how can i increase by poing

[alzami]

in my previous post i posted a problem about my javascript calculator .in my calculator in made a log button which base is 10;it shows a perfect integer result.say it gives accurately 2 if i want to find the log of 100.but if i want log of 104 it doesn't work;how can i solve that

function log10(el){	var m=document.getElementById(el);		for(;{         var d=Math.pow(10,i);				 if(d==m.value){			 m.value=i;			 i=0;			 break;		 }		  i++;	}}	

[Ingolme]

If I remember correctly, you can get the base 10 logarithm out of the natural logarithm by dividing by the logarithm of the base:

function log10(n) {    return Math.log(n)/Math.LN10; //Math.LN10 is the logarithm of 10.}

You can also generalize the function:

function log(n,base) {    return Math.log(n)/Math.log(base);}

[alzami]

the problem is if i want log of 100 it gives 2.its not the problem.problem is if i want log of 104 ,it gives me no value.but normal scientific calculator gives value of log(104) .how can i persuade javascript to to such calculation?

[Ingolme]

The functions I just gave you solve those equations.

function log10(n) {    return Math.log(n)/Math.LN10; //Math.LN10 is the logarithm of 10.}alert(log10(104)); // Outputs 2.0170333392987803

[alzami]

Math.log(n)/Math.LN10;

is it a complete statement.or it meant i can use Math.log or Math.LN10?i have never encountered such kind of statement.

if u see my calculator codes i already used Math.log. it says Math.log is an e based logarithm which is different frm 10 based logarithm

Edited December 11, 2013 by alzami

[Ingolme]

It's a division, which I explained in a previous post.

you can get the base 10 logarithm out of the natural logarithm by dividing by the logarithm of the base

 

If you understand the properties of the logarithm you can obtain this expression.

 

ln(10^N) = N*ln(10)

therefore:

ln(10^N) / ln(10) = N * ln(10)/ln(10) = N

[Hadien]

I've recently used a formula to calculate in log base 10, and you do need to be wary of rounding errors when diving one log by another. for example:

 

log₂(100) / log₂(10) == log₁₀ (100) == 2

 

however in javascript Math.log(100) / Math.log(10) will equal something like 1.99999999...due to those floating-point rounding errors. something to be mindful of

[Ingolme]

My experiments haven't given any rounding errors up to now. Perhaps there might be if you using a floating point number as the argument., like 0.1 or 0.01.