Solve $2^x + 3^{2x + 2} = 5$ Out of curiosity, how would you approach the following exponential equation:
$$2^x + 3^{2x + 2} = 5$$
This resulted from a work on a system of logarithmic equations and I know for a fact that it has a single solution, $x \approx -0.345464$, which I obtained with the help of Wolfram Alpha. Anyhow, I simply can't figure out a way to solve for $x$ in an enclosed form.
My main question would be: is the solution generally expressable as a real number in a closed form? Or is this equation a dead end and maybe I should reconsider my approach to the system of equations?
That also brought me to the following thought. The equation can be rewritten as:
$$2^x + 9\cdot 9^x - 5 = 0$$
which is a general composition of exponential functions. I couldn't find any general approaches to such general exponential equations, but my intuition is that there's not much 'general' stuff to do with them and all depends on the current case.
I hope you could understand me, my terminology is perhaps not very precise and I'm a bit self learned in mathematics, so any clarification is welcome! Thanks in advance!
 A: I don't think one can solve this "nicely". However, the following graph may help! Even if you can't solve it by hand, you can always get an idea of what range of values should the solution fall within.
If you really want to solve it numerically, the Newton-Raphson method is just one of the many methods you could try out to obtain the root of $f(x) = 2^x + 3^{2x+2} - 5$. Note that $f'(x)>0$ so there's only one solution.

A: As you expect, you cannot get a closed form solution.
Consider that ou are looking for the zero of function
$$f(x)=2^x + 3^{2x + 2}-5$$ By inspection, you know that the solution is somewhere between $-1$ and $0$.
Try Taylor expansion
$$f(x)=5+x (\log (2)+18 \log (3))+x^2 \left(\frac{\log ^2(2)}{2}+18 \log ^2(3)\right)+O\left(x^3\right)$$
Using the first term only, you have, as an estimate
$$x=-\frac{5}{\log (2)+18 \log (3)}\approx -0.244282$$ Now, use more terms and use series reversion to get
$$x=t-\frac{ \left(\log ^2(2)+36 \log ^2(3)\right)}{2 (\log (2)+18 \log (3))}t^2 +O\left(t^3\right)\quad \text{with} \quad t=\frac{f(x)-5}{\log (2)+18 \log (3)}$$ Making $f(x)=0$ as we desire, then another estimate
$$x=-\frac{5 \left(7 \log ^2(2)+828 \log ^2(3)+72 \log (2) \log (3)\right)}{2 (\log   (2)+18 \log (3))^3}\approx -0.30832$$
Now, let us polish the root using the so simple Newton method and, since I am lazy, I shall start with the first estimate I wrote. The successive iterates are
$$\left(
\begin{array}{cc}
 0 & -0.2442817505 \\
 1 & -0.3353427850 \\
 2 & -0.3453567313 \\
 3 & -0.3454642504 \\
 4 & -0.3454642627
\end{array}
\right)$$ and we could get as many decimal figures as required. For the fun of it, the solution
$$-0.345464262655186305826853008027166943006071510\cdots$$
