# Identifying when exponential function argument has doubled or tripled

Suppose that I have a function $f(k) = U\exp(k)$. Suppose also that I know $f(k)$, but not $k$ or $U$.
I modify this function as shown below, and take $k= k_0$ and $U$ as fixed constants:

$f(k=k_0,x) = U\exp(k)\exp(x) = U\exp(k + x)$

I evaluate $f(k,x)$ for a number of $x$, so that $f(k=k_0,x)$ is a curve that I can graphically represent. Note that $k+x$ is nothing more than the equation of a line, with $k$ as the y-intercept.

Knowing only $x$, and not knowing $k = k_0$ and $U$, can I identify $f(k,x)$ where the exponential function has doubled or tripled? Why or why not?

Would there be some sort of numerical method (or something simpler) to do this? What about using derivatives? I want to be able to identify function values $f(k_0,x=k)$ and $f(k_0,x=2k)$ such that:

$f(k_0,x=k) = U\exp(2k)$

$f(k_0,x=2k) = U\exp(3k)$

EXAMPLES

Here's some examples to provide context for my question. Initially, take $U = 1$ as a constant, and take $f_1$ as:

$f_1 = U\exp(3) = \exp(3)$

$\log(f_1) = 3$

Now suppose that the exponential function argument doubles, so that:

$f_2 = \exp(2\times3) = \exp(6)$

$\log(f_2) = 6 = 2\log(f_1)$

So if the exponential argument function doubles or triples, and $U = 1$, and I know $\log(f_1)$, it is possible to compute the following:

$\log(f_1) = 3$

$\log(f_2) = 2\log(f_1) = 6$ ........if the argument doubles

$\log(f_3) = 3\log(f_1) = 9$ ........if the argument triples

When $U = 1$, I can predict $\log(f_2)$ and $\log(f_3)$ only by knowing $\log(f_1)$. But if $U \neq 1$, and I don't know $U$, it becomes trickier. For example, take $U = 2$ so that:

$g_1 = 2\exp(3)$

Then:

$\log(g_1) = \log(2) + 3$

$\log(g_2) = \log(2) + 6$ ........if the argument doubles

$\log(g_3) = \log(2) + 9$ ........if the argument triples

So here is the gist of my question: Is there any way to determine $\log(g_2)$ and $\log(g_3)$ by only knowing $\log(g_1)$ and without knowing anything else? Suppose that I don't know the RHS of the equations above, and note that on the RHS the $\log(2)$ term shows up in each of the equations.

Note also that:

$\log(g_1) = \log(2) + 3$

$\log(g_2) \neq 2\log(g_1) = \log(2^2) + 6 \neq \log(2) + 6$

$\log(g_3) \neq 3\log(g_1) = \log(2^3) + 9 \neq \log(2) + 9$

Now say that I take $x = 0,1,2,3,4,5,6,...$ and compute:

$g(x) = 2\exp(3)\exp(x) = 2\exp(3+x)$

From the curve of $g(x)$, can I find when $g(x=3)$ or $g(x=6)$, when the argument has doubled or tripled?

Can I use curve-fitting to identify when the argument has doubled or tripled? Perhaps one way to do this would be to compare two curves and use some sort of optimization algorithm to minimize the sum of squared errors?

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I don't quite understand "identifying points on the curve where the exponential function has doubled or tripled". I think your $f(k = k_0, x) = U \exp(k) \exp(x)$ is just a constant multiple of $\exp(x)$, and the constant is arbitrary. I don't believe you can gain any more information other than what you are given. –  Tunococ Aug 14 '12 at 20:49
Thanks, Tunococ; I've updated the question above and added more information in an attempt to show what I am trying to ask. –  Nicholas Kinar Aug 14 '12 at 23:40

The key is in your first line. $U\exp (k)$ is a constant and nothing you are doing can tease apart $U$ and $k$. In fact, it equals $\exp (k + \ln U)$. You can multiply it by $\exp (x)$ all you want and can't tell it from $\exp((k-\ln 2)+ \ln (2U))$ (for example).
I totally agree. There is NOTHING you can do to split $U$ and $\exp(k)$ apart. –  Tunococ Aug 14 '12 at 23:50
@RossMillikan: Thanks Ross. Is there anything that can be done to estimate $U$ or $k$, given some additional information (bounds on $U$ or $k$), or some change in $U$ or $k$? –  Nicholas Kinar Aug 15 '12 at 0:00
I am reminded again of a curve-fitting problem example: mathworks.com/support/tech-notes/1500/1508.html. The $A$ and $\lambda$ parameters are being found from $f = A\exp(\lambda t)$, where $A$ and $\lambda$ are constants, and $t$ is a variable. Why does the curve-fitting problem work, and what makes the curve-fitting solution unique, unlike the problem that I posted above? –  Nicholas Kinar Aug 15 '12 at 0:10
I agree as well that there is nothing that can be done to exactly split $U$ and $\exp(k)$ apart, but maybe there is a way to estimate $U$ or $k$. –  Nicholas Kinar Aug 15 '12 at 0:20
@NicholasKinar: for the curve fit example, you are able to change the exponent. The $t$ multiplies $\lambda$, so you can tell the two variables apart. Here you don't have that. It's like trying to solve $x+y=2$ with no further information except that you can add $z$ and find $x+y+z=5$ when $z=3$. You still have one equation with two unknowns. –  Ross Millikan Aug 15 '12 at 0:30