A certain food is dropped on the floor and is growing bacteria exponentially according to the model y=2e^kt, where t is the time in seconds. a) If there are 2 cells initially and 8 cells after 20 seconds, find the value of k for the bacteria. b) The "5 second rule" says that if a person who drops food on the floor eats it within 5 seconds, there will be no harm. How much bacteria is on the food after 5 seconds?
You must do a) to be able to do b). Do you have any ideas for a)?
Yes. I know it's \[8=2e ^{k(20)}\] then divide each side by 2 and you're left with \[4=e ^{20k}\] then you take the ln of both sides and get the ln4=\[lne ^{20k}\] and then I got stuck.
Wait I think I figured it out. Then you do ln4=20k and divide ln4 by 20 and get .0693147181
Ok. Great start! \[4= 20e ^{k}\] then \[\ln 4= ?\]
yes. exactly!
Okay so now for part b, do I substitute the k in for the k in the equation along with the 5 in the t spot?
Yep!
Okay so then I got \[y=2e ^{.3465735905}\]
And divide by 2 and get y=2.828427125
You don't have to divide anything by 2 there. You are just subbing in and the equation will give you a value for y. It will be 2 times whatever you get for the power of e.
That's what I meant. That's the answer though correct?
Since \[k = \frac{ \ln 4 }{ 20 }\] You are taking e to the power of ln4 divided by 4, then multiplying that by 2.
Can you just tell me if my answer is right?
Just grabbing a calculator. One second.
Okay thanks
You are correct. Nice job.
Thank you
A certain bacteria is growing exponentially according to the model \[y=80e ^{kt}\] where t is the time in minutes. a) If there are 80 cells initially and 675 cells after 30 minutes, find the value of k for the bacteria. I did that and got .0710895352 for k. Then I'm confused on b b) When will the bacteria reach a population of 6000 cells?
So you got k. This questions is similar to the last one, but also a bit different. What does the 6000 refer to?
Never mind I figured it out myself!
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