Alice Brassey
New member
- Joined
- Mar 15, 2021
- Messages
- 2
When a colony of bacteria is placed in a solution and then left alone for t hours, the remaining number of bacteria will be B(t)=2000+640ln(t+k). If K represents a constant, and the initial number of bacteria was 2930, how many bacteria will there be after 9 hours?
Not having seen an exponential growth problem expressed in this format up until this point, I was initially quite confused. I am sorry to say that my confusion has not abated. I believe the first thing I have to do is find the constant K, so that I can plug 9 into T, simplify, and receive my output. I am not quite sure how to go about this but I will show you what I have tried so far.
For a while I was messing around with the tradition exponential growth formula y=A_0(e)^kt.
Then I abandoned this because I started to believe it made no sense... I was given a different function, and therefore a different set of rules, and the formula above has its own set of rules.
So I stuck with B(t)=2000+640ln(t+k). **note, that "ln" indeed refers to the natural logarithm.
1. What I know is that when 0 time has elapsed, we have our "initial" amount of 2930. Because b(t) represents the number of bacteria as the output of the function, I think I can write it like this: 2930=2000+640ln(0+k) -----> And then solve for K... eventually I get a constant of 4.27645. So K =4.27645.
2. Then, I plug 9 (to answer the actual question) in for T, and the new constant in for K, to receive B(t).
Am I completely off or do I have it?
Not having seen an exponential growth problem expressed in this format up until this point, I was initially quite confused. I am sorry to say that my confusion has not abated. I believe the first thing I have to do is find the constant K, so that I can plug 9 into T, simplify, and receive my output. I am not quite sure how to go about this but I will show you what I have tried so far.
For a while I was messing around with the tradition exponential growth formula y=A_0(e)^kt.
Then I abandoned this because I started to believe it made no sense... I was given a different function, and therefore a different set of rules, and the formula above has its own set of rules.
So I stuck with B(t)=2000+640ln(t+k). **note, that "ln" indeed refers to the natural logarithm.
1. What I know is that when 0 time has elapsed, we have our "initial" amount of 2930. Because b(t) represents the number of bacteria as the output of the function, I think I can write it like this: 2930=2000+640ln(0+k) -----> And then solve for K... eventually I get a constant of 4.27645. So K =4.27645.
2. Then, I plug 9 (to answer the actual question) in for T, and the new constant in for K, to receive B(t).
Am I completely off or do I have it?
