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Math Help - Logarithms Question HELP

  1. #1
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    Logarithms Question HELP

    Solve:

    4 + log(base 3)(2 - x) = 9

    I got:

    log(base 3)(2 - x) = 5
    2 - x = 3^5
    2 - x = 243
    -x = 241
    x = -241

    Don't think this is right can someone help me out? Thanks.
    Last edited by StephenSkinner; November 12th 2013 at 07:31 PM.
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  2. #2
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    Re: Logarithms Question HELP

    Your question was to solve 4+\log_3(2-x) = 5 and you solved \log_3(2-x) = 5. Those are not the same equations. To solve the first equation, subtract 4 from each side:

    \begin{align*}& \log_3(2-x) = 1 \\ \Rightarrow & 2-x = 3^1 = 3 \\ \Rightarrow & x = -1\end{align*}.
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  3. #3
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    Re: Logarithms Question HELP

    Surely 5 - 4 is NOT 5.
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  4. #4
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    Re: Logarithms Question HELP

    Whoops meant to put: 4 + log(base 3)(2-x) = 9

    SORRY.
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  5. #5
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    Re: Logarithms Question HELP

    Quote Originally Posted by StephenSkinner View Post
    Whoops meant to put: 4 + log(base 3)(2-x) = 9

    SORRY.
    Then try plugging in your answer:

    4+\log_3(2-(-241)) = 4+\log_3(243) = 4+\log_3(3^5) = 4+5\log_3(3) = 4+5(1) = 9

    It looks correct to me.
    Thanks from StephenSkinner
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  6. #6
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    Re: Logarithms Question HELP

    Quote Originally Posted by SlipEternal View Post
    Then try plugging in your answer:

    4+\log_3(2-(-241)) = 4+\log_3(243) = 4+\log_3(3^5) = 4+5\log_3(3) = 4+5(1) = 9

    It looks correct to me.
    Cool I appreciate it!
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