Solve the inequality: $(2.5)^{(x+1)^2} cdot (0.4)^{4|x-1|} \ge (\frac{25}{4})^{6.5}$

Alright class, let's solve the inequality: $(2.5)^{(x+1)^2} cdot (0.4)^{4|x-1|} \ge (\frac{25}{4})^{6.5}$. First, we rewrite the numbers as fractions: $2.5 = \frac{5}{2}$ and $0.4 = \frac{2}{5}$. Also, $\frac{25}{4} = (\frac{5}{2})^2$. So, the inequality becomes: $(\frac{5}{2})^{(x+1)^2} cdot (\frac{2}{5})^{4|x-1|} \ge ((\frac{5}{2})^2)^{6.5}$ We can rewrite $(\frac{2}{5})$ as $(\frac{5}{2})^{-1}$, thus: $(\frac{5}{2})^{(x+1)^2} cdot ((\frac{5}{2})^{-1})^{4|x-1|} \ge (\frac{5}{2})^{2 cdot 6.5}$ Simplify the exponents: $(\frac{5}{2})^{(x+1)^2} cdot (\frac{5}{2})^{-4|x-1|} \ge (\frac{5}{2})^{13}$ Now, combine the terms on the left side using the properties of exponents: $(\frac{5}{2})^{(x+1)^2 - 4|x-1|} \ge (\frac{5}{2})^{13}$ Since the base is greater than 1, we can compare the exponents directly: $(x+1)^2 - 4|x-1| \ge 13$ $x^2 + 2x + 1 - 4|x-1| \ge 13$ $x^2 + 2x - 12 \ge 4|x-1|$ Now we consider two cases for the absolute value: Case 1: $x \ge 1$ In this case, $|x-1| = x-1$, so the inequality becomes: $x^2 + 2x - 12 \ge 4(x-1)$ $x^2 + 2x - 12 \ge 4x - 4$ $x^2 - 2x - 8 \ge 0$ $(x-4)(x+2) \ge 0$ The solutions are $x \le -2$ or $x \ge 4$. Since we have the condition $x \ge 1$, the solution for this case is $x \ge 4$. Case 2: $x < 1$ In this case, $|x-1| = -(x-1) = 1-x$, so the inequality becomes: $x^2 + 2x - 12 \ge 4(1-x)$ $x^2 + 2x - 12 \ge 4 - 4x$ $x^2 + 6x - 16 \ge 0$ $(x+8)(x-2) \ge 0$ The solutions are $x \le -8$ or $x \ge 2$. Since we have the condition $x < 1$, the solution for this case is $x \le -8$. Combining both cases, the solution is $x \le -8$ or $x \ge 4$. Final Answer: The final answer is $\boxed{x \le -8 \text{ or } x \ge 4}$