Logorithm/Exponential bacteria growth? Quick yes or no question.

[Matriculator]

Is this question equals to "after 9 hours"? I'm saying this because with each 2 hours, it's growing by a factor of 6 right? So it should normally be "after 9 hours" that the bacteria would have that mass?! I just want to know if it's right or not. I know of another way to solve it but it gives me a different answer. It this isn't right, I'll use that. Thank you.

 

[Mentallic]

Is this question equals to "after 9 hours"? I'm saying this because with each 2 hours, it's growing by a factor of 6 right? So it should normally be "after 9 hours" that the bacteria would have that mass?! I just want to know if it's right or not. I know of another way to solve it but it gives me a different answer. It this isn't right, I'll use that. Thank you.

No.
Exponential growth isn't linear growth, and that's what you're doing. Thriving populations do not grow at a linear scale.

 

[Matriculator]

No.
Exponential growth isn't linear growth, and that's what you're doing. Thriving populations do not grow at a linear scale.

Thank you. By re-doing it using g=e^rt, I got around 11. I think that about right, right?

 

[HallsofIvy]

"Around 11 hours"? You can be more accurate than that. How many minutes or even seconds?

 

[NascentOxygen]

Thank you. By re-doing it using g=e^rt, I got around 11. I think that about right, right?

If your estimate when mistakenly modelling it as linear growth was 9 hours to reach a particular mass, then modelling it as an exponential growth should give an answer less than 9 hours, I'd expect. So an answer of 11 hours sounds way off.

Why don't you show the steps in your working? (Unless maybe you are not wanting to present others in your class with a fully worked answer?)

 

[Matriculator]

If your estimate when mistakenly modelling it as linear growth was 9 hours to reach a particular mass, then modelling it as an exponential growth should give an answer less than 9 hours, I'd expect. So an answer of 11 hours sounds way off.

Why don't you show the steps in your working? (Unless maybe you are not wanting to present others in your class with a fully worked answer?)

Now I'm really confused. Redoing it I got 14.04 or something of the sort. Which was done by doing Ln(11)-Ln(17)=Answer. I took the answer and divided it by 2(because of the 2 hour difference). After doing that I took the Ln of 23 and divided it by the answer that I got for Ln(23). I'd made a mistake but I got somewhere around "14.04 hours". I thought that since the mass was increasing so would the time(the "after how many hours").

 

[Matriculator]

"Around 11 hours"? You can be more accurate than that. How many minutes or even seconds?

I got a decimal number. 14.04, which I guess would be 14 hours and 2.4 minutes.

 

[Mentallic]

Now I'm really confused. Redoing it I got 14.04 or something of the sort. Which was done by doing Ln(11)-Ln(17)=Answer. I took the answer and divided it by 2(because of the 2 hour difference). After doing that I took the Ln of 23 and divided it by the answer that I got for Ln(23). I'd made a mistake but I got somewhere around "14.04 hours". I thought that since the mass was increasing so would the time(after how many hours).

Well that solution is a mess... Too many logical errors to count.

Let's start again from the beginning, shall we? The formula you want to use is

[tex]M=Ae^{rt}[/tex] where M is the mass of bacteria at time t, A is the initial amount of bacteria and r is the growth rate.
We don't know A or r, so we want to find them.

The first piece of info we are given is that after 5 hours, we have 11 grams of bacteria, so we will plug those values into the formula to obtain:

[tex]11 = Ae^{5r}[/tex]

Using the second piece of info, we get

[tex]17 = Ae^{7r}[/tex]

These are two equations in two unknowns, so it's possible to solve for A and r. Can you do this yourself?

p.s.

I thought that since the mass was increasing so would the time(the "after how many hours").

Since it took 2 hours to grow 6 grams more from 11 to 17g, it should take less than 2 hours to grow 6g more from 17 to 23g because of exponential growth (it grows faster as time goes on).

 

[Matriculator]

Well that solution is a mess... Too many logical errors to count.

Let's start again from the beginning, shall we? The formula you want to use is

[tex]M=Ae^{rt}[/tex] where M is the mass of bacteria at time t, A is the initial amount of bacteria and r is the growth rate.
We don't know A or r, so we want to find them.

The first piece of info we are given is that after 5 hours, we have 11 grams of bacteria, so we will plug those values into the formula to obtain:

[tex]11 = Ae^{5r}[/tex]

Using the second piece of info, we get

[tex]17 = Ae^{7r}[/tex]

These are two equations in two unknowns, so it's possible to solve for A and r. Can you do this yourself?

p.s.
Since it took 2 hours to grow 6 grams more from 11 to 17g, it should take less than 2 hours to grow 6g more from 17 to 23g because of exponential growth (it grows faster as time goes on).

I'm at school right now, I've class in a few minutes. But I'm going to print this out and try. Thank you.