The Math.log() static method returns the natural logarithm (base e) of a number. That is

x > 0 , 𝙼𝚊𝚝𝚑.𝚕𝚘𝚐 ( 𝚡 ) = ln ( x ) = the unique  y  such that  e y = x \forall x > 0,;\mathtt{\operatorname{Math.log}(x)} = \ln(x) = \text{the unique } y \text{ such that } e^y = x

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A number greater than or equal to 0.

Return value

The natural logarithm (base e) of x. If x is ±0, returns -Infinity. If x < 0, returns NaN.


Because log() is a static method of Math, you always use it as Math.log(), rather than as a method of a Math object you created (Math is not a constructor).

If you need the natural log of 2 or 10, use the constants Math.LN2 or Math.LN10. If you need a logarithm to base 2 or 10, use Math.log2() or Math.log10(). If you need a logarithm to other bases, use Math.log(x) / Math.log(otherBase) as in the example below; you might want to precalculate 1 / Math.log(otherBase) since multiplication in Math.log(x) * constant is much faster.

Beware that positive numbers very close to 1 can suffer from loss of precision and make its natural logarithm less accurate. In this case, you may want to use Math.log1p instead.


Using Math.log()

Math.log(-1); // NaN
Math.log(-0); // -Infinity
Math.log(0); // -Infinity
Math.log(1); // 0
Math.log(10); // 2.302585092994046
Math.log(Infinity); // Infinity

Using Math.log() with a different base

The following function returns the logarithm of y with base x (i.e. log x y \log_x y ):

function getBaseLog(x, y) {
  return Math.log(y) / Math.log(x);

If you run getBaseLog(10, 1000), it returns 2.9999999999999996 due to floating-point rounding, but still very close to the actual answer of 3.


ECMAScript Language Specification
# sec-math.log

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See also