1. ## Decay and Growth

Im having problems with decay and growth.
The problems states:

Radioactive Decay: They half-life of phosphorous-32 is 14 Days, if there are 5 grams presently, when will there be 1 gram left?

Doubling your Money: How much time is required for a $1000 deposit to double at an interest rate of 2.5% compounded annually. I have no idea what to do. Do I have to find an equation? Please help. Any help would be greatly appreciated. 2. Originally Posted by takuto Im having problems with decay and growth. The problems states: Radioactive Decay: They half-life of phosphorous-32 is 14 Days, if there are 5 grams presently, when will there be 1 gram left? Doubling your Money: How much time is required for a$1000 deposit to double at an interest rate of 2.5% compounded annually.

I have no idea what to do. Do I have to find an equation?
you seem to state this on most every post you've made.

are you not familiar with the equations

$y = y_0 e^{kt}$

and

$A = P\left(1 + \frac{r}{n}\right)^{nt}$

???

3. I have no idea what to do.
That's ridiculous. NEVER say that again.

Do I have to find an equation?
It's not like looking under rocks. Did you study the section? Have you been to class?

Originally Posted by takuto

Radioactive Decay: They half-life of phosphorous-32 is 14 Days, if there are 5 grams presently, when will there be 1 gram left?

Doubling your Money: How much time is required for a \$1000 deposit to double at an interest rate of 2.5% compounded annually.
You need a model. Since we have a post entitled "Decay and Growth", and you seem to be providing exponential examples, might I sugggest:

$f(t) = A_{0}e^{k \cdot t}$

There are many ways to write it. This should look familiar. It MUST be in your book.

Radioactive Decay: The half-life of phosphorous-32 is 14 Days, if there are 5 grams presently, when will there be 1 gram left?
"there are 5 grams presently"

$f(0) = A_{0}e^{k \cdot (0)} = A_{0} = 5$

Now the model looks like this:

$f(t) = 5 \cdot e^{k \cdot (t)}$

"The half-life of phosphorous-32 is 14 Days"

$f(14) = 5 \cdot e^{k \cdot (14)} = 2.5$

See how that 2.5 is 1/2 of 5? That's what a half-life is all about.

Solve for 'k'

$5 \cdot e^{k \cdot (14)} = 2.5$

$e^{k \cdot (14)} = 0.5$

$k \cdot (14) = ln(0.5) = -ln(2)$

$k = \frac{-ln(2)}{14}$

Now the model looks like this:

$f(t) = 5 \cdot e^{\left[\frac{-ln(2)}{14}\right] \cdot (t)}$

"when will there be 1 gram left?"

$1 = 5 \cdot e^{\left[\frac{-ln(2)}{14}\right] \cdot (t_{0})}$

Solve for $t_{0}$.

As you can see, it is more algebra than anything else. Once you have the model, you are on your way.

Finish this one up and let's see the next one.