\( \sqrt { 16 + 3 ^ { 2 x } } = 3 ^ { x ^ { \prime } } + 2 \)

To solve the equation \[ \sqrt{16 + 3^{2x}} = 3^{x'} + 2, \] we start by squaring both sides to eliminate the square root: \[ 16 + 3^{2x} = (3^{x'} + 2)^2. \] Expanding the right side, we have: \[ (3^{x'} + 2)^2 = 3^{2x'} + 4 \cdot 3^{x'} + 4. \] Now, substituting this back into the equation gives us: \[ 16 + 3^{2x} = 3^{2x'} + 4 \cdot 3^{x'} + 4. \] Next, we can rearrange the equation: \[ 3^{2x} = 3^{2x'} + 4 \cdot 3^{x'} + 4 - 16. \] This simplifies to: \[ 3^{2x} = 3^{2x'} + 4 \cdot 3^{x'} - 12. \] Now, we can rewrite \(3^{2x}\) as \((3^x)^2\) and \(3^{2x'}\) as \((3^{x'})^2\). Let \(y = 3^x\) and \(z = 3^{x'}\). The equation becomes: \[ y^2 = z^2 + 4z - 12. \] Rearranging gives us a quadratic equation in \(y\): \[ y^2 - z^2 - 4z + 12 = 0. \] This can be factored or solved using the quadratic formula. The quadratic formula states that for an equation of the form \(ay^2 + by + c = 0\), the solutions for \(y\) are given by: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \] In our case, we can treat \(y\) as a variable and rearrange the equation to: \[ y^2 - (z^2 + 4z - 12) = 0. \] To find \(y\), we can express it in terms of \(z\): \[ y = \sqrt{z^2 + 4z - 12}. \] Now, we need to ensure that the expression under the square root is non-negative: \[ z^2 + 4z - 12 \geq 0. \] To find the roots of the quadratic \(z^2 + 4z - 12 = 0\), we can use the quadratic formula: \[ z = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-12)}}{2 \cdot 1} = \frac{-4 \pm \sqrt{16 + 48}}{2} = \frac{-4 \pm \sqrt{64}}{2} = \frac{-4 \pm 8}{2}. \] This gives us the roots: \[ z = \frac{4}{2} = 2 \quad \text{and} \quad z = \frac{-12}{2} = -6. \] Since \(z = 3^{x'}\) must be non-negative, we only consider \(z \geq 2\). Now, we can substitute back to find \(y\): \[ y = \sqrt{z^2 + 4z - 12}. \] For \(z = 2\): \[ y = \sqrt{2^2 + 4 \cdot 2 - 12} = \sqrt{4 + 8 - 12} = \sqrt{0} = 0. \] This means \(3^x = 0\), which is not possible. For \(z > 2\), we can analyze the behavior of the function \(z^2 + 4z - 12\) and find that it is increasing. Thus, we can conclude that there are no valid solutions for \(x\) and \(x'\) that satisfy the original equation. Therefore, the solution to the equation is: \[ \text{No solution.} \]