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Cheers.

So if I don't make sense, I have a value for $y$, I need to know what $x$ is.

$$y = a \exp(bx) + c \exp(dx)$$

$a = 12.85$, $b = 0.001857$, $c = -54.24$, $d = -0.05316$

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Write $\displaystyle b = B/M$ and $\displaystyle d = D/M$, where $\displaystyle M \gt 0$ and $\displaystyle B,D$ are integers (I am assuming $\displaystyle b,d$ are rational).

Set $\displaystyle z = e^{x/M}$

We get

$\displaystyle y = az^B + c z^D$

This is a polynomial in $\displaystyle z$. You could try using polynomial root finding methods now.

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this is not really "polynomial" -- your assumption that B and D are integers fails with the given numbers. –  Michael Lugo Feb 25 '11 at 23:24
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@MIchael: If $b,d$ are rational, then you can always find such integer $B,D$. $\frac{m}{n}, \frac{s}{t}$ can be written as $\frac{mt}{nt}, \frac{sn}{nt}$. Of course, $B,D$ could be huge, but I believe can always find $B,D$ to be integers... –  Aryabhata Feb 25 '11 at 23:30
    
Sorry, I got $b, d$ confused with $B, D$. –  Michael Lugo Feb 25 '11 at 23:59
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I doubt there is an analytic solution.

With those values of $a,b,c,d$, it appears that $y$ is smoothly increasing and bijective function of $x$ so given any value of $y$ you should be able to find a very close approximation $x$ by standard numerical methods.

For example, if $y=0$ then $x \approx 26.175\ldots$, if $y=100$ then $x \approx 1104.9\ldots$, if $y=-100$ then $x \approx -13.7278\ldots$.

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You have $y=ae^{bx} + ce^{dx}$ with $a, b$ positive and $c, d$ negative. So this means:

  • if $x$ is large and positive, then $y \approx ae^{bx}$

  • if $x$ is large and negative, then $y \approx ce^{dx}$

So for $x$ large and positive, $x \approx (1/b) \log (y/a)$; for $x$ large and negative, $x \approx (1/d) \log (y/c)$ -- these come from solving the above approximations for $x$. If you need to go further I'd say start with these approximations and then use something like Newton's method.

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