Question:

Step 1: Simplify the exponential expression using the logarithmic identity. We use the fundamental property that $a^{\log_a(t)} = t$ for any valid base $a$ and argument $t$. $$9^{\log_9(x^2-4x+5)} = x^2-4x+5$$ Step 2: Set up and solve the resulting equation. Since the left side simplifies to $x^2-4x+5$, we equate it to the right side of the original equation: $$x^2-4x+5 = x-1$$ Rearranging: $$x^2-5x+6 = 0$$ Factoring: $$(x-2)(x-3) = 0$$ This gives us $x = 2$ or $x = 3$. Step 3: Verify the domain constraints. For the original equation to be valid, we need: - The argument of the logarithm must be positive: $x^2-4x+5 > 0$ - The right side must be positive: $x-1 > 0$, which means $x > 1$ For the first condition, we check the discriminant of $x^2-4x+5$: $$\Delta = (-4)^2 - 4(1)(5) = 16 - 20 = -4 < 0$$ Since the discriminant is negative and the leading coefficient is positive, $x^2-4x+5 > 0$ for all real $x$. Both $x = 2$ and $x = 3$ satisfy $x > 1$, so both are valid solutions. Step 4: Identify which root is $x_1$ and which is $x_2$. Since $x_1 > x_2$: $$x_1 = 3 \quad \text{and} \quad x_2 = 2$$ Step 5: Evaluate the trigonometric expression. We need to find $\tan(x_1\pi) + \sec(x_2\pi)$. Substituting the values: $$\tan(3\pi) + \sec(2\pi)$$ Evaluating each term: - $\tan(3\pi) = \tan(\pi) = 0$ (since tangent has period $\pi$) - $\sec(2\pi) = \frac{1}{\cos(2\pi)} = \frac{1}{1} = 1$ Therefore: $$\tan(3\pi) + \sec(2\pi) = 0 + 1 = 1$$ **Final Answer:** The value of $\tan(x_1\pi) + \sec(x_2\pi) = 1$ The correct option is **Option 3: 1** <div class="key-concept"><strong>Key Concept:</strong> Logarithmic equations and trigonometric evaluation at integer multiples of π</div> <div class="trap-box"><strong>Trap:</strong> Using the identity $a^{log_a(f(x))} = f(x)$ to simplify, then checking domain conditions.</div>