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Thread: Can i get help with this logerithm problem

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

    $\displaystyle 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
    $\displaystyle log{\sqrt{3}} $ 9
    what exactly are you supposed to do with this? is $\displaystyle \sqrt{3}$ 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 $\displaystyle \log_{\sqrt{3}}9$? 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 $\displaystyle \log_a b = c$ then $\displaystyle a^c = b$


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



    now, $\displaystyle \log_{\sqrt{3}}9$ is the power we have to raise $\displaystyle \sqrt{3}$ 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$\displaystyle \sqrt{3} $ to equal 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
    I'm guessing it's 2, since we have to raise 2 to$\displaystyle \sqrt{3} $ to equal 9.
    what?! are you confused about something? where did 2 come from?

    first of all, $\displaystyle 2^{\sqrt{3}}$ 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 $\displaystyle \sqrt{3}^x = 9$ for $\displaystyle x$
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    $\displaystyle \sqrt {3}^4=9 $ so the answer is 4.
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    Quote Originally Posted by Darkhrse99 View Post
    $\displaystyle \sqrt {3}^4=9 $ so the answer is 4.
    yes
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    Cool!!
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