Let's solve each problem in order.

Problem 13

Solve the inequality: $\left( \frac{1}{5} \right)^x < \frac{1}{125}$

Solution:

1. Rewrite $\frac{1}{125}$ as $\left( \frac{1}{5} \right)^3$ since $125 = 5^3$. $\left( \frac{1}{5} \right)^x < \left( \frac{1}{5} \right)^3$
2. Since $\frac{1}{5}$ is a positive fraction less than 1, the inequality direction remains the same. Therefore, we get: $x < 3$

Answer: $x < 3$

Problem 14

Solve the equation: $(5^2)^x = (3^x)^2$

Solution:

1. Simplify each side: $(5^2)^x = 5^{2x} \quad \text{and} \quad (3^x)^2 = 3^{2x}$ So the equation becomes: $5^{2x} = 3^{2x}$
2. For this to hold, $5^{2x} = 3^{2x}$ implies that $2x = 0$ (since 5 and 3 are different bases). Therefore: $x = 0$

Answer: $x = 0$

Problem 15

Solve the equation: $2^{2x} - 2^{x+1} = 0$

Solution:

1. Rewrite $2^{2x}$ as $(2^x)^2$ and $2^{x+1}$ as $2 \cdot 2^x$: $(2^x)^2 - 2 \cdot 2^x = 0$
2. Let $y = 2^x$. The equation becomes: $y^2 - 2y = 0$
3. Factor out $y$: $y(y - 2) = 0$ So, $y = 0$ or $y = 2$.
4. Since $y = 2^x$ and $2^x$ cannot be 0 (as it’s an exponential function), we only have $y = 2$.
5. If $y = 2$, then: $2^x = 2 \Rightarrow x = 1$

Answer: $x = 1$

Would you like a more detailed explanation for any of these solutions?

Here are five questions to deepen understanding on these types of problems:

1. How do you determine when to change the inequality sign in exponential inequalities?
2. What strategies can help identify when to use substitution in equations involving powers?
3. How do we know if an exponential equation has no solutions or multiple solutions?
4. Why does factoring work as a method for solving quadratic-like exponential equations?
5. In exponential equations, when is it necessary to compare the bases directly?

Tip: Always double-check if the bases are the same or can be made the same to simplify exponential equations.