1. **State the problem:** We need to determine the values of $x$ for which the expression $$M = \frac{1}{\sqrt{2x - 1}}$$ is real. 2. **Understand the formula and rules:** For $M$ to be real, the denominator $\sqrt{2x - 1}$ must be a real, non-zero number. - The expression inside the square root, called the radicand, must be \textbf{non-negative} for the square root to be real: $$2x - 1 \geq 0$$ - The denominator cannot be zero because division by zero is undefined: $$\sqrt{2x - 1} \neq 0$$ 3. **Solve the inequality for the radicand:** $$2x - 1 \geq 0$$ Add 1 to both sides: $$2x \geq 1$$ Divide both sides by 2: $$\cancel{2}x \geq \frac{1}{\cancel{2}}$$ $$x \geq \frac{1}{2}$$ 4. **Exclude values that make the denominator zero:** Set the denominator equal to zero: $$\sqrt{2x - 1} = 0$$ Square both sides: $$2x - 1 = 0$$ Solve for $x$: $$2x = 1$$ $$x = \frac{1}{2}$$ Since $x=\frac{1}{2}$ makes the denominator zero, it must be excluded. 5. **Final solution:** $$x > \frac{1}{2}$$ This means $M$ is real for all $x$ strictly greater than $\frac{1}{2}$. **Answer:** $\boxed{x > \frac{1}{2}}$