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Math Help - logaithm help.

  1. #1
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    logarithm help.



    How do you solve this? Why does the answer become squareroot(3)-1??
    Last edited by Iamnothere; March 27th 2012 at 01:58 PM.
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  2. #2
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    Re: logaithm help.

    Quote Originally Posted by Iamnothere View Post
    How do you solve this? Why does the answer become squareroot(3)-1??
    Because \sqrt3>1 we must rewwrite \log_{25}(1-\sqrt3)^2=2\log_{25}(\sqrt3-1)

    Note 25^{\log_{25}(\sqrt3-1)}=\sqrt3 -1
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  3. #3
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    Re: logaithm help.

    5= 25^{1/2} so
    5^{log_{25}(a)}= (25^{1/2})^{log_{25}(a)}= 25^{(1/2)log_{25}(a)}= 25^{log_{25}(\sqrt{a})}= \sqrt{a}
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    Re: logaithm help.

    Do you mind me asking where does that rule come from? The one with sqrt(3)>1, I'm a little bit confused.
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    Re: logaithm help.

    Quote Originally Posted by Iamnothere View Post
    Do you mind me asking where does that rule come from? The one with sqrt(3)>1, I'm a little bit confused.
    ??? You will accept 3> 1 won't you? And square root is an increasing function so \sqrt{3}> \sqrt{1}= 1. In fact, it is not at all difficult to use a calculator to see that \sqrt{3}= 1.7320508075688772935274463415059... > 1.

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

    Obviously both (1- \sqrt{3})^2 and (\sqrt{3}- 1)^2 give the same thing and log(a^2)= 2log(a) provided that a is positive becauser log(a) is not defined for negative a.
    Last edited by HallsofIvy; March 27th 2012 at 02:04 PM.
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  6. #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.
    Last edited by Iamnothere; March 27th 2012 at 02:07 PM.
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