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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.

\(
\newcommand{\eqncomment}[2]{\small{\text{ #2}} } 
\newcommand{\ceqns}{\begin{array}{rcll}}
\newcommand{\ceqne}{\end{array}}
\)

\[
\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.