variable.
Best Answer
Hint
You have data points $(t_i,y_i)$. If you plot $y$ as a function of $t$ and suspect an exponential behavior, you then plot $\log(y)$ as a function of $t$ (this is a semi-log plot). If the data look now to be linear, then this confirm the model to be $$\log(y)=a +b t$$ To come back to the classical linear curve fit, define $z=\log(y)$ and use the classical method to obtain the value of $a$ and $b$.
Using your data makes all of the above very obvious.