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Application of Logarithms – finding the half life

Time for the material to decay to a mass of 2 kg

If there is $$10$$ kg of material at the start then $$M_0$$ is equal to $$10$$ kg. Since time is measured in years, $$3$$ months is $$0.25$$ years. This is the time taken for the material to decay to $$6$$ kg.


$\ceqns M &=& M_0 e^{-kt} \\ 6 &=& 10 e^{-k\times 0.25} & \eqncomment{0.5}{substituting the known}\\&&&\eqncomment{0.5}{information into the equation} \\ \frac{6}{10}&=& e^{-k\times 0.25} & \eqncomment{0.5}{dividing throughout by 10} \\ 0.6 &=& e^{-k\times 0.25} \\ \ln(0.6) &=& \ln e^{-k\times 0.25} & \eqncomment{0.5}{taking the natural logarithm}\\&&&\eqncomment{0.5}{of both sides} \\&=& -k\times 0.25 \ln e & \eqncomment{0.5}{using the logarithm of a}\\&&&\eqncomment{0.5}{power rule} \\ &=& -k\times 0.25 & \eqncomment{0.5}{using the logarithm to the }\\&&&\eqncomment{0.5}{same base} \\ k &=& \frac{\ln (0.6)}{-0.25} & \eqncomment{0.5}{dividing though by $$-0.25$$}\\ &\approx& 2.043302495 \ceqne$

Since we know the decay constant, it is possible to find the time taken for the material to reduce to $$2$$ kg.
\begin{eqnarray*}
2 &=& 10 e^{-2.043302495\times t} \\
0.2 &=& e^{-2.043302495\times t} \\
\ln (0.2) &=& -2.043302495\times t \ln e \\
t &=& \frac{ln(0.2)}{-2.043302495} \\
&\approx& 0.788\mbox{ years} \\
\end{eqnarray*}
Therefore, it takes approximately $$0.788$$ years to reduce to $$2$$ kg.