/JavaScript

# Math.log1p()

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

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

## Syntax

Math.log1p(x)


### Parameters

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.

## 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 $lim x → 0 log ⁡ ( 1 + x ) x = 1 \lim_{x \to 0} \frac{\log(1+x)}{x} = 1$. 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).

## Examples

### Using Math.log1p()

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


## Browser compatibility

Desktop Mobile Server
Chrome Edge Firefox Internet Explorer Opera Safari WebView Android Chrome Android Firefox for Android Opera Android Safari on IOS Samsung Internet Deno Node.js
log1p
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No
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8
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3.0
1.0
0.12.0