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Homework Help: Exponents with different bases

  1. Jan 14, 2009 #1


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    1. The problem statement, all variables and given/known data
    This is re-asking the question in https://www.physicsforums.com/showthread.php?t=273275" which mysteriously went quiet before the answer could be given... Anyway I also tried answering the question, but to no avail.

    2. Relevant equations

    3. The attempt at a solution

    Hence, [tex]2^{2x}=0[/tex] (1) or [tex]3^x-2^{4x-2}=0[/tex] (2)
    (1) has no solution.

    (2) [tex]\Rightarrow[/tex] [tex]3^x=2^{4x+2}[/tex]

    How can one go about solving this? different bases? It gives an answer of [tex]x\approx -0.82814[/tex]
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Jan 14, 2009 #2


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    Oh hang on...


    [tex]x\approx -0.82814449[/tex]

    Ahh never mind.
  4. Jan 14, 2009 #3
    12^x = 4X8^(2x)
    take logs of both sides: log(12x) = log(4 X 82x)
    Get rid of the power 2 (althought not essential) log(12x) = log(4 X 64x)
    Split RHS using the laws of logs and bring the powers out: x log(12) = log(4) + x log(64)
    see if u can do the last step (hint collect all terms with x)
  5. Jan 14, 2009 #4
    It would rather use log3.
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