[Algebra]
Problem: Determine all real values of $x$ for which

$$
3^{x+2}+2^{x+2}+2^{x}=2^{x+5}+3^{x}
$$

Solution: Using exponent rules and arithmetic, we manipulate the given equation:

$$
\begin{aligned}
3^{x+2}+2^{x+2}+2^{x} & =2^{x+5}+3^{x} \\
3^{x} 3^{2}+2^{x} 2^{2}+2^{x} & =2^{x} 2^{5}+3^{x} \\
9\left(3^{x}\right)+4\left(2^{x}\right)+2^{x} & =32\left(2^{x}\right)+3^{x} \\
8\left(3^{x}\right) & =27\left(2^{x}\right) \\
\frac{3^{x}}{2^{x}} & =\frac{27}{8} \\
\left(\frac{3}{2}\right)^{x} & =\left(\frac{3}{2}\right)^{3}
\end{aligned}
$$

Since the two expressions are equal and the bases are equal, then the exponents must be equal, so $x=3$.

Final answer: 3