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Question Number 13412 by tawa tawa last updated on 19/May/17

e^(−kN)  − kN  − 1 = 0  Find the value of  N

$$\mathrm{e}^{−\mathrm{kN}} \:−\:\mathrm{kN}\:\:−\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:\mathrm{N} \\ $$

Answered by mrW1 last updated on 20/May/17

e^(−kN) =kN+1  e^(−kN−1) =(kN+1)e^(−1)   (kN+1)e^((kN+1)) =e  kN+1=W(e)  N=((W(e)−1)/k)=((1−1)/k)=0

$${e}^{−{kN}} ={kN}+\mathrm{1} \\ $$$${e}^{−{kN}−\mathrm{1}} =\left({kN}+\mathrm{1}\right){e}^{−\mathrm{1}} \\ $$$$\left({kN}+\mathrm{1}\right){e}^{\left({kN}+\mathrm{1}\right)} ={e} \\ $$$${kN}+\mathrm{1}={W}\left({e}\right) \\ $$$${N}=\frac{{W}\left({e}\right)−\mathrm{1}}{{k}}=\frac{\mathrm{1}−\mathrm{1}}{{k}}=\mathrm{0} \\ $$

Commented by tawa tawa last updated on 20/May/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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