Show that in case of first order reaction, the time required for 99.9% of the reaction to complete is 10 times that required for half of the reaction to take place?

The integrated rate expression for first order reaction is
t= $\frac{2.303}{K1}$  log ($\frac{a}{a-x}$)
where a= initial concentration of reactants
a-x = concentration of reactants left.

so if the 99.9% of the reaction is complete then 0.1 % of the reaction is left.
Let a= 100 , and if 0.1 % reaction is incomplete, then a-x would be 0.1
now, using formulae given above, we can calculate the time period of the reaction.
t_99.9= $\frac{2.303}{K1}$ log($\frac{100}{0.1}$)

t_99.9=$\frac{2.303}{K1}$ log (1000)............................{log1000=3}


t_99.9  = $\frac{2.303}{K1}$ $\times$3.................. (1)
now, calculating time period for 50% completion of reaction,
t₅₀=  $\frac{2.303}{K1}$ log($\frac{100}{50}$)

t₅₀= $\frac{2.303}{K1}$ log(2)

t₅₀= $\frac{0.3010\times 2.303}{K1}$ ......................(2)
 Dividing equation 2 by 1
 $\frac{t_{50}}{t_{99.9}}$ = $\frac{2.303\times 0.3010}{K1}\times \frac{K1}{2.303\times 3}$ 

 $\frac{t_{50}}{t_{99.9}}=\frac{0.3010}{3}$ 

 $\frac{t_{50}}{t_{99.9}}=0.1003$ 
therefore, t_{99.9 =} 10 $\times$t₅₀