Exponential Growth & Decay Question

[ISITIEIW]

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 !

 

[MarkFL]

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:

\(\displaystyle x_0e^{-k}=3\)

\(\displaystyle x_0e^{-3k}=1\)

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

\(\displaystyle x_0=3e^{k}=e^{3k}\)

Next try dividing through by $e^k$ and then convert from exponential to logarithmic form.

 

[ISITIEIW]

Thanks!
I got k to be 0.549306144
and got a x(not) value of 5.196152423

I got it from here !
Thanks :)

 

[MarkFL]

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:

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

\(\displaystyle x_0=3\sqrt{3}\)

I realize it is possible that you found these values and simply chose to write the approximations. (Angel)

 

[CellCoree]

To find the value of k, we can use the fact that exponential decay follows the function y = ae^(-kt), where a is the initial amount and k is the decay constant. In this case, a = x(not) and t = 1, so we can substitute these values into the first equation:

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

Similarly, for the second equation, we have a = x(not) and t = 3, so we can substitute these values and solve for k:

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

Now we can use algebra to solve for k. We can divide the second equation by the first equation to eliminate x(not):

1/3 = e^(-2k)

Taking the natural logarithm of both sides, we get:

ln(1/3) = -2k

Solving for k, we get:

k = -ln(1/3) ≈ 1.099

Now that we have the value of k, we can use it to find the initial amount, x(not), and the amount remaining after 5 minutes. Substituting k = 1.099 and t = 5 into our original equation, we get:

3 = x(not)e^(-1.099*5)

Solving for x(not), we get:

x(not) = 3/e^(-5.495) ≈ 10.5 grams

So, the initial amount of Kool-Aid powder was approximately 10.5 grams and after 5 minutes, there will be about 0.5 grams remaining. This shows that the Kool-Aid powder is decaying at a rate of approximately 1.099 grams per minute.