Show that $n(t+1)=(1+r)e^{-\alpha n(t)}n(t)$ is equivalent to $n(t+1)=(1+r)^{1-\frac{n(t)}{K}} n(t)$ where $K=\frac{\ln (1+r)}{\alpha}$

Asked Nov 21 '20 at 04:04

Active Nov 21 '20 at 13:19

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I also have to show that $n(t+1)$ will grow when $r<0$ and the initial population is greater than $n=k$. I have also tried to show this but using both variations of the equation I get an increasingly negative value.

edited Nov 21 '20 at 13:19

Martin Sleziak

53,687

asked Nov 21 '20 at 04:04

Alice

What would happen if you substitute $\alpha = \frac {\ln(1+r)}K$ into the LHS? By the way, please type out the question instead of using images. – player3236 Nov 21 '20 at 04:07
please use mathjax, and show your work, writing about failure doesn't help with solving the problem – jimjim Nov 21 '20 at 04:46
There, cleaned up your post, please instead of spending the time writing a sad story of failure use it to provide information of what has been tried instead. Also please learn mathjax – jimjim Nov 21 '20 at 04:58

Show that $n(t+1)=(1+r)e^{-\alpha n(t)}n(t)$ is equivalent to $n(t+1)=(1+r)^{1-\frac{n(t)}{K}} n(t)$ where $K=\frac{\ln (1+r)}{\alpha}$

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