# Solving a sum of exponentials

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 $\log(a + b)$...

• It's equivalent to $v^a+v^b=c$ in which $v=e^x$. Even for $a,b$ integer-valued I wouldn't expect any general formula for this. – anon Apr 9 '12 at 3:26
• Are you simply trying to solve such an equation or it appeared in a context? If you had an homework problem for example, and got stuck on this, perhaps your mistake is elsewhere. If this is just a curiosity then I agree with anon's comment. – Patrick Da Silva Apr 9 '12 at 3:30
• @PatrickDaSilva This arouse from a computer science research problem, so there is no mistake :) – jamaicanworm Apr 9 '12 at 3:35

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 :

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

• Thanks. As I commented on Alex's answer, then is there a known way to estimate $x$, and/or a way to find a lower bound for $x$? – jamaicanworm Apr 9 '12 at 3:41
• @jamaicanworm I'm commenting here because Newton's method is probably 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. – Alex Becker Apr 9 '12 at 3:55

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 $a_0 = a_1 = 1$)

• Any lower bounds on the radius of convergence? Because if there isn't any, it's kind of "risky" to try this solution. Good job though, I expected numerical algorithms would generate solutions such as series solutions, but then again I didn't expect something so explicit! It's great. What surprises me though is that you have only "one root" of the equation, where as when $s$ is an integer, you expect many complex roots. Any explanations on this? – Patrick Da Silva Apr 9 '12 at 20:36
• @Patrick: If I used Stirling's formula correctly, the radius of convergence should be $s^s(1-s)^{1-s}$ assuming $0<s<1$. – robjohn Apr 9 '12 at 21:48
• Bounds on the radius of convergence can be computed using the asymptotics of $a_k$. One that works for all $s$ is $$\left|\frac{a_k}{k!}\right| << \frac{(k(|s|+1))^{k-1}}{k^{k+1/2} e^{-k}}$$ so it converges for $|r| < 1/(e (|s|+1))$. This can be improved, e.g. for $0 < s < 1$. – Robert Israel Apr 9 '12 at 21:49
• Yes, this is just one branch of the solution, the one that approaches $1$ as $r \to 0$. For example, with $s=2$ you have the two solutions $x_{\pm} = \frac{-1 \pm \sqrt{1+4r}}{2r}$; this is the "$+$" solution, and it is analytic in $r$ for $|r| < 1/4$ (note that the singularity at $0$ is removable: $\frac{-1+\sqrt{1-4r}}{2r} = \frac{2}{1+\sqrt{1+4r}}$), but has a branch point at $r=-1/4$, so the radius of convergence is $1/4$. – Robert Israel Apr 9 '12 at 22:01
• @Patrick: seeing as $s$ is not limited to $0<s<1$, I get the radius of convergence for $s\ge1$ to be $(s-1)^{s-1}/s^s$. This is greater than, but limits to, $1/(es)$. – robjohn Apr 9 '12 at 22:21

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.

• Thanks. Is there a way to estimate $x$, and/or a way to find a lower bound for $x$? – jamaicanworm Apr 9 '12 at 3:35

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).