Answer
$t = 1.61 \times 10^3 y$
Work Step by Step
From the equation of radioactive dating, $R = R_0 e^{-\lambda t}$, we
Multiply both sides with natural log (ln)
$(ln) R = R_0 (ln) e^{-\lambda t}$,
Note that $(ln) e = 1$ and the power is brought to become coefficient.
$(ln) R = -R_0 (\lambda t)$,
Now solve for t to find out how old is the charcoal sample.
$t = \frac{1}{\lambda} ln \frac{R_0}{R}$
Where
$\lambda = \frac{ln 2 }{t_{1/2}} $
$ \frac{1}{\lambda} = \frac{t_{1/2}}{ln 2}$
$ \frac{1}{\lambda} = \frac{5730 y}{0.693}$
$ \frac{1}{\lambda} = 8.27 \times 10^{3} y $
So,
$t =(8.27 \times 10^{3} y) ln [\frac{15.3} {63.0} \times \frac{5.0}{1.0}]$
$t = 1.61 \times 10^3 y$
