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Math Help - extraneous roots

  1. #1
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    extraneous roots

    8(3)^2x+3 = 24

    Answer :

    log3^2x+3 = 3

    2x+3 = log3/log3

    2x + 3 = 1

    x = -1

    Is my method correct?
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  2. #2
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    Hi purple dinosaur, i'll kick this one off for you.

    When solving for x you need to undo some of these operations

    You have

    8(3)^{2x}+3 = 24

    Taking 3 from both sides gives

     8(3)^{2x} = 21
    Now diving both sides by 8 you get

    (3)^{2x} = \frac{21}{8}

    Now you need to employ a log

    2x = \log_3\frac{21}{8}
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  3. #3
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    Quote Originally Posted by purpledinosaur View Post
    8(3)^2x+3 = 24

    Answer :

    log3^2x+3 = 3

    2x+3 = log3/log3

    2x + 3 = 1

    x = -1

    Is my method correct?
    it works.

    would be easier to do this ...

    8 \cdot 3^{2x+3} = 24

    3^{2x+3} = 3

    2x+3 = 1

    x = -1
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  4. #4
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    my apologies to you, i think u started answering my question when i posted the original question. I stared it for about 20 mins, and didn't understand it but when i posted the question i sort of had an epiphany, and i edited my post. Thank you for clearing it for me. And for the explanation, much appreciated.
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  5. #5
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    you need to work on making the notation of your equations more clear.

    parentheses help alot.
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  6. #6
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    Quote Originally Posted by purpledinosaur View Post
    8(3)^2x+3 = 24
    Important point: if you do not use LaTex, use parentheses!
    "8(3)^2x+ 3" could be interpreted as 8(3^2)x+ 3, 8(3^{2x})+3 or 8(3^{2x+3}).

    From what you do below, you appear to mean 8(3^(2x+3)).


    Answer :

    log3^2x+3 = 3
    2x+3 = log3/log3

    2x + 3 = 1

    x = -1

    Is my method correct?
    Yes, by taking the logarithm of both sides of 3^{2x+3}= 3 you get (2x+3)log 3= log 3 so 2x+ 3= 1.

    As skeeter pointed out, it is more "fundamental" and easier to argue, from 3^{2x+3}= 3^1= 3 that, because exponentials are "one to one" functions ( a^x= a^y if and only if x= y) that we must have 2x+3= 1. From that, 2x= 1- 3= -2 and so x= -2/2= -1.
    Last edited by HallsofIvy; January 3rd 2010 at 04:01 AM.
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