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

$\displaystyle y = y_0 e^{kt}$

and

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

???

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 $1000 deposit to double at an interest rate of 2.5% compounded annually. You need a model. Since we have a post entitled "Decay and Growth", and you seem to be providing exponential examples, might I sugggest:$\displaystyle f(t) = A_{0}e^{k \cdot t}$There are many ways to write it. This should look familiar. It MUST be in your book. Radioactive Decay: The half-life of phosphorous-32 is 14 Days, if there are 5 grams presently, when will there be 1 gram left? "there are 5 grams presently"$\displaystyle f(0) = A_{0}e^{k \cdot (0)} = A_{0} = 5$Now the model looks like this:$\displaystyle f(t) = 5 \cdot e^{k \cdot (t)}$"The half-life of phosphorous-32 is 14 Days"$\displaystyle f(14) = 5 \cdot e^{k \cdot (14)} = 2.5$See how that 2.5 is 1/2 of 5? That's what a half-life is all about. Solve for 'k'$\displaystyle 5 \cdot e^{k \cdot (14)} = 2.5\displaystyle e^{k \cdot (14)} = 0.5\displaystyle k \cdot (14) = ln(0.5) = -ln(2)\displaystyle k = \frac{-ln(2)}{14}$Now the model looks like this:$\displaystyle f(t) = 5 \cdot e^{\left[\frac{-ln(2)}{14}\right] \cdot (t)}$"when will there be 1 gram left?"$\displaystyle 1 = 5 \cdot e^{\left[\frac{-ln(2)}{14}\right] \cdot (t_{0})}$Solve for$\displaystyle 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.