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Math Help - Logarithms

  1. #1
    Newbie lmschneider's Avatar
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    Logarithms

    14^log14 of 21
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by lmschneider View Post
    14^log14 of 21
    there is a rule: a^{\log_a x} = x
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  3. #3
    Newbie lmschneider's Avatar
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    Do you happen to know what that rule is called?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by lmschneider View Post
    Do you happen to know what that rule is called?
    no, it is just one of the log rules that show up. it is very likely to be in a list of log rules in your text
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  5. #5
    Newbie lmschneider's Avatar
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    I looked and it's not, but that's okay. I will keep looking. So the answer is 21 then?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by lmschneider View Post
    I looked and it's not, but that's okay. I will keep looking. So the answer is 21 then?
    yes

    it is not hard to prove though. i will show you how to prove it. that way, if you are not allowed to use this rule, this is how you would attempt the problem in general.

    Let a^{\log_a x} = y, we will show that y = x

    a^{\log_a x} = y ...........................................take \log to the base a of both sides

    \Rightarrow \log_a a^{\log_a x} = \log_a y

    \Rightarrow \log_a x \cdot \log_a a = \log_a y .................since \log_a (x^n) = n \log_a x

    \Rightarrow \log_a x = \log_a y ...............................since \log_a a = 1

    \Rightarrow x = y ..............................................by taking the antilog of both sides.
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