michifun

isoraqathedh replied to your post: How do you calculate something like 3^2.3?…

Don’t forget that x^(a/b) is the bth root of x^a, so if your exponent is rational then no logarithms are needed.

Ahhhhhh… yes yes, thank you! I was secretly hoping you’d chime in! Also, that explains 4^(5/2).