How can I find the roots of this for k?

Question

$$ln(1-e^{-kx})(1-e^{-kx})+kxe^{-kx}=0$$

I need to find $k$ in this equation , it should be a function of $x$. Any hints on how should I do it ?

Is this $$\ln(1-e^{-kx})^2+kxe^{-kx}=0$$? – Dr. Sonnhard Graubner Apr 12 '19 at 18:01 — Dr. Sonnhard Graubner, Apr 12 '19 at 18:01
no its $ln(x)*x$ where $x = 1 - e^{-kx}$ – oren revenge Apr 12 '19 at 18:04 — oren revenge, Apr 12 '19 at 18:04

score 3 · Accepted Answer · answered Apr 12 '19 at 18:19

3

If $t = e^{-kx}$, your equation says $$ (1-t) \ln(1-t) = t \ln(t) $$ Rather obviously $t=1/2$ is a solution. There are no other real solutions (unless you count $t=0$ and $t=1$).

answered Apr 12 '19 at 18:19

Robert Israel

448,999

How can I find the roots of this for k?

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