7^-|x-3|*log2(6x-x^2-7)>=1

To solve the inequality \( 7^{-|x-3|} \cdot \log_2(6x - x^2 - 7) \geq 1 \), we will break it down step by step. ### Step 1: Analyze the logarithm First, we need to find the domain for the logarithm \( \log_2(6x - x^2 - 7) \). The expression inside the logarithm must be positive: \[ 6x - x^2 - 7 > 0 \] Rearranging gives: \[ -x^2 + 6x - 7 > 0 \] We can factor this quadratic: \[ -(x^2 - 6x + 7) > 0 \implies x^2 - 6x + 7 < 0 \] We can find the roots of the equation \( x^2 - 6x + 7 = 0 \) using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{6 \pm \sqrt{36 - 28}}{2} = \frac{6 \pm \sqrt{8}}{2} = \frac{6 \pm 2\sqrt{2}}{2} = 3 \pm \sqrt{2} \] The roots are \( 3 + \sqrt{2} \) and \( 3 - \sqrt{2} \). The quadratic opens upwards (as the coefficient of \( x^2 \) is positive), so it is negative between its roots: \[ 3 - \sqrt{2} < x < 3 + \sqrt{2} \] ### Step 2: Rewrite the inequality Now we can rewrite the original inequality based on the domain found. The inequality we need to solve is: \[ 7^{-|x-3|} \cdot \log_2(6x - x^2 - 7) \geq 1 \] This can be rewritten as: \[ \log_2(6x - x^2 - 7) \geq \frac{1}{7^{-|x-3|}} \] ### Step 3: Define the critical points Next, we find the critical value for which \( \log_2(6x - x^2 - 7) \) equals \( 1 \): \[ 6x - x^2 - 7 = 2^1 \implies 6x - x^2 - 7 = 2 \] Rearranging: \[ -x^2 + 6x - 9 = 0 \] Using the quadratic formula: \[ x = \frac{-6 \pm \sqrt{36 - 36}}{-1} = \frac{6 \pm 0}{1} = 6 \] ### Step 4: Evaluate \( x = 6 \) Evaluate \( f(x) = 6x - x^2 - 7 \) at \( x = 6 \): \[ f(6) = 6(6) - 6^2 - 7 = 36 - 36 - 7 = -7 \quad \text{(not in the domain)} \] Now we can test the intervals created by our critical points \( 3 - \sqrt{2} \) and \( 3 + \sqrt{2} \). ### Step 5: Test intervals 1. **Interval \( (3 - \sqrt{2}, 3 + \sqrt{2}) \):** Choose a point, e.g. \( x = 3 \): \[ \log_2(6(3) - 3^2 - 7) = \log_2(18 - 9 - 7) = \log_2(2) = 1 \] So at \( x = 3 \), \( 7^{-|x-3|} \cdot 1 \geq 1 \) holds. Testing values in the range of the logarithm will require evaluating \( 7^{-|x-3|} \), which will vary by how far \( x \) is from 3. ### Conclusion The inequality thus holds in: \[ (3 - \sqrt{2}, 3 + \sqrt{2}) \] This is the solution to the given inequality. Always check at the boundaries for inclusion or exclusion.