The n'th (not square) root of a number?

I know the Sqr function, but how do i make a function which retrieves another root than the square root?

I tried searching the help files, but it doesn't seem that there's a built-in function to do the trick.

Anyone knows the algorithm?

Code:

---

    Public Function nSqr(ByVal number as Double, _
                        ByVal n as Double) as Double
        nSqr = number ^ (1 / n)
    End Function
'Usage:
    Label1.Caption = nSqr(81, 4) 'returns 3
    Label1.Caption = nSqr(27, 3) 'returns 3

---

Bit of rough code, but it should do the trick
Hope it helps :)

[Edited by r0ach on 05-19-2000 at 02:13 PM]

Or...

Code:

---

Public Function NthRoot(num As Double, RootOrdinal As Double) As Double
   
    NthRoot = Exp(Log(num) / RootOrdinal)
   
End Function

---

:cool:

Dan

Thanks for the quick answer, guys!

For the moment i don't need the function, but i've saved it in a module for future reference. BTW I didn't know Exp(Log(num) / RootOrdinal), but I should have remembered the number ^ (1 / n). (I just learned some of it in highschool.)

Cheers