The problem asks us to find the domain of the function $f(x) = \frac{\sqrt{(\sqrt{x})^x - x^{\sqrt{x}}}}{\ln(4x^2 - 1)}$.

1. Problem Description

The problem asks us to find the domain of the function $f(x) = \frac{\sqrt{(\sqrt{x})^x - x^{\sqrt{x}}}}{\ln(4x^2 - 1)}$.

2. Solution Steps

To find the domain of the function, we need to consider the restrictions imposed by the square root and the logarithm.

First, let's look at the square root. For $\sqrt{(\sqrt{x})^x - x^{\sqrt{x}}}$ to be defined, we must have:

$(\sqrt{x})^x - x^{\sqrt{x}} \ge 0$

$(\sqrt{x})^x \ge x^{\sqrt{x}}$

$(x^{1/2})^x \ge x^{\sqrt{x}}$

$x^{x/2} \ge x^{\sqrt{x}}$

Since the function is defined for $x>0$, we can consider two cases.

Case 1: $x > 1$. In this case, $x/2 \ge \sqrt{x}$ or $x \ge 2\sqrt{x}$ or $\sqrt{x} \ge 2$, meaning $x \ge 4$.

Case 2: $0 < x < 1$. In this case, $x/2 \le \sqrt{x}$ or $x \le 2\sqrt{x}$ or $\sqrt{x} \le 2$, meaning $x \le 4$. Since $0 < x < 1$, this condition is satisfied. However, we must also have $x^{x/2}$ and $x^{\sqrt{x}}$ well defined. Thus, we need $x>0$. Also, when $0 < x < 1$, $x/2 < \sqrt{x}$, therefore, $(\sqrt{x})^x < x^{\sqrt{x}}$. This contradicts the inequality $(\sqrt{x})^x \ge x^{\sqrt{x}}$.

Case 3: $x = 1$. Then $1^{1/2} \ge 1^{1}$. This yields $1 \ge 1$, so $x=1$ is a solution.

Now, let's look at the logarithm. For $\ln(4x^2 - 1)$ to be defined, we must have:

$4x^2 - 1 > 0$

$4x^2 > 1$

$x^2 > \frac{1}{4}$

$|x| > \frac{1}{2}$

$x > \frac{1}{2}$ or $x < -\frac{1}{2}$.

Additionally, since the logarithm is in the denominator, we must have:

$\ln(4x^2 - 1) \ne 0$

$4x^2 - 1 \ne 1$

$4x^2 \ne 2$

$x^2 \ne \frac{1}{2}$

$x \ne \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$.

Finally, we also require $x \ge 0$ because of $\sqrt{x}$ terms. Combining all conditions, we have:

1. $x \ge 4$ or $x=1$

2. $x > \frac{1}{2}$

3. $x \ne \frac{\sqrt{2}}{2}$

Since $x \ge 4$ satisfies $x > \frac{1}{2}$, and $4 > \frac{\sqrt{2}}{2}$, we consider the case $x \ge 4$.

Now, we check $x=1$, which satisfies $x > \frac{1}{2}$ and $x \ne \frac{\sqrt{2}}{2}$. However $(\sqrt{1})^1 - 1^{\sqrt{1}} = 1-1 = 0$. So, $x=1$ yields $f(1) = \frac{0}{\ln(3)} = 0$.

Then the domain is $[4, \infty) \cup \{1\}$.

3. Final Answer

$[4, \infty) \cup \{1\}$