Solve, for $x,y \in \mathbb{R}$: $$\begin{cases} 2^x + 3^y = 72\\ 2^y + 3^x = 108 \end{cases}$$
Attempt: I could write the following:
From the second equation $$y=\frac{\ln (108-{{3}^{x}})}{\ln 2}=\frac{\ln 3}{\ln 2}\ln (108-{{3}^{x}})$$ and replacing in the first equation we get: $${{2}^{x}}+{{(108-{{3}^{x}})}^{\frac{\ln 3}{\ln 2}}}=72$$
The LHS is monotonically decreasing, so we have only one solution, and indeed $x=4,150...$ is a solution (using WA).
How could I solve it without using a calculator?
I cheated since I used my phone.
From the first equation $$2^x+3^y=72 \quad \implies \quad y=\frac{\log \left(72-2^x\right)}{\log (3)}$$ Replace in the second equation and define the recursion $$x=\frac 1 {\log(3)}\,\log \left(108-2^{\frac{\log \left(72-2^{x}\right)}{\log(3)}}\right) $$
$$x_0=0 \implies x_1=4.12845\implies x_2=4.15026\implies x_3=4.15062$$
But, if we do it using the other way $$2^y+3^x=108 \quad \implies \quad y=\frac{\log \left(108-3^x\right)}{\log (2)}$$ $$x=\frac 1 {\log(2)}\,\log \left(72-3^{\frac{\log \left(108-3^{x}\right)}{\log (2)}}\right)$$ $$x_0=0 \implies x_1=10.6205 + 4.53236\,i \qquad\color{red}{ \text{bad news}}$$