For a first order reaction $A⟶B$, find $\left[\frac{t_{7/8}}{t_{1/2}}\right]$

$\begin{array}{r}t\frac{1}{2}=\text{ sor completia }\\ \frac{1}{2}\times 100=80\\ t_{1/8}=\frac{7}{8}\times 100=\end{array}$

$t_{xy}=\frac{1}{k}\ln \frac{100}{100-x}$

$\frac{t_{18}}{t_{1/2}}=\frac{t_{67:5\mathrm{\%}}^{t50\mathrm{\%}}}{t_0}=\frac{\frac{1}{k}\ln \frac{100}{100-87.5}}{\frac{1}{k}\ln \frac{100}{100-50}}$

$=\frac{\ln \frac{100}{12.5}}{\ln \frac{100}{50}}$

$\to \frac{\ln 8}{\ln 2}$ $\frac{\ln 2^3}{{\ln }^2}=3=\frac{(3\ln 2)}{\ln 2}$

= 3