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Question Number 31954 by mondodotto@gmail.com last updated on 17/Mar/18

Answered by mrW2 last updated on 18/Mar/18

A=A_0 e^(−λt)   0.5A_0 =A_0 e^(−15λ)   ⇒λ=((ln 2)/(15))  0.4A_0 =A_0 e^(−λt)   ⇒λt=ln 0.4^(−1) =ln 2.5  ⇒t=((ln 2.5)/(ln 2))×15=19.8≈20 hours

$${A}={A}_{\mathrm{0}} {e}^{−\lambda{t}} \\ $$$$\mathrm{0}.\mathrm{5}{A}_{\mathrm{0}} ={A}_{\mathrm{0}} {e}^{−\mathrm{15}\lambda} \\ $$$$\Rightarrow\lambda=\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{15}} \\ $$$$\mathrm{0}.\mathrm{4}{A}_{\mathrm{0}} ={A}_{\mathrm{0}} {e}^{−\lambda{t}} \\ $$$$\Rightarrow\lambda{t}=\mathrm{ln}\:\mathrm{0}.\mathrm{4}^{−\mathrm{1}} =\mathrm{ln}\:\mathrm{2}.\mathrm{5} \\ $$$$\Rightarrow{t}=\frac{\mathrm{ln}\:\mathrm{2}.\mathrm{5}}{\mathrm{ln}\:\mathrm{2}}×\mathrm{15}=\mathrm{19}.\mathrm{8}\approx\mathrm{20}\:{hours} \\ $$

Commented by mondodotto@gmail.com last updated on 19/Mar/18

thanx

$$\mathrm{thanx} \\ $$

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