Solve \(2^{2x+1} - 6^x + 16 = 0.\)

To solve this equation, we can rewrite it in terms of powers of 2 and 3:

\(2^{2x+1} - 6^x + 16 = 0\)

\(2 \cdot 2^{2x} - 2^x \cdot 3^x + 2^4 = 0\)

\(2^{2x}(2 - 3^x) + 2^4 = 0\)

Now, let's substitute \(y = 2^x\):

\(y^2(2 - 3y) + 16 = 0\)

\(2y^3 - 3y^2 + 16 = 0\)

This is a cubic equation in terms of \(y\), which can be difficult to solve directly. However, we can see that \(y = 2\) is a solution to this equation:

\(2(2)^3 - 3(2)^2 + 16 = 16 - 12 + 16 = 20 \neq 0\)

Therefore, \(y = 2\) is not a solution to the equation. This means that the original equation \(2^{2x+1} - 6^x + 16 = 0\) does not have a real solution.

$$2^{2x 1} - 6^x 16 = 0$$

$$(2^{2x+1} = 6^x)(4^2 - 16 = 0)$$

$$(2^{2x}2 = 6^x)(4 - 4 = 0)$$

$$(2^{2x} = 6^x)(0 = 0$$

All real numbers make this true, so the solution set is all real numbers, or \((-\infty,\infty)\).