Peter planted a tree. Its growth is given by $h(t) = \frac{a}{1+be^{kt}}$ for $t$ years. Find $a$,$b$ and $k$ 
Peter planted a tree on his backyard. Over the course of some years he
  measured the height of the tree. first directly, and later on using
  his knowledge of trigonometry. When he thought he had enough data, he
  defined as the model for the growth of the tree in meters, after $t$
  years of being planted,  the following function:
$$h(t) = \frac{a}{1+be^{kt}}$$
Determine $a$,$b$ and $k$, knowing that:
  
  
*
  
*the tree was 50 cm tall when it was planted
  
*the maximum height the tree can have is 5 meters (approximate value)
  
*after 2 years, the tree has twice the height it had when it was planted

(Note: " 5 meters (approximate value)" is an exact quote)
I did:
$50 cm = 0.5m$,
$$0.5 = \frac{a}{1+be^{-k\cdot0}} \Leftrightarrow 0.5 = \frac{a}{1+b}$$
so I conlcude that
$a = \frac{b+1}{2}$ and $b = 2a-1$
Then I used the information on the third item,
$h(2) = 1$
and I tried to isolate $k$
$$1 = \frac{\frac{b+1}{2}}{1+be^{-k\cdot2}} \Leftrightarrow 1 = \frac{b+1}{2(1+be^{-k2})} \Leftrightarrow 1 = \frac{b+1}{2+2be^{-k2}} \Leftrightarrow b+1 = 2+2be^{-2k} \Leftrightarrow \frac{b-1}{2} = be^{2k} \Leftrightarrow \frac{\frac{b-1}{2}}{b} = e^{-2k} \Leftrightarrow \frac{b-1}{2b} = e^{-2k} \Leftrightarrow \frac{b-1}{2b} = (\frac{1}{e^2})^k \Leftrightarrow k = \log_{\frac{1}{e^2}}{\frac{b-1}{2b}} \Leftrightarrow k = -\frac{1}{2} \ln(\frac{b-1}{2b}) \Leftrightarrow k = \ln(\sqrt{\frac{2b}{b-1}})$$
Then I tried the third sentence with $k = \ln(\sqrt{\frac{2b}{b-1}})$
$$1 = \frac{\frac{b+1}{2}}{1+be^{-2\ln(\sqrt{\frac{2b}{b-1}})}}$$
This took me an $\infty$ of time to solve only to find out it was wrong, so I will skip the steps and tell you that both sides are always equal to 1 except for $x \in [-1;1]$ which show "ERROR"
How do I solve this?
 A: Let us consider the function  $$h(t) = \frac{a}{1+be^{kt}}$$ and let us use the different given informations

  
*
  
*the tree was 50 cm tall when it was planted
  
*the maximum height the tree can have is 5 meters (approximate value)
  
*after 2 years, the tree has twice the height it had when it was planted

As you wrote, the first one gives $$h(0)=\frac{a}{1+be^{k\times 0}}=\frac{a}{1+b}=\frac 12\tag 1$$ Assuming that $k$ is negative, the second one write $$h(\infty)=\frac{a}{1+be^{k\times \infty}}=\frac{a}{1}=5\tag 2$$ The third one gives $$h(2)=\frac{a}{1+be^{k\times 2}}=\frac{a}{1+b e^{2k}}=2h(0)=1.0\tag 3$$  So, from $(2)$ $a=5$; from $(1)$ $b=9$. Plugging these numbers in $(3)$ $$\frac 5 {1+9e^{2k}}=1\implies 4=9 e^{2k}\implies 2=3e^k\implies k=-\log \left(\frac{3}{2}\right)$$ All of that makes that we can rewrite the model as $$h(t)=\frac 5{1+9 e^{-\log \left(\frac{3}{2}\right)\,t}}$$
Using the formula, this would give the following table
$$\left(
\begin{array}{cc}
 t & h(t) \\
 0 & 0.50 \\
 1 & 0.71 \\
 2 & 1.00 \\
 3 & 1.36 \\
 4 & 1.80 \\
 5 & 2.29 \\
 6 & 2.79 \\
 7 & 3.27 \\
 8 & 3.70 \\
 9 & 4.05 \\
 10 & 4.32 \\
 11 & 4.53 \\
 12 & 4.68 \\
 13 & 4.78 \\
 14 & 4.85 \\
 15 & 4.90 \\
 16 & 4.93 \\
 17 & 4.95 \\
 18 & 4.97 \\
 19 & 4.98 \\
 20 & 4.99 \\
\infty & 5.00
\end{array}
\right)$$
A: We have that
$$
h(t) = \frac{a}{1+be^{kt}}
$$
First information gives us.
$$
h(0) = \frac{a}{1+be^{k\cdot0}} = \frac{a}{1+b} = 0.5 \Leftrightarrow a = \frac{1+b}{2} 
$$
Second information gives us the height of the tree after long time.
$$
\lim_{t \to \infty} h(t) \Rightarrow a = 5 \Rightarrow b = 9
$$
Third information gives $h(2) = 2h(0)$
$$
h(2) = 2h(0) = 1 \Leftrightarrow \frac{5}{1+9e^{2k}} = 1 \Leftrightarrow 5 = 1 + 9e^{2k} \\
\Leftrightarrow\\
4 = 9e^{2k} \\
\Leftrightarrow \\
2k = \ln\left(\frac{4}{9}\right) \Leftrightarrow 
k = \ln\left(\frac{2}{3}\right)
$$
You wrote that $k = \ln\left(\sqrt{\frac{2b}{b-1}}\right)$, with $b=9$ you get the same $k$ as me.
