Plug C {\displaystyle C} and k {\displaystyle k} obtained from part (a) into the expression of P {\displaystyle P} ;
P ( t ) = 300 e 3 5 ln 6 e ln 6 25 t {\displaystyle P(t)={\frac {300}{e^{{\frac {3}{5}}\ln 6}}}e^{{\frac {\ln 6}{25}}t}}
Now, we solve P ( t ) = 3000 {\displaystyle P(t)=3000} ;
300 e 3 5 ln 6 e ln 6 25 t = 3000 ⟺ e ln 6 25 t = 10 e 3 5 ln 6 ⟺ ln 6 25 t = ln 10 + ln ( e 3 5 ln 6 ) = ln 10 + 3 5 ln 6 ⟺ t = 25 ln 6 ( ln 10 + 3 5 ln 6 ) . {\displaystyle {\begin{aligned}{\frac {300}{e^{{\frac {3}{5}}\ln 6}}}e^{{\frac {\ln 6}{25}}t}=3000&\iff e^{{\frac {\ln 6}{25}}t}=10{e^{{\frac {3}{5}}\ln 6}}\\&\iff {\frac {\ln 6}{25}}t=\ln 10+\ln(e^{{\frac {3}{5}}\ln 6})=\ln 10+{\frac {3}{5}}\ln 6\\&\iff t={\frac {25}{\ln 6}}\left(\ln 10+{\frac {3}{5}}\ln 6\right).\end{aligned}}}
Therefore, we have 3000 bacteria in 25 ln 6 ( ln 10 + 3 5 ln 6 ) m i n {\displaystyle \color {blue}{\frac {25}{\ln 6}}\left(\ln 10+{\frac {3}{5}}\ln 6\right)min}
and 
obtained from part (a) into the expression of 
;
;
