How to solve ratio between t_1/2 and t_99.9%?

The integrated rate expression for the first order reaction is
t = $\frac{2.303}{\mathrm{k}}\log \frac{\mathrm{a}}{(\mathrm{a}-\mathrm{x})}$
where a = initial concentration of reactants
a-x = concentration of reactants left

t_99.9 means that 99.9% of the reaction is complete. Then, 0.1% reactant is left
Let a = 100, then a-x = 0.1

t_99.9 = $\frac{2.303}{\mathrm{k}}\log \frac{100}{0.1}$

t_99.9 = $\frac{2.303}{\mathrm{k}}\times 3 \left(1\right)$

Time period for 50% completion of reaction,
t₅₀ = $\frac{2.303}{\mathrm{k}}\log \frac{100}{50}=\frac{2.303}{\mathrm{k}}\log 2$

t₅₀ = $\frac{2.303\times 0.3040}{\mathrm{k}} \left(2\right)$

Dividing equation 2 by 1 
$$\begin{gathered} \frac{{\mathrm{t}}_{50}}{{\mathrm{t}}_{99.9}}=\frac{2.303\times 0.3010}{\mathrm{k}}\times \frac{\mathrm{k}}{2.303\times 3} \\ \frac{{\mathrm{t}}_{50}}{{\mathrm{t}}_{99.9}}=\frac{0.3010}{3} \\ \frac{{\mathrm{t}}_{50}}{{\mathrm{t}}_{99.9}}=\frac{1}{10} \end{gathered}$$

Therefore, t_99.9:t₅₀ = 1:10