Question

How do you find when rate is zero (t) for the following derived equation?

Answer 1

Please type in the equation!

Answer 2

sorry, hold on, my network is having connection issues every time i try to attach the image

Answer 3

\[C'(t)=k(5e ^{-5t}-2e ^{-2t})\]

Answer 4

this is a "derivative." My understanding is that you want to determine the t values for which C'(t) = 0. Is that correct?

Answer 5

k is a positive constant btw

Answer 6

yes

Answer 7

I know that it should be 0= the stuff in the bracket but idk how to solve for t

Answer 8

sorry for the slow response...I'll get back to you asap!

Answer 9

Set\[5e ^{-5t}-2e ^{-25}~equal~\to~zero.\]
Move the 2nd term to the right.
Think of what those two terms have in common. Can you reduce the equation by cancelling out the common term?

Answer 10

*it's -2e^-2t
um... is it the e^t that's common?

Answer 11

I got t=ln(2/3) / -3 which is 0.30543?

Answer 12

but the book says the answer is 7.32

Answer 13

\[Set ~5e ^{-5t}-2e ^{-2t}~equal~\to~zero.\]

\[e ^{-2t}\] is common to both sides and can be factored out.

I trust you've gotten the messages I've sent you when OpenStudy was really slow.

Answer 14

Let me look at our results quickly.

Answer 15

I get:\[5e ^{-3t}=2\]

Answer 16

and to solve this, I take the natural log of both sides (do you see why?)

Answer 17

then ln 5 + ln [e^(-3t)] = ln 2

Answer 18

Does this agree with your result?
Can you simplify my equation?
Find t?

Answer 19

Some more info on the question is the function C(t), where K is a positive constant, can be used to model the concentration at time t, in days, of a drug injected into the bloodstream.

I found that 0.3045 * 24 = 7.33 which is close to the book's answer of 7.32 but in the question t is already in days so idk if it's an error in the book...

Answer 20

I think your 7.33 is well within the allowable difference from the book's answer. Excellent1

How else could I help you with this particular problem?

Answer 21

So t = ln(2/3) / -3 which is 0.30543 right?

Answer 22

I'll need just a minute to check that. Bear with me, please.

Answer 23

My results are different.

I do get -3t = ln (2/3).
ln(2/3)
this results in t = - --------
-3

Answer 24

and for that I get


t = -2.97/3 = 0.990.

Answer 25

does this make any sense to you?

Answer 26

I'm sorry: It appears that you and I are getting one (different) result after the other!!

Let's back up a bit. What do you need to do to complete the solution of this problem?

Answer 27

find t when C'(t) = 0

Answer 28

You've typed: " t = ln(2/3) / -3 which is 0.30543 right? "

Wouldn't that be t = ln (2/5) / (-3), not ln (2/3)?

If you want t when C'(t) = 0, we're on the right track! I think you had a typo, typing 2/3 instead of 2/5.

Sorry, I have to leave this conversation, at least for a few minutes. But I'll see your reply.

Answer 29

ln(2/3)
If I put in t = ---------
-3
t= 0.135155

Answer 30

Yes, but again I think you meant ln (2/5), NOT ln (2/3). Please check that out.

Answer 31

Look what we had before:\[5e ^{-3t}=2\]

Answer 32

oh yah whoops.

Answer 33

I'll be back...but at the moment I have a number of other tasks to take care of. You're doing well... keep on going!!

Answer 34

Okay so ln(2/5)
If I put in t = ---------
-3
t= 0.30543

Answer 35

I get the same result. Looks as tho' we're on the right track now. Does this discussion answer your question adequately?

Answer 36

Yes, thank you so much for your help and patience! :)

Answer 37

My great pleasure, Carmen! Hope to see you again here on OS. Good night!!

Answer 38

Good night!