How long to reach 75% of carrying capacity $P(t)=\frac{14250}{1+29e^{-0.62t}}$ Given the following logistic growth equation:
$$P(t)=\frac{14250}{1+29e^{-0.62t}}$$
I am to find out how many time periods $t$ it takes to reach 3/4 of the carrying capacity. The solution is provided as ~7.2 whereas I got 0.061.
My working:
The carrying capacity is the numerator 14250 so I want to calculate how long it takes to reach 75% of this which is 10,687.5
$$10687.5=\frac{14250}{1+29e^{-0.62t}}$$
$$1+29e^{-0.62t}(10678.5)=14250$$
$$29e^{-0.62t}=\frac{14250}{10687.5}-1$$
$$e^{-0.62t}=\frac{\frac{14250}{10687.5}-1}{29}$$
$$-0.62t=ln(\frac{\frac{14250}{10687.5}-1}{29})$$
$$t=\frac{ln(\frac{\frac{14250}{10687.5}-1}{29})}{-0.62}$$
$$t=0.061$$
Where did I go wrong and how can I arrive at 7.2?
 A: How did you go from
$$10687.5=\frac{14250}{1+29e^{-0.62t}}$$
to
$$1+29e^{-0.62t}(10678.5)=14250$$
?
Don't you think you need to distribute?
By the way you can simplify things by just taking $\frac 34 =\frac{1}{1+29e^{-0.62t}}$ and cross multiplying (the numerator $14250$ is the theoretical carrying capacity, the supremum of the function, so it doesn't matter). The numbers are smaller and maybe there's less chance of confusion.
I've reviewed your working. Let me replace with symbols to make things a bit clearer (hopefully).
Let $a = 10687.5, b = 1, c = 29e^{0.62t}, d = 14250$.
You start with $a = \frac{d}{b+c}$.
When you multiply both sides by $(b+c)$ ("bring over" the $b+c$), what do you get? $a(b+c) = d$.
What happens now? Shouldn't it be $ba+ ca = d$?
But you wrote $b + ca = d$. You didn't distribute.
And then you propagated the error by continuing $ca = d - b$ followed by $c = \frac da - b$, again failing to distribute on division. The correct continuation from this (incorrect) step would've been c = \frac da - \frac ba$.
But (coincidentally), you actually ended up with the correct expression $c = \frac da - b$, because you made two errors of distribution that "cancelled out".
This is how the thing should have gone:
$a = \frac{d}{b+c}$
$a(b+c) = d$
$ab + ac = d$
$ac = d - ab$
$c = \frac da - b$
ending up where you left off.
So I was mistaken - that's not why you're not getting the right result. It doesn't change the fact that your working is still mathematically wrong, though.
Your error seems to occur later, possibly when you compute this expression:
$$t=\frac{ln(\frac{\frac{14250}{10687.5}-1}{29})}{-0.62}$$
to get to this:
$$t=0.061$$ (?)
Because when I punched in that exact expression into my calculator I got the right answer of approximately $7.20$.
To sum up, you have made algebraic errors that "cancelled out" at the start, you ended up with the correct expression, but you somehow still calculated it wrongly. How you did that, I can't tell. Try that computation again.
A: Quicker would be:
$$1 + 29e^{-0.62t} = \frac{4}{3} \implies e^{-0.62t} = \frac{1/3}{29}$$
$$\implies t = \ln \left(\frac{1/3}{29} \right) \cdot \frac{1}{-0.62} \approx 7.20$$
