Put a weak acid in water and set up an ICE table — Initial, Change, Equilibrium. Call the formal concentration $C$: the amount you actually made up, in mol/L, before any proton moved. Call $x$ the concentration that ionizes, in mol/L, which is also the equilibrium $[\mathrm{H^+}]$, since each molecule that lets go releases exactly one proton. Then $K_a = \dfrac{x^2}{C - x}$ — products over reactants, straight off the ICE table's bottom row.

Here is the move that makes it examinable. A weak acid ionizes barely at all, so $x$ is tiny beside $C$, and $C - x \approx C$. That collapses the algebra to $x = \sqrt{K_a C}$, and therefore $\mathrm{pH} = -\tfrac{1}{2}\log(K_a C)$ — read aloud: pH equals minus a half log of K-a C. The half IS the square root, and forgetting it doubles your pH, the single most-marked slip in this unit. Check the approximation afterwards with the percent ionization, $\%\,\mathrm{ion} = \dfrac{[\mathrm{H^+}]}{C} \times 100\%$: under 5 %, the shortcut was honest; over it, go back and solve the quadratic properly.