Find the domain of the following function: `3^(y)+2^(x^(4))=2^(4x^(2)-1)`

Explanation

To find the domain of the function given by the equation \(3^y + 2^{x^4} = 2^{4x^2 - 1}\), we will follow these steps:

### Step 1: Rearranging the Equation
We start by isolating \(3^y\):
\[ 3^y = 2^{4x^2 - 1} - 2^{x^4} \]

### Step 2: Understanding the Range of Exponential Functions
Since \(3^y\) is an exponential function, it is always greater than 0. Therefore, we need to ensure that the right-hand side is also greater than 0:
\[ 2^{4x^2 - 1} - 2^{x^4} > 0 \]

### Step 3: Setting Up the Inequality
This inequality can be rewritten as:
\[ 2^{4x^2 - 1} > 2^{x^4} \]

### Step 4: Comparing Exponents
Since the base (2) is greater than 1, we can compare the exponents:
\[ 4x^2 - 1 > x^4 \]

### Step 5: Rearranging the Inequality
Rearranging gives us:
\[ x^4 - 4x^2 + 1 < 0 \]

### Step 6: Letting \(u = x^2\)
Let \(u = x^2\). Then the inequality becomes:
\[ u^2 - 4u + 1 < 0 \]

### Step 7: Finding Roots of the Quadratic
We can find the roots of the quadratic equation \(u^2 - 4u + 1 = 0\) using the quadratic formula:
\[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3} \]

### Step 8: Analyzing the Quadratic
The roots are \(u_1 = 2 - \sqrt{3}\) and \(u_2 = 2 + \sqrt{3}\). The quadratic opens upwards (since the coefficient of \(u^2\) is positive), so it is less than 0 between the roots:
\[ 2 - \sqrt{3} < u < 2 + \sqrt{3} \]

### Step 9: Converting Back to \(x\)
Since \(u = x^2\), we have:
\[ 2 - \sqrt{3} < x^2 < 2 + \sqrt{3} \]

### Step 10: Finding the Domain for \(x\)
Taking square roots gives us:
\[ -\sqrt{2 + \sqrt{3}} < x < -\sqrt{2 - \sqrt{3}} \quad \text{or} \quad \sqrt{2 - \sqrt{3}} < x < \sqrt{2 + \sqrt{3}} \]

### Final Domain
Thus, the domain of the function is:
\[ x \in \left(-\sqrt{2 + \sqrt{3}}, -\sqrt{2 - \sqrt{3}}\right) \cup \left(\sqrt{2 - \sqrt{3}}, \sqrt{2 + \sqrt{3}}\right) \]