How to solve $ax + b \log x =c$ or $\frac{a}{x}+b\log x=c$?

Question

Here $a,b,c$ are any real numbers. We can use graphical methods using Mathematical tools, but what are the other techniques ?

Answer 1

The equation can potentially have two solutions if $a$ and $b$ have different signs - then the slope of the function $g(x)=ax+b\log x$ is both positive and negative at some points, so, for some $c$, the equation has two solutions.

If $a,b$ have the same sign, then $g(x)$ is strictly monotonic, and approaches the infinities as $x\to 0,+\infty$, so every $c$ has exactly one solution.

If $a$ and $b$ have different signs, then $g$ has has either a strict maximum or minimum at $x=-\frac{b}{a}$, and so for some $c$ there are no solutions.

The exact solution is in terms of the "Lambert $W$-function."

Divide the first by $b$ and exponentiate, and you've got:

$$xe^{\frac{ax}{b}}=e^{c/b}$$

Multiply both sides by $\frac{a}{b}$ and let $z=\frac{ax}{b}$ and you have:

$$ze^z = \frac{a}{b}e^{c/b}$$

So $z=W\left(\frac{a}{b}e^{c/b}\right)$ where $W$ is the Lambert $W$-function, and thus:

$$x=\frac{b}{a}W\left(\frac{a}{b}e^{c/b}\right)$$

That doesn't really help you compute $x$, however, unless your calculator and/or computer has a method for computing the $W$-function.

You might try something iterative. Let $x_0$ be an approximate solution.

Define: $$x_{n+1}=\frac{c-b\log x_n}{a}$$

If that converges, then it converges to a solution, I'd guess fairly quickly.

Otherwise, Newton-Raphson. Let $f(x)=ax+b\log x-c$ then $$x_{n+1}=x_n -\frac{f(x_n)}{f'(x_n)} = x_n-\frac{ax_n+b\log x_n-c}{a+b/x_n}=\frac{x_n(b+c-b\log x_n)}{ax_n+b}$$

(The second class of equations is the same, essentially, as the first - letting $y=1/x$ you have $ay-b\log y = c$.)