We are tasked with solving the inequality:

$\left( \frac{1}{2} \right)^{\log_4 (x^2)} > \frac{1}{4}$

Step 1: Express $\frac{1}{4}$ as a power of $\frac{1}{2}$

Since $\frac{1}{4} = \left( \frac{1}{2} \right)^2$, the inequality becomes:

$\left( \frac{1}{2} \right)^{\log_4 (x^2)} > \left( \frac{1}{2} \right)^2$

Step 2: Take the logarithm base $\frac{1}{2}$ on both sides

We now take the logarithm base $\frac{1}{2}$ of both sides. The logarithm of a power of $\frac{1}{2}$ simplifies the left side:

$\log_{\frac{1}{2}} \left( \left( \frac{1}{2} \right)^{\log_4 (x^2)} \right) > \log_{\frac{1}{2}} \left( \left( \frac{1}{2} \right)^2 \right)$

This gives us:

$\log_4 (x^2) < 2$

Step 3: Solve the inequality $\log_4 (x^2) < 2$

To solve this inequality, we can first convert the logarithmic expression to an exponential form. The inequality $\log_4 (x^2) < 2$ is equivalent to:

$x^2 < 4^2$

$x^2 < 16$

Step 4: Solve for $x$

Now solve for $x$:

$-4 < x < 4$

Step 5: Consider the domain restrictions

We must ensure that $x^2$ is defined and positive for real $x$. Since $x^2$ is valid for all $x$, we do not have additional domain restrictions, but we do exclude $x = 0$ because $\log_4 (x^2)$ would be undefined when $x = 0$.

Final Solution:

Thus, the solution is:

$-4 < x < 4 \quad \text{and} \quad x \neq 0$

So, the solution set for $x$ is:

$x \in (-4, 0) \cup (0, 4)$

Would you like further clarification or have any other questions?

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Here are 5 related questions to help expand the understanding:

1. What is the general method for solving inequalities involving logarithms?
2. How can we express a logarithmic inequality in an exponential form?
3. What is the effect of a negative base in logarithmic functions?
4. How do you solve quadratic inequalities?
5. How does the domain of logarithmic functions affect the solution?

Tip: When solving inequalities with logarithms, always check for restrictions on the domain (e.g., logarithms of negative numbers or zero are undefined).