Answer
1964 years
Work Step by Step
Consider the exponential decay function, $P\left( t \right)={{P}_{0}}{{e}^{-kt}}$ …… (1)
Hence, the seed lost 21% carbon-14 from its initial, so the remaining carbon-14 is 79%.
Therefore, put $k=0.00012$ and $0.79{{P}_{0}}$ at the place of $P\left( t \right)$in equation-1 as,
$\begin{align}
& P\left( t \right)={{P}_{0}}{{e}^{-kt}} \\
& 0.79{{P}_{0}}={{P}_{0}}{{e}^{-0.00012t}} \\
& 0.79={{e}^{-0.00012t}}
\end{align}$
$\begin{align}
& 0.79={{e}^{-0.00012t}} \\
& \ln 0.79=\ln {{e}^{-0.00012t}} \\
& -0.2357=-0.00012t \\
& 0.2357=0.00012t
\end{align}$
Further simplified,
$\begin{align}
& 0.2357=0.00012t \\
& \frac{0.2357}{0.00012}=t \\
& 1964.3=t \\
& 1964\approx t
\end{align}$
