Solve equation with exponentials

Question

I'm trying to obtain $x$ in the following equation:

$$ 1= ae^{bx}+ce^{dx} $$

with a,b,c,d known.

What I did was to take the ln of all the equation, so I have:

$$ \log\frac{1/ac}{b+d} = x $$

But when I put numbers, I am getting something completely wrong. Is my solution incorrect then?

ps:

   a =       2.321
   b =     -0.0529
   c =      0.3172
   d =   -0.003212

and the "correct $x$" it is:

$x\sim 22.526$

Answer 1

There is no nice analytic solution for your case. You need to solve it numerically.

Set $e^{bx}=y$, then $e^{dx}=y^{d/b}$, you equation becomes:

$$y=\frac{1}{a} \left(1-cy^{d/b} \right)$$

Solve by iterations:

$$y_1=\frac{1}{a}$$

$$y_2=\frac{1}{a}\left(1-cy_1^{d/b} \right)$$

$$y_3=\frac{1}{a}\left(1-cy_2^{d/b} \right)$$

$$ \dots $$

You will get something like $y=0.30372$ after 3-4 steps, which means:

$e^{bx}=0.30372$

$$x \approx 22.526$$

Which is very close to your 'correct' value.

Notice, that my choice of $y$ was based on the fact that $a>c$ and $b>d$. If I chose $e^{dx}$ as $y$ the iterations would work much worse.

---

But as it was said in the comments, with your values you better use Taylor series for the exponent. It will be much easier (actually, linear approximation doesn't work very well here, so maybe iterations are better)

Answer 2

Beside a few cases where $b$ and $d$ would be in a ratio of $2$, $3$ or $4$ (in such a case, the equation would reduce to a polynomial), there is no analytical solution to such an equation and numerical methods need to be considered.

In your problem, you are lucky in the sense that $a$ and $c$ have the same sign, as well as $b$ and $d$; so, you are sure that there is only one solution to the equation.

Let us apply Newton method to $$f(x)= ae^{bx}+ce^{dx}-1$$ $$f'(x)=abe^{bx}+cde^{dx}$$ Using your numbers, let us start iterating at $x_0=0$ (I am very lazy). So, the method will generate the following iterates $$x_1=13.2327$$ $$x_2=20.6029$$ $$x_3=22.4337$$ $$x_4=22.5261$$ $$x_5=22.5263$$ which is the solution for six significant figures.

We could solve it faster looking for the zero of function $$g(x)=\log(f(x))=\log( ae^{bx}+ce^{dx})$$ $$g'(x)=\frac{a b e^{b x}+c d e^{d x}}{a e^{b x}+c e^{d x}}$$ Starting again with $x_0=0$ (I am still lazy), the method will generate the following iterates $$x_1=20.6730$$ $$x_2=22.5043$$ $$x_3=22.5263$$ which is the solution for six significant figures.