# Methods for solving equations with exponents?

In the following equation, capital letters represent arbitrary real numbers that are constant with respect to $x$:

$$A\left(x+B\right)\left(1 + \frac{C}{x+D}\right)^E + Fx + G = 0$$

I'm trying to solve for $x$; however, my algebra is failing me—especially when it comes to the exponential part.

Is it even possible to solve this equation? If so, how should one approach problems of this sort?

• What interval is $x$ in? We could potentially replace the exponenetial part with a good polynomial approximation. Commented Jun 14, 2014 at 1:00
• @SandeepSilwal: All I can guarantee is that $0\le x<B$; depending on the instance parameters, $B$ could be of the order of anything from around $10^5$ to $10^8$. Commented Jun 14, 2014 at 1:07
• It would be extremely nice if $E$ were known to be integral. Then this reduces to a polynomial equation.
– MPW
Commented Jun 14, 2014 at 1:25

If $E$ is an integer, as MPW told, the equation will reduce to a polynomial which could be quite unpleasnt to solve if $E$ is large.

For the most general case, I think that the most efficient will be numerical methods for solving nonlinear equations $f(x)=0$, Newton being the simplest to use for a good efficiency (it only requires a "reasonable" starting estimate $x_0$ for each solution to be searched.

For illustration purposes (and since I am lazy), I used $A=1$,$B=2$,$C=3$,$D=4$,$E=5$,$F =6$,$G=7$ and I plotted the function. There are solutions close to $-5$ and $-2$.

So, let us search the first one using $x_0=-5$. Newton method updates the guess according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ So, for this case, the successive iterates will be $-5.10519$,$-5.17494$,$-5.19850$,$-5.20058$,$-5.20060$ which is the solution for six significant digits.

If we do the same using $x_0=-2$, the successive iterates will be $-1.95176$, $-1.94819$,$-1.94817$ which is the solution for six significant digits.

• Thanks. Unfortunately, as mentioned in the question, all of the constants are arbitrary reals: so there is no guarantee that $E$ is integral (indeed, it almost certainly isn't). I was aware of numerical methods, including Newton-Raphson, but was very much hoping for an algebraic solution. Am I to take from your answer that it is impossible to express $x$ algebraically? Commented Jun 14, 2014 at 6:35
• Yes indeed ! If $E$ was rational, it would also lead to a polynomial (which could be a pure nightmare). Commented Jun 14, 2014 at 6:40
• If they diverge, it means that your starting point is not good enough. Plot your function first. Commented Jun 14, 2014 at 14:18