If 10% of a radioactive material decays in 5 days, then the amount of original material left after 20 days is approximately

$$\begin{gathered} \mathrm{N}={\mathrm{N}}_0{\mathrm{e}}^{-\mathrm{\lambda t}} \\ \therefore 0.9{\mathrm{N}}_0={\mathrm{N}}_0{\mathrm{e}}^{-\mathrm{\lambda } \mathrm{x} 5}\Rightarrow 5\mathrm{\lambda }={\log }_{\mathrm{e}}\frac{1}{0.9} .........\left(\mathrm{i}\right) \\ \mathrm{and} {\mathrm{xN}}_0={\mathrm{N}}_0{\mathrm{e}}^{-\mathrm{\lambda } \mathrm{x} 20}\Rightarrow 20\mathrm{\lambda }={\log }_{\mathrm{e}}\left(\frac{1}{\mathrm{x}}\right) .........\left(\mathrm{ii}\right) \end{gathered}$$

Dividing (i) by (ii) we get

$$\begin{gathered} \frac{1}{4}=\frac{{\log }_{\mathrm{e}}(1/0.9)}{{\log }_{\mathrm{e}}(1/\mathrm{x})}=\frac{{\log }_{10}(1/0.9)}{{\log }_{10}(1/\mathrm{x})}=\frac{{\log }_{10} 0.9}{{\log }_{10} \mathrm{x}} \\ \Rightarrow {\log }_{10 }\mathrm{x}=4 {\log }_{10 }0.9\Rightarrow \mathrm{x}=0.658=65.8\% \end{gathered}$$