\( ( \frac { 1 } { 4 } ) ^ { 3 - 2 x } < 8 - x \)

To solve the inequality \[ \left( \frac{1}{4} \right)^{3 - 2x} < 8 - x, \] we start by rewriting \(\frac{1}{4}\) as \(4^{-1}\): \[ \left( \frac{1}{4} \right)^{3 - 2x} = (4^{-1})^{3 - 2x} = 4^{-(3 - 2x)} = 4^{2x - 3}. \] Thus, the inequality becomes: \[ 4^{2x - 3} < 8 - x. \] Next, we express \(8\) as a power of \(4\): \[ 8 = 2^3 = (2^2)^{3/2} = 4^{3/2}. \] Now we can rewrite the inequality: \[ 4^{2x - 3} < 4^{3/2} - x. \] To solve this inequality, we need to analyze the behavior of both sides. We can take the logarithm base \(4\) of both sides, but first, we need to ensure that \(8 - x > 0\) (i.e., \(x < 8\)) for the inequality to be valid. Assuming \(x < 8\), we can rewrite the inequality as: \[ 2x - 3 < \frac{3}{2} - \log_4(8 - x). \] Now, we can solve for \(x\) by isolating \(x\): 1. Rearranging gives: \[ 2x < \frac{3}{2} + 3 - \log_4(8 - x), \] which simplifies to: \[ 2x < \frac{9}{2} - \log_4(8 - x). \] 2. Dividing by \(2\): \[ x < \frac{9}{4} - \frac{1}{2} \log_4(8 - x). \] This inequality is complex to solve directly, so we can analyze it graphically or numerically. To find the critical points, we can check specific values of \(x\): - For \(x = 0\): \[ 4^{2(0) - 3} = 4^{-3} = \frac{1}{64} < 8, \] which is true. - For \(x = 1\): \[ 4^{2(1) - 3} = 4^{-1} = \frac{1}{4} < 7, \] which is true. - For \(x = 2\): \[ 4^{2(2) - 3} = 4^{1} = 4 < 6, \] which is true. - For \(x = 3\): \[ 4^{2(3) - 3} = 4^{3} = 64 < 5, \] which is false. Thus, we can see that the inequality holds for \(x\) in the interval \((-\infty, 3)\). Finally, we need to check the endpoint \(x = 3\): At \(x = 3\): \[ 4^{2(3) - 3} = 4^{3} = 64 \quad \text{and} \quad 8 - 3 = 5, \] which does not satisfy the inequality. Thus, the solution to the inequality is: \[ \boxed{(-\infty, 3)}. \]