The Math.log1p() static method returns the natural logarithm (base e) of 1 + x, where x is the argument. That is:

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

Try it





A number greater than or equal to -1.

Return value

The natural logarithm (base e) of x + 1. If x is -1, returns -Infinity. If x < -1, returns NaN.


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) where x is a small positive number, you should get an answer very close to x, because lim x 0 log ( 1 + x ) x = 1 \lim_{x \to 0} \frac{\log(1+x)}{x} = 1 . 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), and 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(-2); // NaN
Math.log1p(-1); // -Infinity
Math.log1p(-0); // -0
Math.log1p(0); // 0
Math.log1p(1); // 0.6931471805599453
Math.log1p(Infinity); // Infinity


ECMAScript Language Specification
# sec-math.log1p

Browser compatibility

BCD tables only load in the browser

See also