Answer
$$K_w = [H_3O^+][OH^-]$$
At room temperature, we know that: $K_w = 1.0 \times 10^{-14}$ for that reaction. Therefore: $$1.0 \times 10^{-14} = [H_3O^+][OH^-]$$
$$[OH^-] = \frac{1.0 \times 10^{-14} }{[H_3O^+]}$$
$$[H_3O^+] = \frac{1.0 \times 10^{-14} }{[OH^-]}$$
Work Step by Step
At any equilibrium, we can write an expression that relates a constant to a multiplication of the concentrations of the compounds in this equilibrium reaction.
In the autoionization of water, the process is the same. Following the reaction: $$ H_2O(l) + H_2O(l) \leftrightharpoons H_3O^+(aq) + OH^-(aq)$$
We can write the expression:
$$K_c = \frac{[H_3O^+][OH^-]}{[H_2O][H_2O]}$$
But, since the concentration of water in an aqueous solution does not change, we can simplify it to:
$$K_c = [H_3O^+][OH^-]$$
At room temperature, we know that: $K = 1.0 \times 10^{-14}$ for that reaction. Therefore: $$1.0 \times 10^{-14} = [H_3O^+][OH^-]$$
$$[OH^-] = \frac{1.0 \times 10^{-14} }{[H_3O^+]}$$
$$[H_3O^+] = \frac{1.0 \times 10^{-14} }{[OH^-]}$$
