Solve the following equation:

$$
2^{x+1}+2^{x+2}+2^{x+3}=3^{x+1}-3^{x+2}+3^{x+3}
$$
By factoring out $2^{x+1}$ from the left side and $3^{x+1}$ from the right side, we get

$$
2^{x+1}(1+2+4)=3^{x+1}(1-3+9)
$$

which simplifies to

$$
2^{x+1}=3^{x+1}
$$

This can also be written as

$$
2 \cdot 2^{x}=3 \cdot 3^{x}
$$

or

$$
\left(\frac{2}{3}\right)^{x}=\frac{3}{2}=\left(\frac{2}{3}\right)^{-1}
$$

From (1), we can conclude for $x$: If two equal powers with different positive bases are equal, then the exponent must necessarily be 0, that is,

$$
x+1=0,
$$

from which

$$
x=-1
$$

For the equation under (2), we can refer to the well-known fact that the exponential function - if the base is a positive number less than (greater than) one - takes on all positive values exactly once, monotonically decreasing (increasing). Therefore, the exponents must necessarily be equal, that is,

$$
x=-1
$$

Jónás György (Gyöngyös, Közg. techn. II. o. t.)