Math.Log Method

Definition

Returns the logarithm of a specified number.

Overloads

Log(Double)

Returns the natural (base e) logarithm of a specified number.

Log(Double, Double)

Returns the logarithm of a specified number in a specified base.

Log(Double)

Returns the natural (base e) logarithm of a specified number.

public static double Log (double d);
Parameters
d
Double

The number whose logarithm is to be found.

Returns

One of the values in the following table.

d parameter Return value
Positive The natural logarithm of d; that is, ln d, or log e d
Zero NegativeInfinity
Negative NaN
Equal to NaNNaN
Equal to PositiveInfinityPositiveInfinity

Examples

The following example illustrates the Log method.

using System;
public class Example
{
   public static void Main()
   {
      Console.WriteLine("  Evaluate this identity with selected values for X:");
      Console.WriteLine("                              ln(x) = 1 / log[X](B)");
      Console.WriteLine();
          
      double[] XArgs = { 1.2, 4.9, 9.9, 0.1 };
   
      foreach (double argX in XArgs)
      {
         // Find natural log of argX.
         Console.WriteLine("                      Math.Log({0}) = {1:E16}",
                           argX, Math.Log(argX));

         // Evaluate 1 / log[X](e).
         Console.WriteLine("             1.0 / Math.Log(e, {0}) = {1:E16}",
                           argX, 1.0 / Math.Log(Math.E, argX));
         Console.WriteLine();
      }
   }   
}
// This example displays the following output:
//         Evaluate this identity with selected values for X:
//                                     ln(x) = 1 / log[X](B)
//       
//                             Math.Log(1.2) = 1.8232155679395459E-001
//                    1.0 / Math.Log(e, 1.2) = 1.8232155679395459E-001
//       
//                             Math.Log(4.9) = 1.5892352051165810E+000
//                    1.0 / Math.Log(e, 4.9) = 1.5892352051165810E+000
//       
//                             Math.Log(9.9) = 2.2925347571405443E+000
//                    1.0 / Math.Log(e, 9.9) = 2.2925347571405443E+000
//       
//                             Math.Log(0.1) = -2.3025850929940455E+000
//                    1.0 / Math.Log(e, 0.1) = -2.3025850929940455E+000
Module Example
   Sub Main()
      Console.WriteLine( _
         "  Evaluate this identity with selected values for X:")
      Console.WriteLine("                              ln(x) = 1 / log[X](B)")
      Console.WriteLine()
          
      Dim XArgs() As Double = { 1.2, 4.9, 9.9, 0.1 }
   
      For Each argX As Double In XArgs
         ' Find natural log of argX.
         Console.WriteLine("                      Math.Log({0}) = {1:E16}", _
                           argX, Math.Log(argX))

         ' Evaluate 1 / log[X](e).
         Console.WriteLine("             1.0 / Math.Log(e, {0}) = {1:E16}", _
                           argX, 1.0 / Math.Log(Math.E, argX))
         Console.WriteLine()
      Next
   End Sub 
End Module
' This example displays the following output:
'         Evaluate this identity with selected values for X:
'                                     ln(x) = 1 / log[X](B)
'       
'                             Math.Log(1.2) = 1.8232155679395459E-001
'                    1.0 / Math.Log(e, 1.2) = 1.8232155679395459E-001
'       
'                             Math.Log(4.9) = 1.5892352051165810E+000
'                    1.0 / Math.Log(e, 4.9) = 1.5892352051165810E+000
'       
'                             Math.Log(9.9) = 2.2925347571405443E+000
'                    1.0 / Math.Log(e, 9.9) = 2.2925347571405443E+000
'       
'                             Math.Log(0.1) = -2.3025850929940455E+000
'                    1.0 / Math.Log(e, 0.1) = -2.3025850929940455E+000

Remarks

Parameter d is specified as a base 10 number.

See Also

Log(Double, Double)

Returns the logarithm of a specified number in a specified base.

public static double Log (double a, double newBase);
Parameters
a
Double

The number whose logarithm is to be found.

newBase
Double

The base of the logarithm.

Returns

One of the values in the following table. (+Infinity denotes PositiveInfinity, -Infinity denotes NegativeInfinity, and NaN denotes NaN.)

anewBase Return value
a> 0 (0 <newBase< 1) -or-(newBase> 1) lognewBase(a)
a< 0 (any value) NaN
(any value) newBase< 0 NaN
a != 1 newBase = 0 NaN
a != 1 newBase = +Infinity NaN
a = NaN (any value) NaN
(any value) newBase = NaN NaN
(any value) newBase = 1 NaN
a = 0 0 <newBase< 1 +Infinity
a = 0 newBase> 1 -Infinity
a = +Infinity 0 <newBase< 1 -Infinity
a = +Infinity newBase> 1 +Infinity
a = 1 newBase = 0 0
a = 1 newBase = +Infinity 0

Examples

The following example uses Log to evaluate certain logarithmic identities for selected values.

// Example for the Math::Log( double ) and Math::Log( double, double ) methods.
using namespace System;

// Evaluate logarithmic identities that are functions of two arguments.
void UseBaseAndArg( double argB, double argX )
{
   
   // Evaluate log(B)[X] == 1 / log(X)[B].
   Console::WriteLine( "\n                     Math::Log({1}, {0}) == {2:E16}"
   "\n               1.0 / Math::Log({0}, {1}) == {3:E16}", argB, argX, Math::Log( argX, argB ), 1.0 / Math::Log( argB, argX ) );
   
   // Evaluate log(B)[X] == ln[X] / ln[B].
   Console::WriteLine( "         Math::Log({1}) / Math::Log({0}) == {2:E16}", argB, argX, Math::Log( argX ) / Math::Log( argB ) );
   
   // Evaluate log(B)[X] == log(B)[e] * ln[X].
   Console::WriteLine( "Math::Log(Math::E, {0}) * Math::Log({1}) == {2:E16}", argB, argX, Math::Log( Math::E, argB ) * Math::Log( argX ) );
}

void main()
{
   Console::WriteLine( "This example of Math::Log( double ) and "
   "Math::Log( double, double )\n"
   "generates the following output.\n" );
   Console::WriteLine( "Evaluate these identities with "
   "selected values for X and B (base):" );
   Console::WriteLine( "   log(B)[X] == 1 / log(X)[B]" );
   Console::WriteLine( "   log(B)[X] == ln[X] / ln[B]" );
   Console::WriteLine( "   log(B)[X] == log(B)[e] * ln[X]" );
   UseBaseAndArg( 0.1, 1.2 );
   UseBaseAndArg( 1.2, 4.9 );
   UseBaseAndArg( 4.9, 9.9 );
   UseBaseAndArg( 9.9, 0.1 );
}

/*
This example of Math::Log( double ) and Math::Log( double, double )
generates the following output.

Evaluate these identities with selected values for X and B (base):
   log(B)[X] == 1 / log(X)[B]
   log(B)[X] == ln[X] / ln[B]
   log(B)[X] == log(B)[e] * ln[X]

                     Math::Log(1.2, 0.1) == -7.9181246047624818E-002
               1.0 / Math::Log(0.1, 1.2) == -7.9181246047624818E-002
         Math::Log(1.2) / Math::Log(0.1) == -7.9181246047624818E-002
Math::Log(Math::E, 0.1) * Math::Log(1.2) == -7.9181246047624804E-002

                     Math::Log(4.9, 1.2) == 8.7166610085093179E+000
               1.0 / Math::Log(1.2, 4.9) == 8.7166610085093161E+000
         Math::Log(4.9) / Math::Log(1.2) == 8.7166610085093179E+000
Math::Log(Math::E, 1.2) * Math::Log(4.9) == 8.7166610085093179E+000

                     Math::Log(9.9, 4.9) == 1.4425396251981288E+000
               1.0 / Math::Log(4.9, 9.9) == 1.4425396251981288E+000
         Math::Log(9.9) / Math::Log(4.9) == 1.4425396251981288E+000
Math::Log(Math::E, 4.9) * Math::Log(9.9) == 1.4425396251981288E+000

                     Math::Log(0.1, 9.9) == -1.0043839404494075E+000
               1.0 / Math::Log(9.9, 0.1) == -1.0043839404494075E+000
         Math::Log(0.1) / Math::Log(9.9) == -1.0043839404494075E+000
Math::Log(Math::E, 9.9) * Math::Log(0.1) == -1.0043839404494077E+000
*/
// Example for the Math.Log( double ) and Math.Log( double, double ) methods.
using System;

class LogDLogDD
{
    public static void Main() 
    {
        Console.WriteLine( 
            "This example of Math.Log( double ) and " +
            "Math.Log( double, double )\n" +
            "generates the following output.\n" );
        Console.WriteLine( 
            "Evaluate these identities with " +
            "selected values for X and B (base):" );
        Console.WriteLine( "   log(B)[X] == 1 / log(X)[B]" );
        Console.WriteLine( "   log(B)[X] == ln[X] / ln[B]" );
        Console.WriteLine( "   log(B)[X] == log(B)[e] * ln[X]" );

        UseBaseAndArg(0.1, 1.2);
        UseBaseAndArg(1.2, 4.9);
        UseBaseAndArg(4.9, 9.9);
        UseBaseAndArg(9.9, 0.1);
    }

    // Evaluate logarithmic identities that are functions of two arguments.
    static void UseBaseAndArg(double argB, double argX)
    {
        // Evaluate log(B)[X] == 1 / log(X)[B].
        Console.WriteLine( 
            "\n                   Math.Log({1}, {0}) == {2:E16}" + 
            "\n             1.0 / Math.Log({0}, {1}) == {3:E16}", 
            argB, argX, Math.Log(argX, argB),
            1.0 / Math.Log(argB, argX) );

        // Evaluate log(B)[X] == ln[X] / ln[B].
        Console.WriteLine( 
            "        Math.Log({1}) / Math.Log({0}) == {2:E16}",
            argB, argX, Math.Log(argX) / Math.Log(argB) );

        // Evaluate log(B)[X] == log(B)[e] * ln[X].
        Console.WriteLine( 
            "Math.Log(Math.E, {0}) * Math.Log({1}) == {2:E16}", 
            argB, argX, Math.Log(Math.E, argB) * Math.Log(argX) );
    }
}

/*
This example of Math.Log( double ) and Math.Log( double, double )
generates the following output.

Evaluate these identities with selected values for X and B (base):
   log(B)[X] == 1 / log(X)[B]
   log(B)[X] == ln[X] / ln[B]
   log(B)[X] == log(B)[e] * ln[X]

                   Math.Log(1.2, 0.1) == -7.9181246047624818E-002
             1.0 / Math.Log(0.1, 1.2) == -7.9181246047624818E-002
        Math.Log(1.2) / Math.Log(0.1) == -7.9181246047624818E-002
Math.Log(Math.E, 0.1) * Math.Log(1.2) == -7.9181246047624804E-002

                   Math.Log(4.9, 1.2) == 8.7166610085093179E+000
             1.0 / Math.Log(1.2, 4.9) == 8.7166610085093161E+000
        Math.Log(4.9) / Math.Log(1.2) == 8.7166610085093179E+000
Math.Log(Math.E, 1.2) * Math.Log(4.9) == 8.7166610085093179E+000

                   Math.Log(9.9, 4.9) == 1.4425396251981288E+000
             1.0 / Math.Log(4.9, 9.9) == 1.4425396251981288E+000
        Math.Log(9.9) / Math.Log(4.9) == 1.4425396251981288E+000
Math.Log(Math.E, 4.9) * Math.Log(9.9) == 1.4425396251981288E+000

                   Math.Log(0.1, 9.9) == -1.0043839404494075E+000
             1.0 / Math.Log(9.9, 0.1) == -1.0043839404494075E+000
        Math.Log(0.1) / Math.Log(9.9) == -1.0043839404494075E+000
Math.Log(Math.E, 9.9) * Math.Log(0.1) == -1.0043839404494077E+000
*/
' Example for the Math.Log( Double ) and Math.Log( Double, Double ) methods.
Imports System
Imports Microsoft.VisualBasic

Module LogDLogDD
   
    Sub Main()
        Console.WriteLine( _
            "This example of Math.Log( Double ) and " + _
            "Math.Log( Double, Double )" & vbCrLf & _
            "generates the following output." & vbCrLf)
        Console.WriteLine( _
            "Evaluate these identities with selected " & _
            "values for X and B (base):")
        Console.WriteLine("   log(B)[X] = 1 / log(X)[B]")
        Console.WriteLine("   log(B)[X] = ln[X] / ln[B]")
        Console.WriteLine("   log(B)[X] = log(B)[e] * ln[X]")
          
        UseBaseAndArg(0.1, 1.2)
        UseBaseAndArg(1.2, 4.9)
        UseBaseAndArg(4.9, 9.9)
        UseBaseAndArg(9.9, 0.1)
    End Sub 'Main
       
    ' Evaluate logarithmic identities that are functions of two arguments.
    Sub UseBaseAndArg(argB As Double, argX As Double)

        ' Evaluate log(B)[X] = 1 / log(X)[B].
        Console.WriteLine( _
            vbCrLf & "                   Math.Log({1}, {0}) = {2:E16}" + _
            vbCrLf & "             1.0 / Math.Log({0}, {1}) = {3:E16}", _
            argB, argX, Math.Log(argX, argB), _
            1.0 / Math.Log(argB, argX))
          
        ' Evaluate log(B)[X] = ln[X] / ln[B].
        Console.WriteLine( _
            "        Math.Log({1}) / Math.Log({0}) = {2:E16}", _
            argB, argX, Math.Log(argX) / Math.Log(argB))
          
        ' Evaluate log(B)[X] = log(B)[e] * ln[X].
        Console.WriteLine( _
            "Math.Log(Math.E, {0}) * Math.Log({1}) = {2:E16}", _
            argB, argX, Math.Log(Math.E, argB) * Math.Log(argX))

    End Sub 'UseBaseAndArg
End Module 'LogDLogDD

' This example of Math.Log( Double ) and Math.Log( Double, Double )
' generates the following output.
' 
' Evaluate these identities with selected values for X and B (base):
'    log(B)[X] = 1 / log(X)[B]
'    log(B)[X] = ln[X] / ln[B]
'    log(B)[X] = log(B)[e] * ln[X]
' 
'                    Math.Log(1.2, 0.1) = -7.9181246047624818E-002
'              1.0 / Math.Log(0.1, 1.2) = -7.9181246047624818E-002
'         Math.Log(1.2) / Math.Log(0.1) = -7.9181246047624818E-002
' Math.Log(Math.E, 0.1) * Math.Log(1.2) = -7.9181246047624804E-002
' 
'                    Math.Log(4.9, 1.2) = 8.7166610085093179E+000
'              1.0 / Math.Log(1.2, 4.9) = 8.7166610085093161E+000
'         Math.Log(4.9) / Math.Log(1.2) = 8.7166610085093179E+000
' Math.Log(Math.E, 1.2) * Math.Log(4.9) = 8.7166610085093179E+000
' 
'                    Math.Log(9.9, 4.9) = 1.4425396251981288E+000
'              1.0 / Math.Log(4.9, 9.9) = 1.4425396251981288E+000
'         Math.Log(9.9) / Math.Log(4.9) = 1.4425396251981288E+000
' Math.Log(Math.E, 4.9) * Math.Log(9.9) = 1.4425396251981288E+000
' 
'                    Math.Log(0.1, 9.9) = -1.0043839404494075E+000
'              1.0 / Math.Log(9.9, 0.1) = -1.0043839404494075E+000
'         Math.Log(0.1) / Math.Log(9.9) = -1.0043839404494075E+000
' Math.Log(Math.E, 9.9) * Math.Log(0.1) = -1.0043839404494077E+000

Applies to