Facing a simple, yet frustrating formula like this

\[xe^{ax}=b\]

and the task to solve it for x left me googling around for hours until I found salvation in Wolfram Alpha, Wikipedia, and a nice blogpost with R-syntax to solve a similar equation.

Using the results from Wolfram Alpha I was able to find the solution with the gsl library

# install.packages("gsl")
library(gsl)

# create some example data
dat <- data.frame(a = 0.109861, x = 10)
# a is set so that b is roughly 30. 
# Lazy as I am I used Excel and its solver ability to find numbers

# to check if b is close to 30. Using the initial formula
dat$b <- dat$x * exp(dat$a * dat$x)
dat

# solve for x2 and see if x and x2 are similar and close to 10
dat$x2 <- lambert_W0(dat$a * dat$b)/dat$a
dat
#>  a        x  b2       x2
#>1 0.109861 10 29.99993 10.00001

Hurray!

Sometimes life can be so easy (after a long time searching for the right results….).

library(gsl)
library(rootSolve)
library(microbenchmark)
library(ggplot2)

dat <- data.frame(a = 0.109861, x = 10)
dat$b <- dat$x * exp(dat$a * dat$x)

f <- function(x, a, b) x*exp(a*x) - b

autoplot(microbenchmark(
               lambertW = dat$x2 <- lambert_W0(dat$a * dat$b)/dat$a,
               uniroot = dat$x3 <- uniroot.all(f, interval = c(0, 100),
                                               a = dat$a, b = dat$b),
               times = 10000))