# No idea how to solve this equation using two exponentials

The equation I have is:

$$A = B ( \exp(C x) - \exp(-Dx) )$$

How do I solve for $x$ given $A$, $B$, $C$, $D$?

I have no idea how. The only idea I have is that I could express these in terms of a taylor series, but is there a better way that I do not know? I could also generate $A$ based on known $B, C, D$ and for a given range of $x$ of finite steps, then find my particular $A$ of interest and see what $x$ was required. (I'm thinking that's a graphical way to attack this.) But it isn't very satisfying.

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I see $A$ as a solution of a linear OE with constant solution. – Babak S. May 21 '13 at 3:06
Except for very special choices of $C$ and $D$, one cannot expect a "closed form" formula for $x$. A numerical method, like Newton-Raphson, is always a possibility. – André Nicolas May 21 '13 at 3:18
The equation can be rearranged to be $\alpha^x=\gamma+\beta^x$, where $\alpha=e^C, \beta=e^{-D}, \gamma=\frac{A}{B}$. In this form it is easy to prove that there will always be exactly one solution (so long as $\alpha\neq\beta$). Finding it is probably something that must be done numerically, as @Andre points out. – vadim123 May 21 '13 at 3:18
@AndréNicolas Could you explain how one cannot expect a closed form formula for $x$ ? – sciencenewbie May 21 '13 at 4:03
Experience. By choosing $C=4$, $D=1$ we get essentially a degree $5$ polynomial equation, for whose solutions there is no nice closed form. Choosing $C=6$, $D=1$ makes things worse. And these are nice choices. One may be able to get reasonably quickly converging series solutions. But in general exponential equations, apart from quite special cases, are "algebraically" intractable. – André Nicolas May 21 '13 at 4:11