Question Number 31954 by mondodotto@gmail.com last updated on 17/Mar/18
Answered by mrW2 last updated on 18/Mar/18
$${A}={A}_{\mathrm{0}} {e}^{−\lambda{t}} \\ $$$$\mathrm{0}.\mathrm{5}{A}_{\mathrm{0}} ={A}_{\mathrm{0}} {e}^{−\mathrm{15}\lambda} \\ $$$$\Rightarrow\lambda=\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{15}} \\ $$$$\mathrm{0}.\mathrm{4}{A}_{\mathrm{0}} ={A}_{\mathrm{0}} {e}^{−\lambda{t}} \\ $$$$\Rightarrow\lambda{t}=\mathrm{ln}\:\mathrm{0}.\mathrm{4}^{−\mathrm{1}} =\mathrm{ln}\:\mathrm{2}.\mathrm{5} \\ $$$$\Rightarrow{t}=\frac{\mathrm{ln}\:\mathrm{2}.\mathrm{5}}{\mathrm{ln}\:\mathrm{2}}×\mathrm{15}=\mathrm{19}.\mathrm{8}\approx\mathrm{20}\:{hours} \\ $$
Commented by mondodotto@gmail.com last updated on 19/Mar/18
$$\mathrm{thanx} \\ $$