- #1

TheFallen018

- 52

- 0

Hey guys, I have a couple of questions here. One, I was just wondering if someone could elaborate on, and the second, I worked it out, but more by guessing. I was hoping someone would be able to help explain both.

Here is the first of the two questions

View attachment 7620

So, part a was fairly straightforward. I calculated the differential to be

However, I'm not sure if I broke down the expression as much as I could have with laws of logarithms. I could only think of using the inverse of the second law of logarithms, where

Is there a way to break that up further?

As for the second question, it has to do with carrying capacity of a population model.

View attachment 7621

I got all the parts correct for this one, but I'm not sure how to get the carrying capacity of the function. I figured it to be 10,000, as that was what the numerator was. However, I'm sure that's not how it's meant to work. Despite the fact that I got the right answer, I'm not satisfied with the answer I gave.

So, how would be the correct way about solving that?

Thanks for your time. I really appreciate the help.

Kind regards,

TheFallen018

Here is the first of the two questions

View attachment 7620

So, part a was fairly straightforward. I calculated the differential to be

**(-2x+3)/x(x-1)**However, I'm not sure if I broke down the expression as much as I could have with laws of logarithms. I could only think of using the inverse of the second law of logarithms, where

**log(a) - log(b)= log(a/b)**Is there a way to break that up further?

As for the second question, it has to do with carrying capacity of a population model.

View attachment 7621

I got all the parts correct for this one, but I'm not sure how to get the carrying capacity of the function. I figured it to be 10,000, as that was what the numerator was. However, I'm sure that's not how it's meant to work. Despite the fact that I got the right answer, I'm not satisfied with the answer I gave.

So, how would be the correct way about solving that?

Thanks for your time. I really appreciate the help.

Kind regards,

TheFallen018