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Homework Help: Strangely difficult log question

  1. Nov 4, 2004 #1
    485 + 5 ^ (x + 2) = 12 ^ (2x - 1)

    This question is driving me nuts. I can't take the log until both sides have been reduced to one expression each, right? If 485 were a power of 5 or 12, then I'd be able to solve it. As it is, I'm clueless. Can someone walk me through the steps to solving for x?

    And if you can't help me there, could you please tell me how to use those nifty image-generating tags for mathematical expressions?
  2. jcsd
  3. Nov 4, 2004 #2


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    I think you will have to resort to a numerical approximation.
  4. Nov 4, 2004 #3


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    There is no elementary formula for an exact solution of an equation in which the variable appears both in the exponent and not in the exponent. You might try looking at "Lambert's W function".

    It's defined as the inverse to f(x)= xex.

    Here's a link to the "MathWorld" page on it:

    I just looked at this again and noticed that both occurences of "x" ARE as exponents. The difficulty is really taking the logarithm of a sum. Yes, a numerical method is necessary here.
    Last edited by a moderator: Nov 5, 2004
  5. Nov 4, 2004 #4
    Wow, thanks for the help. I'm beginning to think that this is a typo... the question appears in a Math 12 book on a chapter introducing logs, so I doubt it should be THIS hard.
  6. Nov 4, 2004 #5


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    I may be missing the point, but..
    Take the natural logarithm of both sides:
    [tex]485 + 5^{(x + 2)} = 12^{(2x - 1)} \Rightarrow \ln(485)+(x+2)\ln(5)=(2x-1)\ln(12)[/tex]
    Which is a simple linear equation. The unknown only appears in the exponent.
  7. Nov 4, 2004 #6
    Unfortunately [tex]\ln(485 + 5^{(x+2)})[/tex] != [tex]\ln(485)+(x+2)\ln(5)[/tex] so your solution isn't quite correct.
  8. Nov 4, 2004 #7


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    Unfortunately, I have these moments often after a day of abstract algebra.

    I'm gonna lie down now... :zzz:
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