How do you solve this? Why does the answer become squareroot(3)-1??

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- Mar 27th 2012, 12:57 PM #1

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- Mar 27th 2012, 01:09 PM #2

- Mar 27th 2012, 01:55 PM #3

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- Mar 27th 2012, 01:56 PM #4

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- Mar 27th 2012, 01:59 PM #5

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## Re: logaithm help.

??? You will accept 3> 1 won't you? And square root is an increasing function so $\displaystyle \sqrt{3}> \sqrt{1}= 1$. In fact, it is not at all difficult to use a calculator to see that $\displaystyle \sqrt{3}= 1.7320508075688772935274463415059... > 1$.

Or do you mean "$\displaystyle ln(1- \sqrt{3})^2)= 2ln(\sqrt{3}- 1)$"?

Obviously both $\displaystyle (1- \sqrt{3})^2$ and $\displaystyle (\sqrt{3}- 1)^2$ give the same thing and $\displaystyle log(a^2)= 2log(a)$**provided**that a is positive becauser log(a) is not defined for negative a.

- Mar 27th 2012, 02:00 PM #6

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## Re: logaithm help.

Emm, I think you got me a little bit wrong here, I was just wondering why did we have to write sqrt(3)-1 instead of 1-sqrt(3).

Edit: Ah yeah, I totally forgot about looking at that part. Thank you.