# 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?

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Exponential functions increase to infinity or decay to $0$ – Henry Oct 24 '12 at 15:28

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 $c=39$.

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Why can't you change the exponent, leaving it a pure exponential function? – gt6989b Oct 24 '12 at 16:10
Let $a>1$. Any pure exponential function $a^{kx}$ will blow up as $x\to\infty$ if $k$ is positive, and will approach $0$ as $x\to \infty$ if $k$ is negative. You want neither of these. If we pick $0\lt a\lt 1$, the behaviour is reversed, again not what you want. – André Nicolas Oct 24 '12 at 16:20
Yes, sir, but what about something like $a^{f(x)}$ where $f(x) \to 0$ when $x \to \infty$? – gt6989b Oct 24 '12 at 16:22
That's not usually considered to be an exponential function. And for good reason. Let $f(x)=\ln x$. Then $e^{f(x)}=x$. Shall we consider $x$ an exponential function? If you allow $e^{f(x)}$, then any positive function can be considered an exponential function! – André Nicolas Oct 24 '12 at 16:26
You are right :), but it seems there is still some fundamental difference between something like $x = e^{\ln x}$ (which in its simplest form has no explicit exponent) and $e^{1/x^2}$, which does... Even the shape of the graph in the second case strongly suggests exponential origins, while the first one, of course, looks very different... – gt6989b Oct 24 '12 at 16:55

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.

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Why can't you change the exponent, leaving it a pure exponential function? – gt6989b Oct 24 '12 at 16:11
@gt6989b A decreasing exponential function will halve then halve again etc. and so will continue to do so it will not not stop before approaching zero. – Henry Oct 24 '12 at 18:30

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

$f(x) = e^{\ln(40)/x^2} = 40^{1/x^2}$.

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