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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  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,138 Thanks: 872 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
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 
Math Team Joined: Jul 2011 From: Texas Posts: 2,886 Thanks: 1505  
January 31st, 2015, 04:04 PM  #5  
Senior Member Joined: Apr 2014 From: Europa Posts: 575 Thanks: 176  Quote:
$\displaystyle \color{blue}{\sqrt3\sqrt2=(\sqrt3\sqrt2)\cdot \dfrac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}= \dfrac{(\sqrt3\sqrt2)(\sqrt3+\sqrt2)}{\sqrt3+\sqrt2}=\dfrac{( \sqrt3)^2(\sqrt2)^2}{\sqrt3+\sqrt2} = \dfrac{32}{\sqrt3+\sqrt2} = \dfrac{1}{\sqrt3+\sqrt2}}$  
January 31st, 2015, 04:11 PM  #6  
Senior Member Joined: Apr 2014 From: Europa Posts: 575 Thanks: 176  Quote:
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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