files/en-us/web/javascript/reference/global_objects/math/log1p/index.md

metadata

title: Math.log1p()
short-title: log1p()
slug: Web/JavaScript/Reference/Global_Objects/Math/log1p
page-type: javascript-static-method
browser-compat: javascript.builtins.Math.log1p
sidebar: jsref

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

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

{{InteractiveExample("JavaScript Demo: Math.log1p()")}}

console.log(Math.log1p(1));
// Expected output: 0.6931471805599453

console.log(Math.log1p(0));
// Expected output: 0

console.log(Math.log1p(-1));
// Expected output: -Infinity

console.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 (base e) of x + 1. If x is -1, returns -Infinity. If x < -1, returns {{jsxref("NaN")}}.

Description

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: $\underset{x\to 0}{lim}\frac{\log (1+x)}{x}=1\backslash lim\_\{x\; \backslash to\; 0\}\; \backslash frac\{\backslash log(1+x)\}\{x\}\; =\; 1$. 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 you calculate Math.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 of x is less than -1, the return value is always {{jsxref("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).

Examples

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

Specifications

{{Specifications}}

Browser compatibility

{{Compat}}

See also

• Polyfill of Math.log1p in core-js
• es-shims polyfill of Math.log1p
• {{jsxref("Math.exp()")}}
• {{jsxref("Math.log()")}}
• {{jsxref("Math.expm1()")}}
• {{jsxref("Math.log10()")}}
• {{jsxref("Math.log2()")}}
• {{jsxref("Math.pow()")}}