MHB Exponential Growth & Decay Question

Click For Summary
The discussion revolves around solving exponential decay equations related to Kool-Aid powder in water. Two equations are established based on the remaining amounts after specific time intervals: 3 grams after 1 minute and 1 gram after 3 minutes. The user successfully finds the decay constant k and the initial amount of Kool-Aid powder, x(not), with k calculated as approximately 0.549 and x(not) as approximately 5.196. Another participant suggests expressing these values in exact form rather than decimals. The conversation concludes with a focus on the importance of precise mathematical representation.
ISITIEIW
Messages
17
Reaction score
0
Suppose that there is initially x(not) grams of Kool-Aid powder in a glass of water. After 1 minute there are 3 grams remaining and after 3 minutes there is only 1 gram remaining. Find x(not) and the amount of Kool-Aid powder remaining after 5 minutes…

So, i set up 2 equations…

3=x(not)e^-k(1)

and 1=x(not)e^-k(3)

I know it is decaying ,but i don't know what i have to do with these equations that i made to find the value of k.

Thanks !
 
Mathematics news on Phys.org
Since there is no actual calculus involve in solving this problem, I am going to move the topic to our Pre-Calculus sub-forum.

You're off to a good start:

$$x_0e^{-k}=3$$

$$x_0e^{-3k}=1$$

I think what I would do next is solve both equations for $x_0$ and equate:

$$x_0=3e^{k}=e^{3k}$$

Next try dividing through by $e^k$ and then convert from exponential to logarithmic form.
 
Thanks!
I got k to be 0.549306144
and got a x(not) value of 5.196152423

I got it from here !
Thanks :)
 
ISITIEIW said:
Thanks!
I got k to be 0.549306144
and got a x(not) value of 5.196152423

I got it from here !
Thanks :)

You're welcome! :D

I would get in the habit of obtaining/writing exact values rather than decimal approximations. I find:

$$k=\ln\left(\sqrt{3} \right)$$

$$x_0=3\sqrt{3}$$

I realize it is possible that you found these values and simply chose to write the approximations. (Angel)
 
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K