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If I have an equation of the form $e^{ax} + e^{bx} = c,$ where $a$, $b$, and $c$ are constants, how can I simplify the equation to solve for $x$?

Taking the logarithm of both sides is tricky, since I know $\log(ab) = log(a) + log(b)$, but I don't know how to simplify $\log(a + b)$...

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    @PatrickDaSilva This arouse from a computer science research problem, so there is no mistake `:)`2012-04-09

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If $a/b$ is 2 or 1/2, then it will reduce to a quadratic. This can be useful if you have an experiment and you can control the time at which measurements are taken to be at $t_0$, $t_0+d$, and $t_0+2d$, and the process proceeds exponentially (like Newton's law of cooling).

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Write the equation as $z + r z^s = 1$ where $z = e^{ax}/c$, $r = c^{b/a-1}$, $s = b/a$. There is a series for a solution of this, which should converge for sufficiently small $r$:

$ z = \sum_{k=0}^\infty \frac{(-1)^k a_k}{k!} r^k \ \text{where} \ a_k = \prod_{j=0}^{k-2} (ks-j)$

(taking $a_0 = a_1 = 1$)

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    @Patrick: seeing as $s$ is not limited to 0, I get the radius of convergence for $s\ge1$ to be $(s-1)^{s-1}/s^s$. This is greater than, but limits to, $1/(es)$.2012-04-09
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Following up with Alex's Becker's answer, you can turn your equation into an equation of the form $ y^a + y^b = c, $ which, for $a$ and $b$ distinct positive integers with one of them greater or equal to $5$, we know by Galois theory that there exists no solution in terms of traditional arithmetic (i.e. addition, subtraction, multiplication, division, and taking $n^{\text{th}}$ roots) to this polynomial equation. I've tried to find a website that speaks about it but a quick look over google and wikipedia gave me nothing ; this is a very well known result though. Therefore we expect no general solution to your equation, because it would imply very specific results for which we know there exists no general method to solve.

Hope that helps,

EDIT : There wasn't enough space in the comment box to detail this.

If you want computer accuracy, you can use numerical methods. Find a root of $f(x) = c - e^{ax} - e^{bx}$ using, for instance, Newton's method. But analytically I have not much hope. There is one thing you could do though : using the Taylor expansion of $e^x$, $ 0 = e^{ax} + e^{bx} - c \ge (1 + ax) + (1 + bx) - c = (2-c) + (a+b) x, $ which gives you a rough upper bound on $x$ like this : $ x \le \frac{c-2}{a+b}. $ I have no idea how to get a lower bound though. Note that this bound feels very crappy after you give some though about it ; fix $a=b=1$, which means you're trying to solve $2e^x = c$, which means $e^x = c/2$ and $x = \log(c/2) < \frac{c-2}2$. Here's an idea of how crappy this bound is :

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We see that for $c > 4$, it's already very crappy. Anyway.

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    @jamaicanworm I'm commenting here because Newton's method is probabl$y$ the best and easiest way to go in most cases. However, if you are specifically looking for a rigorous lower bound you may want to look into real semialgebraic methods. Warning: they are very nasty.2012-04-09
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Unfortunately, no elementary solution exists for general $a,b$. This is because solving $e^{ax}+e^{by}=c$ is equivalent to solving $y^{a/b}+y=c$ where $y=e^{bx}$, and even in the case $a/b=5$ the solution is expressed in terms of Bring radicals.

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    Thanks. Is there a way to estimate $x$, and/or a way to find a lower bound for $x$?2012-04-09