The equation \[9^x - 3^{x+1} = k\] has one or more real solutions precisely when

\(k \geq -9/4\),

\(k > 0\),

\(k \leq -1\),

\(k \geq 5/8\).

If we write \(y = 3^x\) and use the fact that \(9^x = (3^2)^x = (3^x)^2\), the equation can be written as \[y^2 -3y - k = 0.\]

This has solutions \[y = \frac{3 \pm \sqrt{9 + 4k}}{2}\] which are real when \(k \geq -9/4\). Since \(y = 3^x\), we need at least one of these solutions to be positive for \(x\) to be real, but the larger root is clearly always positive, so this is not a problem. Hence the answer is (a).