Which of the following are true for the first order reaction

According to hit and trial method for first order reaction,

$\mathrm{t}=\frac{2.303}{\mathrm{K}}\log \left(\frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}\right)$

${\mathrm{t}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{2.303}{\mathrm{K}}\log \left[\frac{\mathrm{a}}{\mathrm{a}-{\displaystyle \raisebox{1ex}{$\mathrm{a}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right]$

${\mathrm{t}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{2.303}{\mathrm{K}}\log \left[\frac{2\mathrm{a}}{\mathrm{a}}\right]$

${\mathrm{t}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{2.303}{\mathrm{K}}\mathrm{x}0.3010$

${\mathrm{t}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{0.693}{\mathrm{K}}$

$\frac{{\mathrm{t}}_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{{\mathrm{t}}_{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}}=\frac{{\displaystyle \frac{2.303}{K}}\log 1}{{\displaystyle \frac{2.303}{K}}x\log 2}$

$\frac{{\mathrm{t}}_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{{\mathrm{t}}_{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}}=\frac{1}{2}\Rightarrow t_{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}=2t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

${\mathrm{t}}_{\raisebox{1ex}{$15$}\!\left/ \!\raisebox{-1ex}{$16$}\right.}=\frac{2.303}{\mathrm{K}}\log \left[\frac{\mathrm{a}}{\mathrm{a}-{\displaystyle \frac{15\mathrm{a}}{16}}}\right]$

$=\frac{2.303}{\mathrm{k}}\log \left[\frac{\mathrm{a}}{{\displaystyle \frac{16\mathrm{a}-15\mathrm{a}}{16}}}\right]$

$=\frac{2.303}{\mathrm{K}}4\log 2$

$=\frac{4\mathrm{x}0.693}{\mathrm{K}}$

$\frac{{\mathrm{t}}_{{\displaystyle \raisebox{1ex}{$15$}\!\left/ \!\raisebox{-1ex}{$16$}\right.}}}{{\mathrm{t}}_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$1$}\right.}}}=\frac{{\displaystyle \frac{2.303}{K}}x4x \log 2}{{\displaystyle \frac{2.303}{K}}x\log 2}$

$\Rightarrow {\mathrm{t}}_{\raisebox{1ex}{$15$}\!\left/ \!\raisebox{-1ex}{$16$}\right.}=4{\mathrm{t}}_{\raisebox{1ex}{$\mathrm{t}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$