For what values of x is the following expression a real number? \( \frac{7}{\sqrt{16-2 x}} \)

For the expression to be a real number, the denominator must be positive, which means \( \sqrt{16 - 2x} \) must be greater than zero. This leads to the inequality \( 16 - 2x > 0 \). Solving this gives \( 16 > 2x \) or \( x < 8 \). Additionally, since we can't take the square root of a negative number, we require \( 16 - 2x \geq 0 \). Thus, the solution to this expression being a real number is \( x < 8 \). In practical terms, you can't have a negative number under the square root, as that would lead to complex numbers, which are not real. Always remember to check the conditions where the function is undefined—like division by zero—which can make a function blow up or become undefined at certain points!