population of bacteria modeled by N = 5500 e^(-0.1t)

[jhawk555]

The population of a bacteria culture with an initial population of 5500 being treated with a new antibiotic can be modeled by:

N = 5500 e^(-0.1t)

where N is the number of bacteria present and t is the time in hours since the treatment began. In how many hours will the culture have a count of 2200?

I am confused as to what the 2200 stands for. Is it N?

I did the problem as:

2200=5500e^-0.1t

Divided both sides by 5500 getting:

ln.4=lne-0.1t then divided both sides by -0.1 to find t getting 9.2 hours.

Am I right?

 

[stapel]

N is the number of bacteria present and t is the time in hours.... In how many hours will the culture have a count of 2200?

I am confused as to what the 2200 stands for. Is it N?

You are given that "N is the number of bacteria present" and that 2200 is the population count (that is, the number of bacteria present) at some time t hours after starting.

Then, according to the exercise's definitions, yes, you are being asked for the time t when N = 2200.

Eliz.

 

[Denis]

ln.4=lne-0.1t then divided both sides by -0.1 to find t getting 9.2 hours.

Next time you post, try and be clearer; above easier to read if like this:
e^(-.1t) = .4
-.1t = ln(.4) / ln(e)
t = [ln(.4) / ln(e)] / -.1
t = 9.1629....