Explanation:
Let \({\mathrm{a=100,(a-x)=80}}\) (Amount left).
\({\mathrm{\therefore}}\) Amount left \({\mathrm{=\dfrac{a}{(2)^{n}} \quad[n=}}\) number of half lives \({\mathrm{]}}\)
\({\rm{80 = }}\frac{{{\rm{100}}}}{{{{{\rm{(2)}}}^{\rm{n}}}}} \Rightarrow {{\rm{(2)}}^{\rm{n}}}{\rm{ = }}\frac{{{\rm{10}}}}{{\rm{8}}}\)
\(\therefore {\rm{n \,log}}\,{\rm{2}}\,{\rm{ = }}\,{\rm{log10}} - {\rm{3\,log}}\,{\rm{2}}\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 - 3 \times 0.3 = 0.1\)
\({\rm{n}} = \frac{{0.1}}{{{\rm{log2}}}} = \frac{{0.1}}{{0.3}} = \frac{1}{3}\)
\[{\rm{n}}\,{\rm{ = }}\frac{{\rm{1}}}{{\rm{3}}}{\rm{ = }}\frac{{{\rm{t}}\left( {\begin{array}{*{20}{c}}
{{\rm{Time}}\,\,{\rm{for}}\,\,{\rm{80\% }}\,\,{\rm{amount}}}\\
{{\rm{left}}\,{\rm{or}}\,{\rm{20\% }}\,{\rm{decomposed}}}
\end{array}} \right)}}{{{{\rm{t}}_{{\rm{1/2}}}}\left( {\begin{array}{*{20}{c}}
{{\rm{Time}}\,{\rm{for}}\,{\rm{50\% }}\,{\rm{amount}}}\\
{{\rm{left}}\,{\rm{order}}\,\,{\rm{composed}}}
\end{array}} \right)}}\]
\( \Rightarrow {{\rm{t}}_{{\rm{1/2}}}}{\rm{ = 3t}}\)