Here is my work as well. 
I always get stuck when I have to do a u-sub and deal with a negative LN. Some help or a strategy for the future please? Thank you!
2026-03-30 06:50:00.1774853400
Lake with capacity of 1000 fish. How long will it take to reach 900?
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you can show that$$N = \frac{1000}{1 + ce^{-0.2t}} $$ solves $$\frac{dN}{dt}= 0.2N \left( 1-\frac N{1000}\right)$$ for arbitrary $c.$ that $N = 20$ at $t = 0$ fixes $c = 49.$ now solve $$ 900= \frac{1000}{1 + 49e^{-0.2t}}$$ for $t.$