TheMath.log1p() static method returns the natural logarithm (basee) of1 + x, wherex is the argument. That is:

Try it

console.log(Math.log1p(1));// Expected output: 0.6931471805599453console.log(Math.log1p(0));// Expected output: 0console.log(Math.log1p(-1));// Expected output: -Infinityconsole.log(Math.log1p(-2));// Expected output: NaN

Syntax

Math.log1p(x)

Parameters

x

A number greater than or equal to -1.

Return value

The natural logarithm (basee) ofx + 1. Ifx is -1, returns-Infinity. Ifx < -1, returnsNaN.

Description

For very small values ofx, 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), wherex is a small positive number, you should get an answer very close tox because:$\underset{x\to 0}{lim}\frac{\log (1+x)}{x}=1$. If you calculateMath.log(1 + 1.1111111111e-15), you should get an answer close to1.1111111111e-15. Instead, you will end up taking the logarithm of1.00000000000000111022 (the roundoff is in binary, so sometimes it gets ugly), and get the answer 1.11022…e-15, with only 3 correct digits. If you calculateMath.log1p(1.1111111111e-15) instead, 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 ofx is less than -1, the return value is alwaysNaN.

Becauselog1p() is a static method ofMath, you always use it asMath.log1p(), rather than as a method of aMath object you created (Math is not a constructor).

Examples

Using Math.log1p()

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

Specifications

Browser compatibility

See also

• Polyfill ofMath.log1p incore-js
• es-shims polyfill ofMath.log1p
• Math.exp()
• Math.log()
• Math.expm1()
• Math.log10()
• Math.log2()
• Math.pow()