Solve the following equation:

$$
\frac{1}{4^{3 x}-13 \cdot 4^{2 x}+51 \cdot 4^{x}-60}+\frac{1}{4^{2 x}-7 \cdot 4^{x}+12}=0
$$
Let's substitute $4^{x}=y$, so the denominators are polynomials in $y$ and are defined for all $y>0$. The zeros of the second denominator, $y^{2}-7 y+12$, are $y_{1}=3$ and $y_{2}=4$, which are within the domain of definition, and the denominator can be factored as $(y-3)(y-4)$.

To bring to a common denominator, we examine whether $N=y^{3}-13 y^{2}+51 y-60$, the first denominator, is divisible by $y-3$ and $y-4$. This is true if and only if the value of $N$ is 0 when $y=3$ and $y=4$. For $y=3$, $N=3$, so $N$ is not divisible by $y-3$. However, for $y=4$, $N=0$, so $N$ can be factored by $y-4$. Indeed,

$$
N=\left(y^{3}-4 y^{2}\right)-\left(9 y^{2}-36 y\right)+(15 y-60)=(y-4)\left(y^{2}-9 y+15\right)
$$

Thus, the common denominator of the left side of (1) after substitution is

$$
(y-3)(y-4)\left(y^{2}-9 y+15\right)
$$

(1) is undefined if any denominator is 0, that is, in the following cases:

$$
y=4^{x}=3, \quad 4, \quad 4.5 \pm \sqrt{5.25}
$$

The last two values are also positive and belong to the domain of definition of the denominators, so we exclude all four values from further examination. According to this, (1) becomes:

$$
\begin{gathered}
\frac{1}{y^{3}-13 y^{2}+51 y-60}+\frac{1}{y^{2}-7 y+12}=\frac{1}{(y-4)\left(y^{2}-9 y+15\right)}+ \\
+\frac{1}{(y-3)(y-4)}=\frac{(y-3)+\left(y^{2}-9 y+15\right)}{(y-3)(y-4)\left(y^{2}-9 y+15\right)}= \\
=\frac{y^{2}-8 y+12}{(y-3)(y-4)\left(y^{2}-9 y+15\right)}=0
\end{gathered}
$$

The numerator is 0 at $y_{1}=2$ and $y_{2}=6$, and (2) is defined at these points, so these are the roots of the transformed equation. The roots of the original equation are then

$$
\begin{array}{ll}
4^{x}=2 \text { gives } \quad x_{1}=\log_{4} 2=0.5 \\
4^{x}=6 \text { gives } \quad x_{2}=\log_{4} 6=\frac{\log 6}{\log 4} \approx \frac{0.7782}{0.6021} \approx 1.292 .
\end{array}
$$

Géza Etényi (Aszód, Petőfi S. g. III. o. t.)

Note. Many solutions took the product of the two denominators as the common denominator, so after simplification, the numerator became $(y-4)\left(y^{2}-8 y+12\right)$. However, they did not exclude the root $y=4$.