d
\(t_{ LAS }=\frac{1}{ K } \ln \left(\frac{ a }{ a - x }\right)=\frac{1}{ K } \ln \left(\frac{100}{90}\right)\)

\(t _{10 s }=\frac{2.303}{ K }(\log 10-\log 9)\)

\(t _{10 s }=\frac{2.093}{ K } \times(0.04)\)

Similarly,

\(t _{90 \% 6}=\frac{1}{ K } \ln \left(\frac{100}{10}\right)\)

\(t _{90 s }=\frac{2.303}{ K }\)

\(\frac{t_{9 rP _6}}{t_{10 \%}}=\frac{1}{0.04}=25\)

\(e^k=\frac{a}{a-x}\)

\(\frac{a-x}{a}=e^{-k t}\)3

\(1-\frac{x}{a}=e^{-k t}\)

\(x=a\left(1-e^{-l a t}\right)\)

\(\alpha=\frac{x}{a}=\left(1-e^{-b}\right)\)