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
- 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.
