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 January 30th, 2015, 04:49 PM #1 Newbie   Joined: Jan 2015 From: Asia Posts: 3 Thanks: 0 logarithms please help ! how do I prove that square root 3- square root 2 = 1/square root 3 + square root 2?
January 30th, 2015, 05:01 PM   #2
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Quote:
 Originally Posted by hi2you how do I prove that square root 3- square root 2 = 1/square root 3 + square root 2?
$\displaystyle \frac{1}{\sqrt{3} + \sqrt{2}}$

Try multiplying the numerator and denominator by the conjugate:
$\displaystyle \frac{1}{\sqrt{3} + \sqrt{2}} \cdot \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}}$

Can you finish it?

-Dan

 January 31st, 2015, 03:44 PM #3 Newbie   Joined: Jan 2015 From: Asia Posts: 3 Thanks: 0 but I'm supposed to find that square root 3 - square root 2 equates to 1/square root three + square root 2, so I can't work from the back..?
January 31st, 2015, 04:03 PM   #4
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Quote:
 Originally Posted by hi2you but I'm supposed to find that square root 3 - square root 2 equates to 1/square root three + square root 2, so I can't work from the back..?
much easier the way topsquark has recommended ... and that is working "from the back"

January 31st, 2015, 04:04 PM   #5
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Quote:
 Originally Posted by hi2you but I'm supposed to find that square root 3 - square root 2 equates to 1/square root three + square root 2, so I can't work from the back..?

$\displaystyle \color{blue}{\sqrt3-\sqrt2=(\sqrt3-\sqrt2)\cdot \dfrac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}= \dfrac{(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)}{\sqrt3+\sqrt2}=\dfrac{( \sqrt3)^2-(\sqrt2)^2}{\sqrt3+\sqrt2} = \dfrac{3-2}{\sqrt3+\sqrt2} = \dfrac{1}{\sqrt3+\sqrt2}}$

January 31st, 2015, 04:11 PM   #6
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Quote:
 logarithms please help ! how do I prove that square root 3- square root 2 = 1/square root 3 + square root 2?

Where is the logarithm ...?

 January 31st, 2015, 06:21 PM #7 Newbie   Joined: Jan 2015 From: Asia Posts: 3 Thanks: 0 sorry! because this question was in my logarithm homework questions so I thought I needed to use logarithms! But thank you for your help !!

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