Answer
${\large Q=2.5e^{0.347t} }$
Work Step by Step
This is an exponential decay problem with,
$\quad Q=Q_{0}e^{+kt}$
where, $Q_{0}=2.5$ cm
From the doubling time formula, we have:
$ \left\{\begin{array}{ll}
t_{d}k & =\ln 2\\
2k & =\ln 2\\
k & =\dfrac{\ln 2}{2}\approx 0.347
\end{array}\right.$
We plug in the $Q_o$ and $k$ values:
$Q=Q_{0}e^{kt}$
${\large Q=2.5e^{0.347t} }$
