Solving $2^x + 3^x = 6^x$ How would I solve $2^x + 3^x = 6^x$?
So far, I've been able to simplify the equation into
$$(2^x - 1)(3^x - 1) = 1,$$
but I'm not quite sure where to go from here. Additionally, letting $y = 2^x$ yielded
$$(y-1)(y^{\log_23} - 1) = 1,$$
however as with before, I'm pretty much stuck at this step.
WolframAlpha says that $x \approx 0.787885$, but I'd like a closed form for $x$ if possible.
 A: You are looking for the zero of function $$f(x)=6^x-(2^x + 3^x)$$ As Quanto suggested, it is better to consider
$$g(x)=x \log(6)-\log(2^x + 3^x)$$ which is almost a straight line.
Using a series expansion of $g(x)$ around $x=1$ and then series reversion, we can get
$$x=1+t+\frac{3  \log ^2\left(\frac{3}{2}\right)}{5 \log (72)}t^2+O\left(t^3\right)\qquad \text{where} \qquad t=\frac{5 \left(g(x)-\log \left(\frac{6}{5}\right)\right)}{\log (72)}$$
Making $g(x)=0$ we end with
$$x=1+\frac{5 \log \left(\frac{6}{5}\right) \left(3 \log \left(\frac{6}{5}\right) \log
   ^2\left(\frac{3}{2}\right)-\log ^2(72)\right)}{\log ^3(72)}\approx 0.7878894$$ while the "exact" solution, obtained using Newton method, would be         $0.7878849$. Adding the next term in the expansion would lead to $0.7878852$.
Inverse symbolic calculators do not find any expression but (just for the fun of it), it is very close (relative error equal to $1.30 \times 10^{-18}$%) to 
$$-\exp\left(-\frac{49}{26}-\frac{3}{26 e}+\frac{3 e}{2}-\frac{11}{13 \pi }-\frac{\pi
   }{26}\right)\frac{ \sec (e \pi )}{\pi ^{\frac{11}{13}+\frac{e}{2}}\sin ^{\frac{7}{26}}(e \pi
   )}$$ This has been obtained by a friend of mine who enjoys this kind of jokes !
$$e^{}$$
A: Here is a good approximate close-form for the root. 
Let $f(x) = \frac{\ln(2^x+3^x)}{\ln 6}-x$ and use the first-order approximation below, 
$$0=f(1)+f'(1)(x-1)$$
which yields
$$x = \frac{5\ln5-2\ln2-3\ln3}{\ln72}=0.7868$$
vs. the exact result 0.7879.
