As you know, in aqueous solution at 298*K, pH = -log_10[H_3O^+], pOH = -log_10[HO^-], AND pH+pOH=14. Back in the day, before the advent of electronic calculators, students and engineers would routinely use log tables to enable complex multiplication and division. If I have a basic solution, I can thus calculate [H_3O^+] on the basis of pOH by means of simple addition or subtraction rather than by more complex multiplication/division.
If you are still unsure, come back here, and someone will help you.
In the meantime, K_w =([H_3O^+][HO^-])=10^(14)) under standard conditions, i.e. 298*K, and 1*atm. If we were to increase the temperature, say to water's normal boiling point, 373*K, how would you predict K_w and pK_w to evolve?
Remember that the autoprotolysis of water is an example of a bond-breaking reaction.
2H_2O(l) rarr H_3O^+ + HO^-
