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.

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.