Exponential Growth & Decay Question

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SUMMARY

The discussion focuses on solving an exponential decay problem involving Kool-Aid powder, where the initial amount, denoted as x(not), is determined through two equations derived from the decay model. The equations established are 3 = x(not)e^(-k) and 1 = x(not)e^(-3k). The value of k was calculated to be approximately 0.5493, leading to an initial amount of Kool-Aid powder, x(not), of approximately 5.1962 grams. The final values can also be expressed in exact terms as k = ln(√3) and x(not) = 3√3.

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ISITIEIW
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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 !
 
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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)
 

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