TITLE: How to solve exponential equations like $2^x+x=5$?
QUESTION [7 upvotes]: I tried the following:
Let $y=5-x$. Then, $2^{5-y}=y \implies y \cdot 2^y=32$
Taking the log of both sides yields
$$\log_2 y + y = 5$$
And that's where I'm stuck.

REPLY [2 votes]: Without using Lambert function (which I am in love with !), you can approximate the function building around $x=2$ (which would be the solution of $2^x+x=6$) the simplest $[1,1]$ Padé approximant of 
$$f(x)=2^x+x-5$$ This would give
$$2^x+x-5 \approx \frac{1+\frac{ 1+8 \log (2)+14 \log ^2(2)}{1+4 \log
   (2)}(x-2)}{1-\frac{2  \log ^2(2)}{1+4 \log (2)}(x-2)}$$ Then
$$x\approx \frac{1+28 \log ^2(2)+12 \log (2)}{1+14 \log ^2(2)+8 \log (2)}\approx 1.71574$$ Better would be the result using instead the $[1,n]$ approximant and  get an approximation still at the price of a linear equation. For example, the next one would give
$$x\approx \frac{3+292 \log ^3(2)+222 \log ^2(2)+48 \log (2)}{3+146 \log ^3(2)+132 \log
   ^2(2)+36 \log (2)}\approx 1.71560$$