How to solve quation $\exp(ax)+bx+c=0$

Question

Anyone help me to solve the equation below: $$\exp(ax)+bx+c=0$$

Answer 1

This can be solved with the Lambert W function. It is the inverse function of $\exp(x)x$.

First, multiply the equation by $\frac{a}{b}$ so the coefficients of $x$ match, then migrate the coefficient of the exponential into the exponent, like so:

$$ \exp(ax+\ln\frac{a}{b})+ax+\frac{a}{b}c=0. $$

Next, subtract $\frac{a}{b}c$ and add $\ln\frac{a}{b}$ so we get matching functions of $x$:

$$ \exp(ax+\ln\frac{a}{b})+(ax+\ln\frac{a}{b})=\ln\frac{a}{b}-\frac{a}{b}c. $$

Now, set $u=\exp(ax+\ln\frac{a}{b})$ and exponentiate both sides to get

$$ \exp(u)u=\frac{a}{b}\exp(-\frac{a}{b}c). $$

Now, take apply $W$ to both sides:

$$ u=W(\frac{a}{b}\exp(-\frac{a}{b}c)). $$

Take $\ln$ of both sides to get

$$ ax+\ln\frac{a}{b}=\ln W(\frac{a}{b}\exp(-\frac{a}{b}c)) $$

which you can now solve for $x$.