Simple algebra

• Mar 10th 2008, 06:05 AM
Altair
Simple algebra
Looks like simple algebra but can't do it.

The numerical vale of the following forms come equal. But I can't convert the 1st form into 2nd :-

1st. $ln {\frac{(\sqrt{1+x^2} - 1)(\sqrt{1+y^2} - 1)}{|xy|}}$

2nd. $-ln {\frac{(\sqrt{1+x^2} + 1)(\sqrt{1+y^2} + 1)}{|xy|}}$

• Mar 10th 2008, 06:15 AM
colby2152
Quote:

Originally Posted by Altair
Looks like simple algebra but can't do it.

The numerical vale of the following forms come equal. But I can't convert the 1st form into 2nd :-

1st. $ln {\frac{(\sqrt{1+x^2} - 1)(\sqrt{1+y^2} - 1)}{|xy|}}$

2nd. $-ln {\frac{(\sqrt{1+x^2} + 1)(\sqrt{1+y^2} + 1)}{|xy|}}$

For this to be true, then:

$\frac{(\sqrt{1+x^2} - 1)(\sqrt{1+y^2} - 1)}{|xy|} = \left[\frac{|xy|}{(\sqrt{1+x^2} + 1)(\sqrt{1+y^2} + 1)}\right]$

$(\sqrt{1+x^2} - 1)(\sqrt{1+y^2} - 1)(\sqrt{1+x^2} + 1)(\sqrt{1+y^2} + 1) = (xy)^2$

$(1+x^2-1)(1+y^2-1) = (xy)^2$

$(xy)^2 = (xy)^2$

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• Mar 10th 2008, 06:33 AM
Altair
I want the conversion from 1st form to the 2nd.Not its proof.
• Mar 10th 2008, 07:17 AM
colby2152
Quote:

Originally Posted by Altair
I want the conversion from 1st form to the 2nd.Not its proof.

Showing the path from one form to another is a proof. It is unfortunate that my answer is not good enough for you.

Why did I setup the first line of my problem like so?

This comes from a log property: $log(a^b)=b*log(a)$
In this case, $b=-1$ hence why the 2nd term in my 1st line is a reciprocal of what is being logged in your 2nd equation.