How can I find the equation for a reverse exponential curve based on three known points? I have a curve such that as $x$ approaches infinity, $y$ approaches $2$, and as $x$ approaches negative infinity, $y$ approaches infinity. I know it contains the following approximate points: 
$$(0, 70),\
(5, 40),\
(25, 11),\
(50, 2.2).$$
How can I find a function that approximates this curve? Neither logarithms nor powers under $1$ have worked for me.
 A: A function with exponential decay and a horizontal asymptote of y=2 will have the form
$$
y= A e^{-Cx} + 2\tag1
$$
If you subtract $2$ from both sides and take the log, this gives
$$
\log(y-2) =\log A - Cx\tag2
$$
Equation (2) says that there is a linear relationship between $\log(y-2)$ and $x$. So one way to fit this curve is to fit a line (either by eye, or by using a software package) to the four pairs
$$
(0, \log(68)),\quad (5,\log(38)),\quad (25, \log (9)),\quad (50, \log(0.2))$$
If the line turns out to have slope $m$ and intercept $b$, then you can solve for $A$ and $C$ in (1) using the relationships
$m=-C$ and $b=\log A$.
A: Here's an exponential "trendline" from Excel, for the data after subtracting $2$ from the $y$ values:
It's just what @grand_chat suggested in his answer.
I wouldn't put much faith in all those decimal places in the two constants.

A: You could have a better fit using, as a model, $$y=a \exp(-b\, x^c)+2$$ which is seriously more difficult to fit compared to the case $c=1$ (in particular because of $x_1=0$).
Trying first for a few values of $c$ we have 
$$\left(
\begin{array}{cc}
 c & SSQ(c) \\
 1.00 & 20.18 \\
 0.90 & 7.468 \\
 0.81 & 2.865 \\
 0.80 & 2.883 \\
 0.79 & 3.023 \\
 0.70 & 10.52
\end{array}
\right)$$
Replacing $x_1=0.00001$, the nonlinear regression works fine and we get $$y=67.9542 \exp\left(-0.157416 \,x^{0.806492}\right)+2$$ corresponding to $SSQ=2.858$.
The data and  predicted values are 
$$\left(
\begin{array}{ccc}
 x & y & y_\text{calc}\\
 0 & 70 & 69.952 \\
 5 & 40 & 40.183 \\
 25 & 11 & 10.231 \\
 50 & 2.2 & 3.694
\end{array}
\right)$$ which is not very good for the last point.
