Natural Logarithms and Exponentials – Roots and Powers

Filed in Mathematics 13 comments

today I was preparing myself for the math exam that I have for the next week

I learned the exponentional exp(x) and logarithms ln(x) functions

Uniqueness (or 1 to 1) Property:
If a > 0, b> 0 and  ln(a) = ln(b) then a = b..

Inversion Properties

  • ln(exp(x)) = x for all real x
  • exp(ln(x)) = x if x > 0

For each real number a,   x = exp(a) is the unique solution of  a =  ln(x).  Solving the latter equation  is one way to define or compute exp(a).

Fundamental property of logarithms

ln(ab) = ln(b) +ln(a)    (proof available in calculus)

Fundamental property of exponentials:

exp(x1) · exp(x2) = exp(x1+x2)

This follows from the uniqueness property of logarithms and the fundamental properties of logarithms.

The fundamental property of logarithms implies

  • ln( 1/a) = (-1) ln(a) as
    0 = ln(1) = ln ( (1/a) a )
  • ln(a m) = m ln (a) for all whole numbers and then for all integers. integers.

Logarithms to base c > 0.

The logarithm of x > 0 to a base c > 0 is given by

logc(x) = ln(x)


ln(c)
·

Here ln(e) = 1 implies loge(x) = ln(x).

The logarithm of x > 0 to a base 10 is given by

log(x) = log10(x) = ln(x)


ln(a)

The button log(x) on a calculator computes log10(x).

The definition of logc(x) in terms of ln(x) implies

  • logc(ab) = logc(b) + logc(a) for a> 0, b > 0 and c >0
  • logc( 1/a) = (-1) logc(a) as  0 = logc(1) = logc( (1/a) a )
  • logc(a m) = m logc (a)

No related posts.

Posted by voshka   @   1 January 2010 13 comments

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13 Comments

Comments
Mar 10, 2010
8:03 pm
#1 Zellcorp :

I really like when people are expressing their opinion and thought. So I like the way you are writing

Author Mar 10, 2010
8:08 pm
#2 voshka :

Thank you:)
hopefully I passed that exam :-p

Apr 15, 2010
12:55 am

works for me

Apr 21, 2010
2:56 pm
#4 free trial :

After reading you site, Your site is very useful for me .I bookmarked your site!

Author Apr 22, 2010
12:32 pm
#5 voshka :

thanks

Nov 24, 2010
11:15 pm
#6 pauk :

how to calculate this function on action script?

http://infamer.ru/kper.png

private function kper(p:Number,i:Number,a:Number) {
var ret:Number = Math.log(p/a+1) / Math.log(1+i);
return ret;
}
work wrong :(

Author Nov 25, 2010
1:24 am
#7 voshka :

Hi,pauk

working on it.
Please be a little patient

Thanks

Nov 25, 2010
1:57 am
#8 pauk :

This is a correct realization of kper; pmt must be positive number :)

private function kper(rate:Number,pmt:Number,n:Number) {
var ret:Number = Math.log(pmt/(pmt – rate * n))/Math.log(1+rate);
//trace(“kper : “+ret);
return ret;
}

Author Nov 25, 2010
2:03 am
#9 voshka :

Hi,pauk

I changed your code like this
and working just fine for me

kper(1,2,3);
var ret:Number
function kper(p:Number, i:Number, a:Number):Number
{
ret = Math.log(p / a + 1) / Math.log(1 + i);

return ret;

}
trace(ret);

is that the thing you want from this function?
and do you want the i or p or a to be incremented each time within your function or it just need one value to calculate

THanks

Author Nov 25, 2010
3:04 am
#10 voshka :

Hi again pauk

I created for this a class that you can modify and reuse it easily

package
{
import flash.display.Sprite;

public class Tests extends Sprite
{
private var ret:Number;

public function Tests(p:Number=1,i:Number=2,a:Number=3)
{
kper(1,2,3);
trace(ret);
}
private function kper(p:Number,i:Number,a:Number):Number
{

ret = Math.log(p / a + 1) / Math.log(1 + i);

return ret;

}

}
}

Thanks

Dec 3, 2010
7:13 am
#11 pauk :

thanx for code, i already do this job.

code Math.log(p/a+1) / Math.log(1+i) do result not like excel :(
this code Math.log(pmt/(pmt – rate * n))/Math.log(1+rate) return results as excel;

Work of this code u can see there – http://ba-finans.ru/cc/.

Good cooding :)

Author Dec 3, 2010
7:23 am
#12 voshka :

Hi,pauk
the coding structure is fine.

i think the rpoblem was calculation with your variables that you didn’t tell me about that,if you did i would suggest you what to do ;)

seems you have found the solution,haven’t you ?

If not i would be happy of any assistance to my russian friend ;)

after all the place you commented had no relation to ActionScript :)

Thanks

Feb 3, 2012
12:24 pm
#13 jfqqhzxcm :

PwEZTY cyvitlmvoqxl

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