Find the position of rightmost set bit of integer in vb.net

Vb program for Find the position of rightmost set bit of integer. Here problem description and explanation.

' Include namespace system
Imports System 
'  Vb.net Program for
'  Find position of rightmost set bit

Public Class BitPosition
    '  Finding the right most active bit position
    Public Sub rightmostActiveBit(ByVal n As Integer)
        if (n <= 0) Then
            Return
        End If
        '    Example 
        '    ———————
        '    n    = 320
        '    n    = (00101000000)   Binary
        '    -n   = (11011000000)   (2s)
        '    Calculate log2 
        '    ——————————————
        '    Formula : log(n & -n) / log(2) + 1
        '    -----------------------------------
        '    log(320 & -320) / log(2) + 1)
        '    Here : log(320 & -320) = log(64) = 4.158883
        '           log(2)  = 0.693147    
        '    (4.158883 / 0.693147) + 1 = (7) position
        '    ————————————————————————————————————————
        '  Calculate rightmost active bits
        Dim result As Integer = CInt((Math.Log(n And -n) / Math.Log(2) + 1))
        Console.WriteLine(" Number : " + 
                          n.ToString() + " Result : " + 
                          result.ToString())
    End Sub
    Public Shared Sub Main(ByVal args As String())
        Dim task As BitPosition = New BitPosition()
        '  Test Cases
        '  320 = Binary(101000000)
        task.rightmostActiveBit(320)
        '  (1000) = Binary(1111101000)
        task.rightmostActiveBit(1000)
        '  (153) = Binary(10011001)
        task.rightmostActiveBit(153)
        '  (354) = Binary(101100010)
        task.rightmostActiveBit(354)
        '  160 = Binary(10100000)
        task.rightmostActiveBit(160)
    End Sub
End Class

Output

 Number : 320 Result : 7
 Number : 1000 Result : 4
 Number : 153 Result : 1
 Number : 354 Result : 2
 Number : 160 Result : 6