The Math.log1p() function returns the natural logarithm (base e) of 1 + a number, that is

x > - 1 , Math.log1p ( x ) = ln ( 1 + x ) \forall x > -1, \mathtt{\operatorname{Math.log1p}(x)} = \ln(1 + x)




A number.

Return value

The natural logarithm (base e) of 1 plus the given number. If the number is less than -1, NaN is returned.


For very small values of x, adding 1 can reduce or eliminate precision.  The double floats used in JS give you about 15 digits of precision.  1 + 1e-15 = 1.000000000000001, but 1 + 1e-16 = 1.000000000000000 and therefore exactly 1.0 in that arithmetic, because digits past 15 are rounded off.  

When you calculate log(1 + x), you should get an answer very close to x, if x is small (that's why these are called 'natural' logarithms).  If you calculate Math.log(1 + 1.1111111111e-15) you should get an answer close to 1.1111111111e-15.  Instead, you will end up taking the logarithm of 1.00000000000000111022 (the roundoff is in binary so sometimes it gets ugly), so you get the answer 1.11022...e-15, with only  3 correct digits.  If, instead, you calculate Math.log1p(1.1111111111e-15) you will get a much more accurate answer 1.1111111110999995e-15 with 15 correct digits of precision (actually 16 in this case).

If the value of x is less than -1, the return value is always NaN.

Because log1p() is a static method of Math, you always use it as Math.log1p(), rather than as a method of a Math object you created (Math is not a constructor).


Using Math.log1p()

Math.log1p(1);  // 0.6931471805599453
Math.log1p(0);  // 0
Math.log1p(-1); // -Infinity
Math.log1p(-2); // NaN


Browser compatibility

BCD tables only load in the browser

See also