An organic compound undergoes first order decomposition. The time taken for its  decomposition to 1/8 and 1/10 of its initial concentration are $t_{1/8}$  and  $t_{1/10}$  respectively.
What is the value of  $\frac{\left[t_{1/8}\right]}{\left[t_{1/10}\right]}\times 10$ ?  $({\log }_{10}2=0.3)$

For a first order process,  $kt=\ln \frac{{\left[A\right]}_0}{\left[A\right]}$
Where  ${\left[A\right]}_0$   = initial concentration 
$\left[A\right]$  = concentration of reactant remaining at time t.
$k{t}_{1/8}=\ln \frac{{\left[A\right]}_0}{{\left[A\right]}_0/8}=\ln 8$  ……………. (1) 
 $k{t}_{1/10}=\ln \frac{{\left[A\right]}_0}{{\left[A\right]}_0/10}=\ln 10$  ………… (2)
Therefore,  $\frac{t_{1/8}}{t_{1/10}}=\frac{\ln 8}{\ln 10}=\log 8=3\log 2$
 $\frac{t_{1/8}}{t_{1/10}}=3\times 0.3=0.9$
 $\frac{t_{1/8}}{t_{1/10}}\times 10=0.9\times 10=9.0$