[tex]\qquad \textit{Amount for Exponential Decay} \\\\ A=P(1 - r)^t\qquad \begin{cases} A=\textit{current amount}\dotfill &\stackrel{40\%~of~P}{0.4P~~~~}\\ P=\textit{initial amount}\dotfill &P\\ r=rate\to 22.5\%\to \frac{22.5}{100}\dotfill &0.225\\ t=\textit{elapsed time}\\ \end{cases} \\\\\\ 0.4P=P(1-0.225)^t\implies \cfrac{0.4P}{P}=(1-0.225)^t\implies 0.4=0.775^t[/tex]
[tex]\log(0.4)=\log(0.775^t)\implies \log(0.4)=t\log(0.775) \\\\\\ \cfrac{\log(0.4)}{\log(0.775)}=t\implies \stackrel{\textit{about 3years, 215days and 8hrs}}{3.59\approx t}[/tex]
so at that time the value is 40% of its original value, I guess is less than that a minute or an hour later.
