B1. Find all real numbers $x$ that satisfy the equation

$$
6^{\log _{10} x}+x^{\log _{10} 6}=72
$$
B1. The logarithm is defined only for positive numbers, so $x>0$. The number $x=1$ is obviously not a solution to the equation, so we can assume that $x \neq 1$. For a positive real number $x \neq 1$, it holds that $\log _{10} x=\frac{1}{\log _{x} 10}$, so $\log _{10} 6=\frac{\log _{x} 6}{\log _{x} 10}=\log _{10} x \cdot \log _{x} 6$. Using this relationship, the given equation can be transformed to

$$
6^{\log _{10} x}+\left(x^{\log _{x} 6}\right)^{\log _{10} x}=72
$$

and then considering the equality $x^{\log _{x} 6}=6$, we get

$$
2 \cdot 6^{\log _{10} x}=72
$$

From this, it follows that $6^{\log _{10} x}=36=6^{2}$, so $\log _{10} x=2$ or $x=100$.

1. method:
Using the equation - change of base of the logarithm ..... 1 point
Transforming to the equation $2 \cdot x^{\log _{10} 6}=72$ ..... 5 points
Simplified result $x=100$ ..... 1 point
2. method:
Verified solution $x=100$ ..... 1 point
Full justification that the solution is unique, as the left side of the equation is a strictly monotonic function
6 points