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12^x = 18

  1. Jun 2, 2005 #1
    Hi everyone. Here's a simple problem I need help with:

    Find x such that [itex]12^x = 18[/itex]

    From one point of view, [itex]x = log(18)/log(12)[/itex] and the problem is solved.

    However, if we write 12 as [itex]3*2^2[/itex] and 18 as [itex]3^2*2[/itex] then,

    [itex](2 * 3^2) = 3^x * 2^{2x}[/itex]

    and hence by the uniqueness of prime factorization (in particular that of the exponents of the prime factors),

    x = 2
    and 2x = 1

    but these equations do not have a consistent solution. I think the error is in the second reasoning.

    Can someone help please?

    Last edited: Jun 2, 2005
  2. jcsd
  3. Jun 3, 2005 #2


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    You can only guarantee unique factorization if you have integral exponents. When you try to invoke the fundamental theorem of arithmetic you would also need the assumption that the exponents x and 2x are integers. Your contradiction only tells you that there is no integer x that satisfies the original equation.
  4. Jun 3, 2005 #3
    Thanks Shmoe.
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