exact roots of $e^{ax}-x=0$

Question

How can I find the general solution to (not a numerical approximation) $e^{ax}-x=0$ as a function of $a$. I thought maybe something like $\frac{ln(x)}{a}$.

Answer 1

Depends on the value of $a$. For instance, if $a \geq 1 $, then there is no solution. Why?

well, $e^{ax} - x = 0 \iff e^{ax} = x $. Hence, your equation is asking where this two functions intersect. IF $a \geq 1 $, then

$$ e^{ax} \geq ax + 1 > ax > x $$

Hence, there is no solution. Now, there is indeed a solution if $a \leq \frac{1}{e} $. To see this, notice you can rewrite your equation as

$$ ax = \ln x \iff a = \frac{\ln x }{x} $$

Next, put $f(x) = \frac{ \ln x }{x } $. Doing some calculus, you will find out that $f$ has a max at $( e, f(e) ) $. Now, I dont have the power to find explicitly the root of that equation.

Answer 2

• $e^{ax}=x$
• $e^a=x^{\frac{1}{x}}$

Now $x^{\frac{1}{x}}$ is a transcendental function, see here under the examples section, which renders $y=x^{\frac{1}{x}}$ a transcendental equation, which means it cannot be rearranged in subject to $x$ using the elementary functions. So there is no general solution, only numerical approximations.

Answer 3

As Steve Kass commented, there is no anlytical solution to $$e^{ax}-x=0$$ except using Lambert function; in such a case, if the solution exists, it is given by $$x=-\frac{W(-a)}{a}$$ which, as Horacio Oliveira answered, is real only if $a \leq \frac{1}{e}$.

Otherwise, the problem of solving $f(x)=0$ can be addressed using numerical methods such as Newton (probably the simplest). Starting from a "reasonable" guess $x_0$, Newton scheme will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ From a practical point of view, the equation you are interested will give $$x_{n+1}=\frac{e^{a x_n} (a x_n-1)}{a e^{a x_n}-1}$$

For illustration purposes, let us use $a=\frac{1}{4}$ and start at $x_0=0$; the give formula will lead to the following iterates $1.33333$, $1.42899$, $1.42961$ which is the solution for six significant figures.