Question

Let $S=\left\{\alpha: \log _2\left(9^{2 \alpha-4}+13\right)-\log _2\left(\frac{5}{2} \cdot 3^{2 \alpha-4}+1\right)=2\right\}$ Then the maximum value of $\beta$ for which the equation $x^2-2\left(\displaystyle\sum_{\alpha \in s} \alpha\right)^2 x+\displaystyle\sum_{\alpha \in s}(\alpha+1)^2 \beta=0$ has real roots, is _____

Answer 1

Correct Answer: 25

The correct answer is 25.
${\log }_2\left(9^{2\alpha -4}+13\right)-{\log }_2\left(\frac{5}{2}\cdot 3^{2\alpha -4}+1\right)=2$
$\Rightarrow \frac{9^{2\alpha -4}+13}{\frac{5}{2}{3}^{2\alpha -4}+1}=4$
$\Rightarrow \alpha =2\text{or}$
$\sum _{\alpha \in S}\alpha =5\text{and}\sum _{\alpha \in S}(\alpha +1{)}^2=25$
$\Rightarrow x^2-50x+25\beta =0\text{has real roots}$
$\Rightarrow \beta \le 25$
$\Rightarrow {\beta }_{\max }=25$

Answer 2

1. Simplify the logarithmic equation: Start with: \[ \log_2 \left(9^{2\alpha-4} + 13\right) - \log_2 \left(\frac{5}{2} \cdot 3^{2\alpha-4} + 1\right) = 2. \] Using the logarithmic property \( \log_a(x) - \log_a(y) = \log_a\left(\frac{x}{y}\right) \), rewrite: \[ \log_2 \left(\frac{9^{2\alpha-4} + 13}{\frac{5}{2} \cdot 3^{2\alpha-4} + 1}\right) = 2. \] Exponentiating both sides: \[ \frac{9^{2\alpha-4} + 13}{\frac{5}{2} \cdot 3^{2\alpha-4} + 1} = 4. \] 2. Solve for \( \alpha \): Multiply through by the denominator: \[ 9^{2\alpha-4} + 13 = 4\left(\frac{5}{2} \cdot 3^{2\alpha-4} + 1\right). \] Simplify: \[ 9^{2\alpha-4} + 13 = 10 \cdot 3^{2\alpha-4} + 4. \] Using \( 9^{k} = (3^k)^2 \), substitute \( k = 2\alpha-4 \): \[ (3^k)^2 - 10 \cdot 3^k + 9 = 0. \] Let \( y = 3^k \), so: \[ y^2 - 10y + 9 = 0. \] 3. Solve the quadratic equation: Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ y = \frac{10 \pm \sqrt{100 - 36}}{2} = \frac{10 \pm 8}{2}. \] \[ y = 9 \quad \text{or} \quad y = 1. \] 4. Back-substitute to find \( \alpha \): Recall \( y = 3^k \) and \( k = 2\alpha-4 \). For \( y = 9 \), \( 3^k = 9 \), so \( k = 2 \). \[ 2\alpha - 4 = 2 \implies \alpha = 3. \] For \( y = 1 \), \( 3^k = 1 \), so \( k = 0 \). \[ 2\alpha - 4 = 0 \implies \alpha = 2. \] Thus, \( S = \{2, 3\} \). 5. Substitute \( S \) into the equation: Compute: \[ \sum_{\alpha \in S} \alpha = 2 + 3 = 5, \quad \sum_{\alpha \in S} (\alpha + 1)^2 = (2+1)^2 + (3+1)^2 = 9 + 16 = 25. \] The equation becomes: \[ x^2 - 2(5)x + 25\beta = 0. \] 6. Find the maximum \( \beta \): For real roots, the discriminant must be non-negative: \[ \Delta = b^2 - 4ac = (-10)^2 - 4(1)(25\beta) \geq 0. \] \[ 100 - 100\beta \geq 0 \implies \beta \leq 1. \] The maximum value of \( \beta \) is \( 25 \).