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Math Help - Can i get help with this logerithm problem

  1. #1
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    Can i get help with this logerithm problem

     log{\sqrt{3}} 9
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Darkhrse99 View Post
     log{\sqrt{3}} 9
    what exactly are you supposed to do with this? is \sqrt{3} the base?
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    I'm suppose to find the exact value of this logerithm. I believe that is the base.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Darkhrse99 View Post
    I'm suppose to find the exact value of this logerithm. I believe that is the base.
    does it look like \log_{\sqrt{3}}9? if so, it is the base.

    remember what a logarithm is, logarithms are powers.


    Definition: the logarithm of a number to a given base is the power to which the base must be raised to give the number.

    that is, if \log_a b = c then a^c = b


    example, \log_28 = 3 since we have to raise the base 2 to the power 3 to get 8



    now, \log_{\sqrt{3}}9 is the power we have to raise \sqrt{3} to to get 9. so what number is that?
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    I'm guessing it's 2, since we have to raise 2 to  \sqrt{3} to equal 9.
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    Quote Originally Posted by Darkhrse99 View Post
    I'm guessing it's 2, since we have to raise 2 to  \sqrt{3} to equal 9.
    what?! are you confused about something? where did 2 come from?

    first of all, 2^{\sqrt{3}} is not 9

    second of all, there are no 2's here

    third of all, we want to know what power we have to raise the base to, that is we want to solve \sqrt{3}^x = 9 for x
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     \sqrt {3}^4=9 so the answer is 4.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Darkhrse99 View Post
     \sqrt {3}^4=9 so the answer is 4.
    yes
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    Cool!!
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