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I've empiricaly produced this exponential equation to express a graphical representation :

$$ y = \left(a^x + (bx)²\right) \left((1-10^{-x}) x\right) $$

I know the constants $a$ and $b$.

Now, i would like to extract the formula to be able to calculate x plots separatively.

How can i factor this equation to solve x ?

$$ x = ?$$

I didnt practice maths for so many years, i almost completely forgot all of the factorisation rules so im likely stucked...
Some help would be greatly appreciated.

EDIT :

I forgot to say that A and B values can be restricted to a range allowing to find an acceptable solution. For example, a = 1 and b = 10

You can view the representation result in this spreadsheet : https://docs.google.com/spreadsheet/ccc?key=0AiLgphtsXoERdDN2Y2RTeUlEa1FaNFdSM3dsT0I5V3c

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For $y$ in a limited range, there may be a formula $F(y)$ such that $x$ is well-approximated by $F(y)$. An exact formula for $x$ in terms of $y$ is hopeless. –  André Nicolas Jan 14 '12 at 16:35
    
In fact i have a fixed range of X for which i would like to be able to find y. With theses y values i would like to be able to find the corresponding x. You can check the spreadsheet link i posted to fully understand my problem. –  Puls Jan 14 '12 at 17:01
    
Unfortunately, I do not have the specialized skills to produce a good formula. –  André Nicolas Jan 14 '12 at 17:05

1 Answer 1

up vote 1 down vote accepted

That is too complicated to solve explicitly, but it may be possible to solve numerically, if there is a solution (assuming $a$ is positive, there may be some negative values of $y$ for which there is no real solution).

For example if $y=2$, $a=3$ and $b=4$ then $x \approx -0.403132$ and $x \approx 0.503299$ are solutions.

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How do you solve it numerically. What formula do you use to find x in your example ? –  Puls Jan 14 '12 at 16:57
    
It looked to me as if $(0,0)$ was the minimum point on the curve, so I just found positive and negative $x$ which gave $y$ too high and then used bisection methods to find solutions: there are many others ways of doing it. An alternative (at least to start) would to draw the curve. –  Henry Jan 14 '12 at 20:50

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