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Exponents with different bases

  1. Jan 14, 2009 #1

    Mentallic

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    1. The problem statement, all variables and given/known data
    This is re-asking the question in Logarithms! which mysteriously went quiet before the answer could be given... Anyway I also tried answering the question, but to no avail.

    2. Relevant equations
    [tex]a^m.a^n=a^{m+n}[/tex]
    [tex](a^m)^n=a^{mn}[/tex]
    [tex]log(ab)=loga+logb[/tex]
    [tex]log(a^b)=bloga[/tex]

    3. The attempt at a solution
    [tex]12^x=4.8^{2x}[/tex]
    [tex]3^x.(2^2)^x=2^2(2^3)^{2x}[/tex]
    [tex]3^x.2^{2x}-2^{6x+2}=0[/tex]
    [tex]2^{2x}(3^x-2^{4x-2})=0[/tex]

    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]
    [tex]xlog3=(4x+2)log2[/tex]

    How can one go about solving this? different bases? It gives an answer of [tex]x\approx -0.82814[/tex]
     
  2. jcsd
  3. Jan 14, 2009 #2

    Mentallic

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    Oh hang on...

    [tex]xlog3=(4x+2)log2[/tex]
    [tex]xlog3=4xlog2+2log2[/tex]
    [tex]x(log3-4log2)=2log2[/tex]
    [tex]x=\frac{2log2}{log3-4log2}[/tex]

    [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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