# Thread: Exponential Pattern Problem - Part A

1. ...

2. Yeah, another way to set it up is

$\frac{106}{250} = \frac{x}{400}$

Then you do the math

$42,400=250x \rightarrow x=169.6$

3. Oh, I see! That make sense, too. We've been doing a very similar thing in Chemistry lately.

Thank you for the alternate method, and I am also glad I did this correctly.

4. Hello, Jaydee!

You are correct.
I would write it: . $C(t) \:=\:\frac{169.6}{2^{\frac{t}{5.5}}}$

5. Remember the equation...

$A = A_0e^{kt}$ with $A_0$ being the original amount and k is the rate of growth/decay and t is time.

You know half life means after 5.5 hrs there will be half the substance left so you can....

$.5 = e^{k5.5}$

Because if you put in the values before like....

$84.8 = 169.6e^{k5.5}$ or even for the original in your part A $53 = 106e^{k5.5}$ they both will reduce to...

$.5 = e^{k5.5}$

now you need to know the rate, k.

You could make $.5 = e^{k5.5}$ even shorter to begin with, but I think it's important to understand how this came about in case you are given other problems where it's not so cut and dry around half life or exponential growth or decay.

Oh, I forgot to include that once you find k, you can get the equation, with k filled in...

$f(t) = 169.6e^{kt}$