Find an exponential function with given condition How can I have an example of an exponential function defined in the X range 1 - infinity, with values starting at 40 and converging to 1?
 A: It can't be a pure exponential, since a decaying exponential function decays to $0$. But we can look for a function of the kind $1+ke^{-x}$. Then our condition of having value $40$ at $1$ becomes the equation
$$1+ke^{-1}=40.$$
Solve. We get $ke^{-1}=39$, so $k=39e$.
Slightly more naturally, we can look for a function of the type $1+ce^{-(x-1)}$. Then we find that $c=39$.
We have freedom in adjusting the rate of decay, by looking for functions of the shape
$$1+ce^{-\lambda(x-1)}.$$
Pick any positive $\lambda$ that you like, and let $c=39$.
There is no need to use $e$ as the base. Let $a$ be your favourite base.  We can look for a function of shape  $1+c a^{-(x-1)}$. Again we will get $c=39$.  
A: Since this cannot be a pure exponential function, let's try some others.  
$1+39k^{x-1}$ is one possibility for any $0 \lt k \lt 1$ and and produces functions $1$ more than an exponential function.
$40 x^{1/x^k}$ for any $k \gt 0$ may have the sort of shape you are looking for, but I think it is not an exponential function. 
A: I would like to propose an exponential function which would converge to $1$. To do that, we must have the exponent converge to $0$ as $x \to \infty$, say $1/x^2$ would do just fine. So consider now $f(x) = e^{A/x^2}$ for some $A \in \mathbb{R}$. Note that
$\lim_{x \to \infty} f(x) = \lim_{x \to \infty} e^{A/x^2} = 1$
as desired, so it remains to pick $A$ so that $f(1) = 40$. But $f(1) = e^A$, so we can let $A = \ln 40$ to get
$f(x) = e^{\ln(40)/x^2} = 40^{1/x^2}$.
