If the pH is 7.3 what is the (OH-) and the (H30+)

I'm having trouble understanding what to type into my calculator. Thanks very much!

pH means -log[H+] (or H3O+ in this case)

so to 'unlog' you need to type into ur calculator SHIFT then LOG to give a 10^

so to find [H3O+]:
-log[H3O+] = 7.3
log[H3O+] = -7.3
[H3O+] = 10^ (-7.3)
=5.012x10^-8

to find [OH-] the easiest way is to say that pH + pOH = 14
so 7.3 + -log[OH-] = 14
-log[OH-] = 6.7
log[OH-] = -6.7
[OH-] = 1.995x10^-7

pH = 7.3
[H3O+] = 10^-7.3 = 5.0E-8

pOH = 14 - 7.3 = 6.7
[OH-] = 10^-6.7 = 2.0E-7

or

[H3O+] x [OH-] = 1E-14
[OH-] = 1E-14 / 5E-8 = 2E-7