Question

How to use the function Math.log1p() to get the natural logarithm (base e) of 1 + a number in JavaScript

Answer 1

document.write("Math.log1p(0) = " + Math.log1p(0) + "<br />");
document.write("Math.log1p(1) = " + Math.log1p(1) + "<br />");
document.write("Math.log1p(-1) = " + Math.log1p(-1) + "<br />");
document.write("Math.log1p(2) = " + Math.log1p(2) + "<br />");
document.write("Math.log1p(-2) = " + Math.log1p(-2) + "<br />");
document.write("Math.log1p(5) = " + Math.log1p(5) + "<br />");
document.write("Math.log1p(10) = " + Math.log1p(10) + "<br />");
document.write("Math.log1p(100) = " + Math.log1p(100) + "<br />");
document.write("Math.log1p(100000) = " + Math.log1p(100000) + "<br />");
document.write("Math.log1p(125) = " + Math.log1p(125) + "<br />");
document.write("Math.log1p(125) / Math.log1p(5) = " + Math.log1p(125) / Math.log1p(5) + "<br />");

 
/*
run

Math.log1p(0) = 0
Math.log1p(1) = 0.6931471805599453
Math.log1p(-1) = -Infinity
Math.log1p(2) = 1.0986122886681098
Math.log1p(-2) = NaN
Math.log1p(5) = 1.791759469228055
Math.log1p(10) = 2.3978952727983707
Math.log1p(100) = 4.61512051684126
Math.log1p(100000) = 11.51293546492023
Math.log1p(125) = 4.836281906951478
Math.log1p(125) / Math.log1p(5) = 2.6991803252671502
 
*/

 

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