The solution to this problem is -0.51975 = ln (B(14)/B(7)) B(14)/B(7) = e^-0.519
ID: 2844576 • Letter: T
Question
The solution to this problem is
-0.51975 = ln (B(14)/B(7))
B(14)/B(7) = e^-0.51975
B(14) = (0.5947) * B(7)
0.5947 * 100 = 59.47%
100 - 59.47 = 40.53% net decrease
I can't for the life of me figure out how the book gets the number -0.51975. Any help would be much appreciated. I understand the rest of the problem.
The relative rate of change of the number of bacteria present in a lab culture is given by Where t is measured in days and 0 t 16. Initially, there were 24,000 bacteria in the culture. Calculate the net percentage change in culture population between t = 7 and t = 14 days. Give your result as a percentage, correct to two decimal places. Indicate whether this percentage gives an increase or decrease in population. Determine the lowest population that the culture reaches.Explanation / Answer
The relative rate of change of bacteria = 0.001t^3 - 0.027t^2 + 0.179t - 0.153
The no. of bacteria = 0.00025t^4 - 0.009t^3 + 0.0895t^2 - 0.153t + C
At t=0, no. of bacteria = 24000, thus C = 24000.
No. of bacteria at t = 7
= 0.08575 - 0.441 + 1.253 + 24000 = 24000.90
No. of bacteria at t = 14
= 0.343 - 1.764 + 5.012 + 24000 = 24003.59
Percentage change = {24003.59 - 24000.90} / 24000.90 = 1.12 * 10^-2 %
OR the percentage change is 0.0112 %.
The lowest population is when the rate becomes 0.
The rate becomes 0 at t = 1, t = -9 and t = -17.
But since t is +ve, the time when population becomes least is at t=1.
The lowest population = 0.00025 - 0.009 + 0.0895 - 0.153 + 24000
= 23999.9275 bacteria............
THANK YOU. DO RATE.