Half-life of Am-$241$, $3$ micrograms decays over $9$ years, how much if left? $3$ micrograms of Americium-$241$, which has a half life of $432$ years. After $9$ years how much will remain?
I'm not sure of the formula to use or how to calculate it. I'm assuming it's exponential decay since it's a radioactive substance.
 A: Denote mass of substance by $M(t)$ at any given time $t$. 
So $M(t) = $ (initial mass) $ \times 2 ^{\frac{-t}{\text{(half life)}}}$.
From your question, $M(9) = 3 \times 2^{\frac{-9}{432}} \approx 2.957$. 
Edit for further explanation: 
In case the above formula for $M(t)$ isn't clear, observe than when $t=0$, $M(t)$ equals the initial mass - obviously (since $2^{0}=1$). 
When $t=$(half life), $M(t)$ equals the initial mass times $\frac{1}{2}$, i.e. it has divided by one half - exactly what "half-life" means. 
When $t = 2 \times $(half life), $M(t)$ equals the initial mass times $\frac{1}{4}$ (since $2^{-2} = \frac{1}{4}$), i.e. it has halved twice. Make sense? 
It is exponentially decaying because of the exponent $2^{-t/432}$, as $t$ gets bigger this quantity gets smaller exponentially fast.  
A: HINT:
After $432x$ years, you are left with $\frac{1}{2^x}$ of the original amount.
So calculate the value of $x$ in the equation $432x=9$, and plug it into $[\frac{1}{2^x}\times\text{original amount}]$.
A: Another way to derive the formula is as follows.
Suppose that the initial amount of the element is $N_0$ micrograms.


*

*In $1\times 432$ years there will be left $N_0\cdot\frac{1}{2}=\frac{N_0}{2} $ micrograms.

*In $2 \times 432$ years there will be left $\frac{N_0}{2}\cdot \frac{1}{2}  = N_0\cdot \left(\frac{1}{2}\right)^2$ micrograms.
..................


*

*In $n\times 432$ years there will be left $N_0\cdot \left(\frac{1}{2}\right)^n  $ micrograms.


Setting $x = n\times 432\implies n = \frac{x}{432}$. Substituting to the last formula we have:


*

*In $x$ years there will be left $N_0\cdot \left(\frac{1}{2}\right)^{\frac{x}{432}}$ micrograms.


Substitute $x$ and $N_0$ and you will reach the answer.
A: You start wit 
$$
m(0) = m_0
$$
where $m_0 = 3$ micrograms. After the time $t = T$ has passed. with $T=432$y you get
$$
m(T) = (1/2) m_0
$$
After $t = 2T$
$$
m(2T) = (1/2) m(T) = (1/2)^2 m_0
$$
After $t = 3T$
$$
m(3T) = (1/2) m(2T) = (1/2)^2 m(T) = (1/2)^3 m_0
$$
and so on. This leads to
$$
m(kT) = m_0 (1/2)^k
$$
and with $t = k T \iff k = t / T$ finally to
$$
m(t) = m_0 (1/2)^{t / T}
$$
You are looking for $m(9)$.
