Question Number 206302 by ayietaamos last updated on 11/Apr/24
Answered by A5T last updated on 11/Apr/24
$${k}=\frac{{ln}\left(\frac{{T}}{\mathrm{3}{y}}\right)−\mathrm{1}}{\mathrm{2}} \\ $$
Answered by MATHEMATICSAM last updated on 11/Apr/24
$${T}\:=\:\mathrm{3}{ye}^{\mathrm{2}{k}\:+\:\mathrm{1}} \\ $$$$\Rightarrow\:{e}^{\mathrm{2}{k}\:+\:\mathrm{1}} \:=\:\frac{{T}}{\mathrm{3}{y}} \\ $$$$\Rightarrow\:\mathrm{2}{k}\:+\:\mathrm{1}\:=\:\mathrm{ln}\left(\frac{{T}}{\mathrm{3}{y}}\right) \\ $$$$\Rightarrow\:\mathrm{2}{k}\:=\:\mathrm{ln}\left(\frac{{T}}{\mathrm{3}{y}}\right)\:−\:\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{2}{k}\:=\:\mathrm{ln}\left(\frac{{T}}{\mathrm{3}{y}}\right)\:−\:\mathrm{ln}{e} \\ $$$$\Rightarrow\:\mathrm{2}{k}\:=\:\mathrm{ln}\left(\frac{{T}}{\mathrm{3}{ye}}\right)\: \\ $$$$\Rightarrow\:{k}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\frac{{T}}{\mathrm{3}{ye}}\right)\: \\ $$