Solve the equation $x^{2}-x-1=2^{x}-\log _{2}\left(x^{2}+2^{x}\right)$ for positive numbers.
Solution. Let the new unknown be $y=\log _{2}\left(x^{2}+2^{x}\right)$, then $x^{2}+2^{x}=2^{y}$; thus, our equation becomes $2^{y}-2^{x}-x-1=2^{x}-y$, and after rearrangement, it takes the form

$$
2^{y}+y=2^{x+1}+(x+1)
$$

The functions $t \rightarrow t$ and $t \rightarrow 2^{t}$ are strictly increasing, so their sum is also strictly increasing. The left and right sides of the equation are the values of this strictly increasing sum function at $y$ and $x+1$, respectively; these are equal precisely when $y=x+1$, that is,

$$
\log _{2}\left(x^{2}+2^{x}\right)=x+1
$$

Raising 2 to the power of both sides and rearranging:

$$
x^{2}+2^{x}=2^{x+1}, \quad x^{2}=2^{x}
$$

Thus, for $x>0$, $x=2^{\frac{x}{2}}$. Here, the graph of the function $x \rightarrow 2^{\frac{x}{2}}$ is strictly convex, while the graph of the function $x \rightarrow x$ is a straight line. Therefore, the two curves can have at most two common points, so the equation can have at most two solutions. However, we can easily find two solutions: $x_{1}=2$ and $x_{2}=4$ are clearly solutions, and according to this, there are no other solutions.