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Question-206302




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

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