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Math Help - log base 2 help

  1. #1
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    log base 2 help

    I have to find x in this problem

    logbase2(2^4x) = 20

    The 2 is raised to the power 4x.

    I started off by doing (2^4x)/ln2 = 20
    then got 2^4x = 20ln2 and got stuck too.
    Thanks.. I am trying to get ready for my quiz tommorow in calc 2.
    Thanks for all the help.
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  2. #2
    MHF Contributor red_dog's Avatar
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    We have \log_aa^n=n.
    So \log_22^{4x}=20\Leftrightarrow 4x=20\Leftrightarrow x=5
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  3. #3
    Bar0n janvdl's Avatar
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    Quote Originally Posted by red_dog View Post
    We have \log_aa^n=n.
    So \log_22^{4x}=20\Leftrightarrow 4x=20\Leftrightarrow x=5
    I solved it using the same method, but i thought i may have missed something because it's supposed to be university maths.
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  4. #4
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    Quote Originally Posted by red_dog View Post
    We have \log_aa^n=n.
    So \log_22^{4x}=20\Leftrightarrow 4x=20\Leftrightarrow x=5

    How did you get from logbase2 2^(4x) = 20 to 4x = 20?
    Last edited by davecs77; July 11th 2007 at 02:18 PM. Reason: oh i figured it out..need to know the rule. My bad thanks.
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  5. #5
    Bar0n janvdl's Avatar
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    Quote Originally Posted by davecs77 View Post
    How did you get from logbase2 2^(4x) = 20 to 4x = 20?
     log_{2} 2^{4x} = 20

     4x \ log_{2} 2 = 20

     But \ log_{2} 2 = 1

     4x (1) = 20

     x = 5
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