If 75% of radioactive material remains, how many half-lives have elapsed?
@gerrit provided a formula, but without stating the reasoning behind it.

Radioactive decay is an exponential function.  After $n$ half-lives, the amount of the original material remaining is

$$\textrm{amount remaining after}\ n\ \textrm{half-lives} = \left(\frac{1}{2}\right)^n$$

Therefore, you want to solve

$$\begin{align*}
\left(\frac{1}{2}\right)^n &amp;= \frac{3}{4}\\
\log_\frac{1}{2} \left(\frac{1}{2}\right)^n &amp;= \log_\frac{1}{2} \frac{3}{4}\\
n &amp;= \frac{\log \frac{3}{4}}{\log \frac{1}{2}} = \frac{\log \frac{3}{4}}{- \log 2}
   \approx 0.415
\end{align*}$$