(0,3)^x^2-8x+5>100/9

To solve the inequality \( (0.3)^{x^2 - 8x + 5} > \frac{100}{9} \), first, we can rewrite it in a more manageable form. Since \( 0.3 < 1 \), the inequality flips when we take the logarithm. First, let's take the logarithm of both sides: \[ x^2 - 8x + 5 < \log_{0.3}\left(\frac{100}{9}\right) \] Now, we need to calculate \( \log_{0.3}\left(\frac{100}{9}\right) \): \[ \frac{100}{9} \approx 11.11 \quad \text{(you can use a calculator for more precision)} \] Calculate \( \log_{10}(11.11) \) and \( \log_{10}(0.3) \): 1. \( \log_{10}(0.3) \approx -0.5228787 \) 2. \( \log_{10}(11.11) \approx 1.045757 \) Now, applying the change of base formula for logarithms, we can write: \[ \log_{0.3}(11.11) = \frac{\log_{10}(11.11)}{\log_{10}(0.3)} \approx \frac{1.045757}{-0.5228787} \approx -2.000 \] So the inequality becomes: \[ x^2 - 8x + 5 < -2 \] Simplify this inequality: \[ x^2 - 8x + 7 < 0 \] Next, we'll factor the quadratic equation: \[ (x - 7)(x - 1) < 0 \] Now, we need to find the critical points and test the intervals. The critical points are \( x = 1 \) and \( x = 7 \). We test the intervals \( (-\infty, 1) \), \( (1, 7) \), and \( (7, \infty) \): 1. For \( x < 1 \) (e.g., \( x = 0 \)): \((0 - 7)(0 - 1) = 7 > 0\) 2. For \( 1 < x < 7 \) (e.g., \( x = 5 \)): \((5 - 7)(5 - 1) = (-2)(4) = -8 < 0\) 3. For \( x > 7 \) (e.g., \( x = 8 \)): \((8 - 7)(8 - 1) = (1)(7) = 7 > 0\) The solution to the inequality \( (x - 7)(x - 1) < 0 \) is the interval where the product is negative: \[ \boxed{(1, 7)} \] So, the solution to the inequality \( (0.3)^{x^2 - 8x + 5} > \frac{100}{9} \) is \( x \in (1, 7) \).