---
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](/en-US/docs/Web/JavaScript/Reference/Global_Objects/Math/E)) of `1 + x`, where `x` is the argument. That is:
{{InteractiveExample("JavaScript Demo: Math.log1p()")}}
```js interactive-example
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
```js-nolint
Math.log1p(x)
```
### Parameters
- `x`
- : A number greater than or equal to -1.
### Return value
The natural logarithm (base [e](/en-US/docs/Web/JavaScript/Reference/Global_Objects/Math/E)) of `x + 1`. If `x` is -1, returns [`-Infinity`](/en-US/docs/Web/JavaScript/Reference/Global_Objects/Number/NEGATIVE_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: . 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 1.11022β¦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()
```js
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`](https://github.com/zloirock/core-js#ecmascript-math)
- [es-shims polyfill of `Math.log1p`](https://www.npmjs.com/package/math.log1p)
- {{jsxref("Math.exp()")}}
- {{jsxref("Math.log()")}}
- {{jsxref("Math.expm1()")}}
- {{jsxref("Math.log10()")}}
- {{jsxref("Math.log2()")}}
- {{jsxref("Math.pow()")}}