Please I need help in knowing how to solve this. I know I need to take the antilog of something but don't know what.What is the values of [H3O+] for a solution with [OH−]=1×10−5M?

that isn't helpful

No, you don't need the antilog to do this one.

(H3O^+)(OH^-) = Kw = 1E-14.
You know OH^- and Kw, solve for (H3O^+.

To find the value of [H3O+] for a solution with [OH−]=1×10−5M, you can use the relationship between [H3O+] (the concentration of hydronium ions) and [OH−] (the concentration of hydroxide ions) in water.

In water, the product of [H3O+] and [OH−] is always constant and equal to the ion product of water (Kw), which is 1.0 × 10^14. This relationship is described by the equation:

[H3O+] × [OH−] = Kw

Given that [OH−] = 1 × 10^−5 M, we can substitute this value into the equation:

[H3O+] × (1 × 10^−5) = 1.0 × 10^14

To solve for [H3O+], divide both sides of the equation by (1 × 10^−5):

[H3O+] = (1.0 × 10^14) / (1 × 10^−5)

Simplifying the expression on the right side gives you:

[H3O+] = 1 × 10^9 M

Therefore, the value of [H3O+] for a solution with [OH−] = 1 × 10^−5 M is 1 × 10^9 M.