If the radioactive isotope strontium $240$ has a half life of $120$ years, how long until it decays to only $60\%$ of its original radioactivity? I been trying to solve this problem for hours and the only thing i came up with, was a formula for their relationship.
$1/2A = A_0 e^{120r}$
$\ln(1/2 = e^{120}r)$
$\ln(1/2) = 120r$
$r = \ln(0.5)/120 $
$r = -0.0057762265$
So, when the isotope is $120$ years old the percentage is $-0.0057762265$. But I don't know how to find how long until the isotope decays to $60\%$? 
I really need on understanding this problem, I don't only want the answer I want an explanation of how to do it? please help, I have an exam tomorrow base on problems like this. 
 A: Your approach is the hard way.  Do it like this:
Every time $120$ years pass, you multiply the amount by $1/2$.
If $t$ is the number of years that have passed, then $t/120$ is the number of $120$-year periods that have passed.  That is therefore how many times you multiply by $1/2$.  Hence the amount remaining after $t$ years is
$$
\text{original amount} \times \left( \frac 1 2 \right)^{t/120}.
$$
So you need to know what $t$ is when $\left(\dfrac 1 2 \right)^{t/120}\!\!\!\!\! = 0.6$.
You have $$\frac t {120} = \log_{1/2} 0.6.$$  And then you can find $t$.
The reason for bringing in base-$e$ logarithms and base-$e$ exponential functions is to take about instantaneous rates of change, i.e. derivatives.  As long as the problem doesn't involve those, then bringing in the number $e$ is just a pointless complication.
However, by your method one should say
$$
A = A_0 e^{rt}
$$
$$
\frac 1 2 = e^{r120}
$$
$$
r = \text{a particular negative number}
$$
$$
e^{rt} = 0.6
$$
$$
rt = \ln 0.6 \quad \text{(This is negative.)}
$$
$$
t = \frac{\ln 0.6} r \quad \text{(This is positive.)}
$$
One thing that happens when this method is used is that rounded values are used where exact values are available.  The happens especially when a student see something like $e^{t\ln 0.7}$ and instead of realizing that that is exactly $0.7^t$, gets a rounded value of $0.7$ from a calculator and goes on from there.
A: You know that the basic exponential growth/decay equation is
\begin{equation}
A=A_0e^{rt}
\end{equation}
You are told that when $t=120$ that
\begin{equation}
A=\tfrac{1}{2}A_0e^{120r}
\end{equation}
which you solved correctly for $r$.
Now you wish to know the value of $t$ which causes
\begin{equation}
 0.60A_0=A_0e^{-0.005762265\,t}
\end{equation}
so you must solve for $t$ the following equation:
\begin{equation}
e^{-0.005762265\,t}=0.60
\end{equation}
which you should have no trouble doing.
