Solve the following equation in the set of real numbers:

$$
8^{x}+27^{x}+2 \cdot 30^{x}+54^{x}+60^{x}=12^{x}+18^{x}+20^{x}+24^{x}+45^{x}+90^{x}
$$
Solution. For better clarity, let $2^{x}=a, 3^{x}=b$ and $5^{x}=c$. With these notations, the equation to be solved is:

$$
a^{3}+b^{3}+2 a b c+a b^{3}+a^{2} b c=a^{2} b+a b^{2}+a^{2} c+a^{3} b+b^{2} c+a b^{2} c
$$

We rearrange the right side to zero and group the terms on the left side to factorize it:

$$
\begin{aligned}
a^{3}-a^{2} b+b^{3}-a b^{2}-a^{2} c+2 a b c-b^{2} c+a b^{3}-a^{3} b+a^{2} b c-a b^{2} c & =0 \\
a^{2}(a-b)-b^{2}(a-b)-c(a-b)^{2}-a b(a-b)(a+b)+a b c(a-b) & =0
\end{aligned}
$$

Now we can factor out $(a-b)$ and continue grouping the terms:

$$
\begin{aligned}
(a-b)\left(a^{2}-b^{2}-c(a-b)-a b(a+b)+a b c\right) & =0 \\
(a-b)[(a-b)(a+b-c)-a b(a+b-c)] & =0 \\
(a-b)(a+b-c)(a-b-a b) & =0
\end{aligned}
$$

Returning to the original meaning of the variables and using the fact that a product is zero if and only if one of its factors is zero, we need to examine the following cases:
i)

$$
2^{x}-3^{x}=0 \Longleftrightarrow\left(\frac{3}{2}\right)^{x}=1 \quad\left(\text { since } 2^{x} \neq 0\right)
$$

We know that the function $x \mapsto\left(\frac{3}{2}\right)^{x}$ is strictly increasing, so it takes the value 1 exactly once, at $x=0$.

ii)

$$
2^{x}+3^{x}-5^{x}=0 \Longleftrightarrow\left(\frac{2}{5}\right)^{x}+\left(\frac{3}{5}\right)^{x}=1 \quad\left(\text { since } 5^{x} \neq 0\right)
$$

The functions $x \mapsto\left(\frac{2}{5}\right)^{x}$ and $x \mapsto\left(\frac{3}{5}\right)^{x}$ are strictly decreasing. Therefore, their sum, the function $x \mapsto\left(\frac{2}{5}\right)^{x}+\left(\frac{3}{5}\right)^{x}$, is also strictly decreasing, so it can take the value 1 exactly once, which happens at $x=1$. This is the second solution to the equation.

iii)

$$
2^{x}-3^{x}-2^{x} \cdot 3^{x}=0 \Longleftrightarrow\left(\frac{2}{3}\right)^{x}-1=2^{x} \quad\left(\text { since } 2^{x} \neq 0\right)
$$

We know that the function $x \mapsto\left(\frac{2}{3}\right)^{x}-1$ is strictly decreasing, and the function $x \mapsto 2^{x}$ is strictly increasing, so this equation can have only one solution, and $x=-1$ is indeed a solution.

The equation has three solutions: $x=0, x=1$, and $x=-1$.

Remark. Most people managed to get to the point where they realized that one of the following must hold: $2^{x}=3^{x}, 2^{x}+3^{x}=5^{x}$, or $2^{x}=3^{x}+6^{x}$. However, many only stated the solution to these equations without justifying why there could not be other solutions - this resulted in a point deduction. It also resulted in a point deduction if someone graphed the two sides as functions of $x$ and read off from the graph that there could only be one solution - this can be a good guess but not a proof.