Answer:
A) 2033%
B) 75 bacteria
C) n(t) = 75(1 + 1.83%)^t R = 1.83%
D) 81.4 bacteria
E) 348.83 seconds
Step-by-step explanation:
Given that the count in a culture of bacteria was 600 after 2 hours and 38,400 after 6
A) growth (600 - 0)/2 = 300 bacteria/s
38400/6 = 6400 bacteria/s
Relative growth=(6400-300)/300 ×100
= 6100/300 × 100
= 2033%
B) initial size of the culture
Using exponential equation
P = I( 1 + R)^t
Where I = initial size
R = rate
600 = I(1 + R)^2
Log both sides
Log600 = log I(1+R)^2
2.778 = logI + 2log(1 +R)
2log (1+R) = 2.778 - logI ..... (1)
Also,
38400 = I(1 + R)^6
Log both sides
Log 38400 = logI + 6log(1+R)
6Log(1+R) = 4.584 - logI .... (2)
Divide equation 2 by 1
6/2 = (4.584 - logI)/(2.778 - logI)
Cross multiply
16.668 - 6logI = 9.168 - 2logI
6logI - 2logI = 16.668 - 9.168
4logI = 7.5
LogI = 7.5/4
LogI = 1.875
I = 74.98 = 75 bacteria
C) A function that models the number of bacteria n(t) after t hours.
If I = 75 bacteria
Then n(t) = 75(1 + R)^t
600 = 75(1+R)^2
8 = (1+R)^2
Log both sides
Log8 = 2log(1+R)
0.903/2 = log(1+R)
0.45 = log(1+R)
1 + R = 2.83
R = 1.83%
The model function is therefore
n(t) = 75(1 + 1.83%)^t
D) the number of bacteria after 4.5 hours
n(t) = 75(1.02)^4.5
n(t) = 81.4 bacteria
E) After how many hours will the number of bacteria reach 75,000
n(t) = 75(1.02)^t
75000 = 75(1.02)^t
1000 = 1.02^t
Log both sides
Log 1000 = tlog 1.02
3 = 0.0086t
t = 348.83 seconds
