## C# Math.ILogB() – Examples

In this tutorial, we will learn about the C# Math.ILogB() method, and learn how to use this method to compute base 2 integer logarithm of a specified number, with the help of examples.

## ILogB(Double)

Math.ILogB(x) returns the base 2 integer logarithm of a specified number `x`.

### Syntax

The syntax of ILogB() method is

`Math.ILogB(Double x)`

where

Return Value

The method returns Int32 value.

### Example 1 – ILogB(x)

In this example, we will take some double-precision floating-point numbers and find their integer logarithm value with base value of 2, using Math.ILogB() method.

C# Program

```using System;

class Example {
static void Main(string[] args) {
Double x;
Double result;

x = 2;
result = Math. ILogB(x);
Console.WriteLine(\$"ILogB({x}) = {result}");

x = 3;
result = Math. ILogB(x);
Console.WriteLine(\$"ILogB({x}) = {result}");

x = 4;
result = Math. ILogB(x);
Console.WriteLine(\$"ILogB({x}) = {result}");
}
}```

Output

```ILogB(2) = 1
ILogB(3) = 1
ILogB(4) = 2```

Please note that ILogB() returns an integer value. Therefore, the precision or decimals points are trimmed out after computing the logarithm.

### Conclusion

In this C# Tutorial, we have learnt the syntax of C# Math.ILogB() method, and also learnt how to use this method with the help of C# example programs.