Question:

Step 1: Identify the general conditions for $a^b = 1$. For an equation of the form $a^b = 1$ to hold, we need one of the following cases: - Case 1: Base $a = 1$ (and exponent $b$ can be anything) - Case 2: Exponent $b = 0$ (and base $a \neq 0$) - Case 3: Base $a = -1$ and exponent $b$ is an even integer Step 2: Apply Case 1 — Base equals 1. We need $|x-3| = 1$. This gives us: $$x - 3 = 1 \quad \text{or} \quad x - 3 = -1$$ $$x = 4 \quad \text{or} \quad x = 2$$ Both values are valid since the base is well-defined at these points. Step 3: Apply Case 2 — Exponent equals 0 with non-zero base. We need the exponent to equal zero: $$3x^2 - 10x + 3 = 0$$ Factoring: $$(3x - 1)(x - 3) = 0$$ $$x = \frac{1}{3} \quad \text{or} \quad x = 3$$ However, we must verify that the base $|x-3| \neq 0$ at these values. At $x = 3$: $|3-3| = 0$, which violates the condition that the base must be non-zero. So $x = 3$ is **not valid**. At $x = \frac{1}{3}$: $\left|\frac{1}{3} - 3\right| = \left|-\frac{8}{3}\right| = \frac{8}{3} \neq 0$ ✓ So $x = \frac{1}{3}$ is **valid**. Step 4: Apply Case 3 — Base equals $-1$ with even exponent. Since the base is $|x-3|$, which is always non-negative (absolute value), it can never equal $-1$. This case does **not apply**. Step 5: Compile all valid solutions. The valid solutions are: $$x = \frac{1}{3}, \quad x = 2, \quad x = 4$$ This gives us exactly **three real solutions**. The answer is **Option 2: exactly three**.