Solving $x + 3^x < 4$ analytically 
I am solving the problems in Michael Spivak's Calculus book. In the Prologue chapter, there is the following problem:
$$x + 3^x < 4$$
I can solve this using graphical means, but not analytically. Is it possible to do the latter?

I can arrive at the solution as follows:
$$3^x < 4 - x$$
$$x < \log_3 (4-x)$$
I then draw the graphs of the functions $\log_3 (4-x)$ and $x$, and I can determine that $x$ needs to be smaller than 1 to be smaller than the $\log_3 (4-x)$.


How do I solve this purely analytically?


$$x+3^x-4=0$$
1.) Solution as series
To solve the equation analytically as a series, you can use Lagrange inversion.
2.) No solution with elementary inverses
The equation is a zeroing equation of an elementary function.
Powers with irrational exponents are transcendental functions. Therefore the equation isn't an algebraic equation, it's a transcendental equation:
$$x+e^{\ln(3)x}-4=0$$
Because $x+e^{\ln(3)x}-4$ is a polynomial of two algebraically independent monomials, the function $\mathbb{R}\to\mathbb{R}, x\mapsto x+e^{\ln(3)x}-4$ seems to have no elementary inverse. Therefore we possibly cannot solve the equation by rearranging it only by applying elementary functions (elementary operations) which we can read from the equation.
3.) Solution in closed form with Lambert W
The equation is solvable by applying Lambert W. By some simple rearrangings, we can bring the equation into a form that is solvable by Lambert W.
The only real solution is:
$$x=\frac{4\ln(3)-W(81\ln(3))}{\ln(3)}.$$
That's equal to $1$, but I don't know the methods to show this.
