Answer
1065
Work Step by Step
Consider the exponential decay function, $P\left( t \right)={{P}_{0}}{{e}^{-kt}}$ …… (1)
Here, the value of $k$ is 0.00012; since the loss of carbon-14 is 12%, there is 88 percent left.
So, put the values in equation-1:
$\begin{align}
& P\left( t \right)={{P}_{0}}{{e}^{-kt}} \\
& 0.88{{P}_{0}}={{P}_{0}}{{e}^{-0.00012t}} \\
& 0.88={{e}^{-0.00012t}}
\end{align}$
$\begin{align}
& 0.88={{e}^{-0.00012t}} \\
& \ln 0.88=\ln {{e}^{-0.00012t}} \\
& -0.12783=-0.00012t \\
& 0.12783=0.00012t
\end{align}$
Further simplified,
$\begin{align}
& 0.12783=0.00012t \\
& \frac{0.12783}{0.00012}=t \\
& 1065.27=t \\
& 1065\approx t
\end{align}$
