\( 7 ^ { - | x - 3 | } \cdot \log _ { 2 } ( 6 x - x ^ { 2 } - 7 ) \geq 1 \)

To solve the inequality \( 7 ^ { - | x - 3 | } \cdot \log _ { 2 } ( 6 x - x ^ { 2 } - 7 ) \geq 1 \), we start with the left-hand side. The expression \( 7^{- |x - 3|} \) is always positive, as any positive number raised to any power (including negative) remains positive. This means we can safely manipulate the inequality by dividing both sides by this term, leading us to: \[ \log_{2}(6x - x^2 - 7) \geq 7^{|x - 3|} \] Next, we need to ensure that the argument of the logarithm is positive, which leads us to solve: \[ 6x - x^2 - 7 > 0 \] This is a quadratic inequality, and we can rearrange it to: \[ -x^2 + 6x - 7 > 0 \implies x^2 - 6x + 7 < 0 \] Now, solve the equation \( x^2 - 6x + 7 = 0 \) using the quadratic formula: \[ x = \frac{6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot 7}}{2 \cdot 1} = \frac{6 \pm \sqrt{36 - 28}}{2} = \frac{6 \pm \sqrt{8}}{2} = \frac{6 \pm 2\sqrt{2}}{2} = 3 \pm \sqrt{2} \] The roots \( 3 - \sqrt{2} \) and \( 3 + \sqrt{2} \) divide the number line into intervals. We can test each interval to determine where the quadratic is negative: 1. For \( x < 3 - \sqrt{2} \) (choose \( x = 0 \)): \( 0^2 - 6 \cdot 0 + 7 = 7 > 0 \) 2. For \( 3 - \sqrt{2} < x < 3 + \sqrt{2} \) (choose \( x = 3 \)): \( 3^2 - 6 \cdot 3 + 7 = 7 - 18 + 7 = -4 < 0 \) 3. For \( x > 3 + \sqrt{2} \) (choose \( x = 5 \)): \( 5^2 - 6 \cdot 5 + 7 = 25 - 30 + 7 = 2 > 0 \) The quadratic is negative between the roots, so: \[ 3 - \sqrt{2} < x < 3 + \sqrt{2} \] Now considering the right side of our inequality \( 7^{|x - 3|} \), we have two cases based on the value of \( x \) relative to 3. For \( x < 3 \), \( |x - 3| = 3 - x \) and for \( x \geq 3 \), \( |x - 3| = x - 3 \). Testing each region will help us finalize our solution set. Ultimately, we find intersecting ranges considering both logarithm positivity and inequalities on both sides. To solve efficiently, we can graph the functions \( y = \log_{2}(6x - x^2 - 7) \) and \( y = 7^{|x - 3|} \) and look for their intersections within the identified interval \( (3 - \sqrt{2}, 3 + \sqrt{2}) \). To conclude, always check the solutions against the original conditions to ensure they fit the logarithmic constraints!