what is the ph for a 7.4*10^-2M KOH solution

KOH --> K⁺ + OH^-

pH + pOH = 14

pOH = -log₁₀([OH^-]) = -log₁₀(7.4*10^-2)

pH = 14 - (-log₁₀(7.4*10^-2))

$${\mathtt{pH}} = {\mathtt{14}}{\mathtt{\,\small\textbf+\,}}{log}_{10}\left({\mathtt{7.4}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{-{\mathtt{2}}}\right) \Rightarrow {\mathtt{pH}} = {\mathtt{12.869\: \!231\: \!719\: \!730\: \!976\: \!2}}$$

or pH ≈ 12.9