1. Originally Posted by mankvill
So A(10) = 4(a)^10?
yes, what you want is to find A(10) when A_0 = 4. now plug in the value for a and solve

you need to know what each term is. that way, you can translate from the word problem to the math equation and know what they are asking you for and what you are given.

$\displaystyle A(t)$ is the amount left after a time $\displaystyle t$ has passed

$\displaystyle A_0$ is the initial amount. this is what we have when time is zero. nothing has started to decay yet

$\displaystyle t$ is the the amount of time that has passed.

2. Blech, I think I'm still confused.

Set up all together, is it:

1/2 (4) = (4)(.981925)^10
?

Because I don't think that works. D:

3. Originally Posted by mankvill
Blech, I think I'm still confused.

Set up all together, is it:

1/2 (4) = (4)(.981925)^10
?

Because I don't think that works. D:
where is the 1/2 coming from, and why is the 4 on the left there? note, the $\displaystyle \frac 12 A_0 = A_0a^t$ equation was ONLY for half-life, nothing else. we did it to solve for a, it is NOT the equation we are working with. our equation is:

$\displaystyle \boxed{A(t) = A_0a^t}$

now you found a in the first part, you are now given $\displaystyle A_0 \mbox{ and }t$ and you are asked for how much is left, namely $\displaystyle A(t)$. plug in what you know, and solve for what you don't

4. Originally Posted by Jhevon
no! where is the 1/2 coming from, and why is 4 there? note, the $\displaystyle \frac 12 A_0 = A_0a^t$ equation was ONLY for half-life, nothing else. we did it to solve for a, it is NOT the equation we are working with. our equation is:

$\displaystyle \boxed{A(t) = A_0a^t}$

now you found a in the first part, you are now given $\displaystyle A_0 \mbox{ and }t$ and you are asked for how much is left, namely $\displaystyle A(t)$. plug in what you know, and solve for what you don't
OOHHH, oh oh oh. Okay.

A(10) = 4(.981925)^10
?

that seems right.

5. Originally Posted by mankvill
OOHHH, oh oh oh. Okay.

A(10) = 4(.981925)^10
?

that seems right.