Convert the logarithmic equation into its exponential equivalent: $x^2 - 3x = 2^2$ $x^2 - 3x = 4$ $x^2 - 3x - 4 = 0$.

Factorizing the quadratic equation: $(x - 4)(x + 1) = 0 \implies x = 4 \text{ or } x = -1$.

Now, verify both solutions in the original domain of the logarithm, which requires $x^2 - 3x > 0$:

• For $x = 4$: $4^2 - 3(4) = 16 - 12 = 4 > 0$ (Valid).
• For $x = -1$: $(-1)^2 - 3(-1) = 1 + 3 = 4 > 0$ (Valid).

Since both yield positive arguments, there are exactly $2$ valid real solutions.