# Applications

• November 15th 2010, 11:02 AM
Mathematics
Applications
Greetings MHF users!

In my Algebra class we went over some applications. (It Was very quick, a few examples for homework purposes and then it was over.) Anyways, with the abundance of time on my hands I wanted to make an Exponential Growth/Decay solver. It will be able to find a missing variable (Any of them as long as the user has the majority of the other variables). Here is my question.

Will both of these formula's work for Decay?

A(t) = A(1/2)^(t/h)
Where...
A = The start amount.
A(t) = The amount at time 't'. (or the end amount)
t = Time
h = Half-life

&

A(t) = Ae^(kt)
Where...
k = Decay rate (K < 0)
A = The start amount.
A(t) = The amount at time 't'. (or the end amount)
e = the base of natural log.

To conclude, Will be able to use both of these? Or will I be better of sticking to one?
Also, If i can use these, I have a few more questions!

Thanks, Tyler
• November 15th 2010, 11:10 AM
Unknown008
Actually, those two are equations used with half life...

Both are used and they are used depending on what is known.
• November 15th 2010, 11:25 AM
Mathematics
Quote:

Originally Posted by Unknown008
Actually, those two are equations used with half life...

Both are used and they are used depending on what is known.

Both are used with half-life?
• November 15th 2010, 12:34 PM
e^(i*pi)
Quote:

Originally Posted by Mathematics
Both are used with half-life?

Yep, it depends on what you want to know. I personally prefer the second equation because it works easily in all cases although in some cases not as quick.

Another relationship is between the decay constant (k, k>0) and the half life: $t_{1/2} = \dfrac{\ln(2)}{k}$ which can be shown using the second equation.
• November 15th 2010, 07:04 PM
Mathematics
Quote:

Originally Posted by e^(i*pi)
Yep, it depends on what you want to know. I personally prefer the second equation because it works easily in all cases although in some cases not as quick.

Another relationship is between the decay constant (k, k>0) and the half life: $t_{1/2} = \dfrac{\ln(2)}{k}$ which can be shown using the second equation.

Oh Alright, I totally see it now! Cool cool.
I am going to work on it tonight. Hope it works!