How do you solve $2^x-x=3$?

Question

Maybe it's a simple question, but I can't figure it out. How do you solve $2^x-x=3$? Using logarithms? I could write $\log_2(x-3)=x$, but then what?

Thank you!

Answer 1

Here is a step-by-step Lambert W solution (see Robert's answer).
Recall the definition: $ae^a = b\Longleftrightarrow a = W(b)$.
So we try to get our equation into the form $ae^a = b$.
$$ 2^x-x=3 \\ 2^x=x+3 \\ 2^{x+3}=8(x+3) \\ e^{(x+3)\log 2}=8(x+3) \\ \frac{-\log 2}{8}=-(x+3)\log 2\; e^{-(x+3)\log 2} \\ W\left(\frac{-\log 2}{8}\right)=-(x+3)\log 2 \\ \frac{1}{\log 2}W\left(\frac{-\log 2}{8}\right)=-(x+3) \\ x = -3-\frac{1}{\log 2}W\left(\frac{-\log 2}{8}\right) $$

Answer 2

The solutions are $$ x = -3 -{\frac {{\rm W} \left(-\ln \left( 2 \right)/8 \right) }{\ln \left( 2 \right) }} $$ where $W$ is a branch of the Lambert W function. There are two real solutions, approximately $-2.862500372$ and $2.444907554$, corresponding to the principal and the "$-1$" branch respectively.