I'm trying to code an equation that calculates the nth root of an expression. For example, I want to isolate the variable 'r' in the expression r ^ n = x. ( i.e. "r to the power of n = x"). If n=5 for example, the expression could be written as r = 5th root ( x ). Is there a specific function that does this?
I dont think theres anything specific in the system.math namespace so I did a quick search and found the following formula which when placed in a function may be what you looking for.
Public Function NthRoot(ByVal y As ValueType, ByVal RootIndex As Integer) As Single
Return System.Math.Exp(1 / RootIndex * (System.Math.Log(y)))
End Function
So if I call it
NthRoot(49,2) I get 7 as 7^2 is 49
NthRoot(625,4) I get 5 as 5^4 is 625
So I think this may be close to what you looking for.
Hi,
What is 2.5 ^ 6.7 though ? Well it equals
463.65832088157585126331596260465
according to windows calculator.
You can stick DOUBLE number TYPEs into this version.
E.G.
'Returns 2.5
MessageBox.Show(NthRoot(6.7, 463.65832088157583).ToString)
More about LOGarithms.
Log(2) + Log(3) =Log(6) 'Like 2 * 3 = 6
( just ADD the LOG values ).
Log(x) + Log(y) = Log(z)
Log(2) * 3 = LOG(8)
Log(x) * y = Log(z) ' Where x ^ y = z
EXP in VB.Net is the equivent of ALOG or the Arc-Log ( Inverse Log ) if you like.
Public
Class Form1
'This should be one line of code in your code window.>>>>
Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click
'Returns 2. as 2 is the 3rd root of 8
MessageBox.Show(NthRoot(3, 8).ToString)
End Sub
Public Function NthRoot(ByVal nthRootToFind As Double, ByVal ofThisNumber As Double) As Double
Return Math.Exp(Math.Log(ofThisNumber) / nthRootToFind)End FunctionEnd Class
Regards,
S_DS
Hey Groovers,
Sorry to re-hash this old thread.
I need to be able to find out what x is in the formula below using vb.net. can anyone offer any ideas
2^x = 10 (ie. 2 to the power of x = 10. alternatively, 2 is the xth root of 10).
Any thoughts.
Greg
Hi,
2^x = 10
<=>
log(2^x) = log(10)
<=>
X log(2) = log(10)
->
X= Math.Log(10) / Math.Log(2)
Regards
Csaba
