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FoolForMath

Easy yet cute problem, if \( p^{12}=q^6=r^3=s^2, p \neq 1 \). Find the value of \(\log_p pqrs \). Lets see who can solve this the fastest. Pleas read on for the extra question (posted later): How many values of \(x \) satisfy the equation \( \log (2x)=\frac 14 \log (x-15)^4 \)?

  • one year ago
  • one year ago

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  1. matricked
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    is it 13

    • one year ago
  2. apoorvk
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    23/12

    • one year ago
  3. apoorvk
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    13

    • one year ago
  4. Limitless
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    q=p^2 r=p^4 s=p^6 2+4+6+1=13

    • one year ago
  5. apoorvk
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    I think too much actually :/ and act too hastily. :/ It is 13.

    • one year ago
  6. FoolForMath
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    I like posting sitters for a change ... :P

    • one year ago
  7. asnaseer
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    I think it can have several values:\[1+4\pm2\pm6\]

    • one year ago
  8. asnaseer
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    since q and s can be negative

    • one year ago
  9. Limitless
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    Asnaseer, we do not know if the even number radical function is defined for strictly positive or negative. Positive is assumed.

    • one year ago
  10. asnaseer
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    so are you saying, for example, that:\[s=-p^6\]is not a valid solution to this?

    • one year ago
  11. Limitless
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    Hmm... Rather hard to say exactly what I mean. It's true that \(s=-p^6\). But, I assumed we were using the positive roots only because that's conventional. Otherwise, as you see, our function is actually a multivalued relation rather than a function and thus the logarithm becomes murkily defined. (I am potentially abusing terminology here.)

    • one year ago
  12. asnaseer
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    maybe FFM can clarify?

    • one year ago
  13. Limitless
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    My main point is that if we allow ourselves to take both of the roots, the logarithm is no longer a function. This can't be the case, since the logarithm is always a function.

    • one year ago
  14. asnaseer
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    BTW: You are most probably correct Limitless. I am not a /true/ mathematician as such - I only follow maths as a hobby.

    • one year ago
  15. Limitless
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    I hope you realize that I am still impressed by your solution, critique aside.

    • one year ago
  16. asnaseer
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    :) thx

    • one year ago
  17. FoolForMath
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    Asnaseer is technically right, but I am happy with the less pedantic solution for this one.

    • one year ago
  18. asnaseer
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    Is there some article on this area that I could read about to help me understand this better?

    • one year ago
  19. Limitless
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    asnaseer, precisely what confuses you? This is more of a notion of convention rather than of serious mathematical concepts.

    • one year ago
  20. asnaseer
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    your description of log being a function and therefore it can only have one value - I was just wondering if there was any formal definition of this concept somewhere?

    • one year ago
  21. apoorvk
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    Well. I seem to think that 'q' and 's' can be negative, but their resultant powers would definitely be positive - so how does the question of the powers being 'negative' arise? s= (-p)^6 ---> alrighty! but the exponent can only be '6' right? of the exponent changes sign, then the whole thingy changes. Or am I totally wrong once again? -_-

    • one year ago
  22. Limitless
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    http://en.wikipedia.org/wiki/Function_(mathematics)#Formal_definition :) I rather love formal definitions if they are also intuitively beautiful. This is one such case. The reason that functions are considered injective and surjective (i.e. one-to-one, which is what I have been saying) is that this allows us to suppose the existence of their inversion via another function! It's rather interesting stuff. :)

    • one year ago
  23. asnaseer
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    good point apoorvk!

    • one year ago
  24. asnaseer
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    does that mean that:\[\log_p(-p)^6\]is undefined?

    • one year ago
  25. asnaseer
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    thx for the link Limitless.

    • one year ago
  26. asnaseer
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    sorry I meant:\[\log_p-(p)^6\]

    • one year ago
  27. FoolForMath
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    In real yes.

    • one year ago
  28. Limitless
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    hmm.. Good question. \(\log_{p}(-p^6)=\log_{p}(-1)+\log_{p}(p^6)=\log_{p}(-1)+6\) Here's an intuitive way of thinking: What in the world is \(\log_{p}(-1)\)???

    • one year ago
  29. FoolForMath
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    I mean it is not defined in \( \mathbb{R}\)

    • one year ago
  30. asnaseer
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    It always amazing me how even /seemingly/ simple problems can generate a whole raft of new learning! :)

    • one year ago
  31. asnaseer
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    *amazes

    • one year ago
  32. apoorvk
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    Yup^!

    • one year ago
  33. FoolForMath
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    There is a tricky problem on this topic that I encountered last month,I will post it some time later.

    • one year ago
  34. asnaseer
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    BTW: The wolf says \(\log(-1)=i\pi\)

    • one year ago
  35. asnaseer
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    so FFM - you are right - it does not belong in the reals

    • one year ago
  36. Limitless
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    You will find it very intriguing if you ask yourself why we restrict ourselves to certain domains on popular function. It taught me many deep ideas to ponder this. A rather elementary reason is that much of a function cannot even be graphed on a real 2-dimensional plane. Consider \(f(x)=x^2\) at \(x=i\). It is clear that we can graph \(f(i)=-1\), however how do we plot \(i\)??? This is where it becomes very intriguing: Complex numbers require special plotting techniques or special geometric spaces. :)

    • one year ago
  37. asnaseer
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    You just melted by brain Limitless! (in a good way) :D

    • one year ago
  38. Limitless
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    Well, I am glad I could! :) May someone more enlightened than me correct me if I am misrepresenting things.

    • one year ago
  39. FoolForMath
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    How many values of \(x\) satisfy the equation \( \log (2x)=\frac 14 (x-15)^4 \)?

    • one year ago
  40. lgbasallote
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    42 =_=

    • one year ago
  41. asnaseer
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    Limitless: That little plotting problem you posed up there is indeed very intriguing. Looks like my weekend is going to be filled with reading material :)

    • one year ago
  42. Limitless
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    Good lord, FFM. Surely this does not require numerical methods. D: I hope so, Asn. I had much spare time in class, LOL. You ask yourself questions like this out of boredom.

    • one year ago
  43. FoolForMath
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    Well for complex numbers we need a completely new reference axes, The argand plane.

    • one year ago
  44. FoolForMath
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    *plotting

    • one year ago
  45. asnaseer
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    FFM: Can this be solved analytically?

    • one year ago
  46. FoolForMath
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    Oh yes, 1 line if you start sufficiently from left ;)

    • one year ago
  47. asnaseer
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    ah! - two values?

    • one year ago
  48. Limitless
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    This whole factor \((x-15)^5\) is turning me off.

    • one year ago
  49. FoolForMath
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    No. Btw consider \(\mathbb{R} \)

    • one year ago
  50. asnaseer
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    I was thinking in \(\log(2x)\), 2x cannot be negative, therefore x must be positive. And I /assumed/ that \(\frac 14 (x-15)^4\) would have 2 negative and 2 positive solutions, so only 2 solutions possible.

    • one year ago
  51. Limitless
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    \(x\) is not an integer. Is this true?

    • one year ago
  52. FoolForMath
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    No.

    • one year ago
  53. experimentX
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    \( \log(2x)^4 = (x-15)^4\)

    • one year ago
  54. Limitless
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    I missed something, then.

    • one year ago
  55. experimentX
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    |dw:1338633080637:dw|

    • one year ago
  56. experimentX
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    can't find the solution analytically though!!

    • one year ago
  57. asnaseer
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    that graph makes it look like there are indeed only 2 solutions

    • one year ago
  58. Limitless
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    Which are symmetric.

    • one year ago
  59. asnaseer
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    symmetric in which sense?

    • one year ago
  60. experimentX
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    yes ... expoential function >> polynomial function >> log function

    • one year ago
  61. Limitless
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    Oops. I was wrong, sorry.

    • one year ago
  62. FoolForMath
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    Then the two solutions are x =? and x = ? .. :)

    • one year ago
  63. experimentX
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    lol ... i hope you don't mind if i use guessing technique

    • one year ago
  64. asnaseer
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    FFM: you are changing the goal posts here - your question was: How many values of x satisfy the equation...

    • one year ago
  65. FoolForMath
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    Yes, but can't we find the solutions if there are two solutions?

    • one year ago
  66. experimentX
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    i asked the same type of question of MSE they showed me Newton's method

    • one year ago
  67. experimentX
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    lol :D

    • one year ago
  68. Limitless
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    ^yes.

    • one year ago
  69. Limitless
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    Exact same experience. I don't even think Lambert W can be applied.

    • one year ago
  70. asnaseer
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    is the log using base 10 or base e?

    • one year ago
  71. FoolForMath
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    common logarithm base 10 I would say.

    • one year ago
  72. FoolForMath
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    Well x=5 is the only solution. If you find the other one please let me know :)

    • one year ago
  73. Limitless
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    \(x=5\) isn't a solution. Unless my math skills are horribly wrong right now.

    • one year ago
  74. asnaseer
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    x=5 does not seem to satisfy the equation?

    • one year ago
  75. asnaseer
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    this is the equation right?:\[\log (2x)=\frac 14 (x-15)^4\]

    • one year ago
  76. FoolForMath
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    I just checked in mathematica: x = 5; 1/4 Log[10, (x - 15)^4] = Log [10, 2 x]=1

    • one year ago
  77. FoolForMath
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    OMG!! I didn't mention the log carp :(

    • one year ago
  78. experimentX
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    what am i getting?? http://www.wolframalpha.com/input/?i=solve+%28x-15%29%5E4+-+log%5B10%2C+%282x%29%5E4%5D+%3D+0

    • one year ago
  79. FoolForMath
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    \[ \log (2x)=\frac 14 \log (x-15)^4 \]

    • one year ago
  80. asnaseer
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    bad FFM - bad FFM - go stand in a corner now! :)

    • one year ago
  81. FoolForMath
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    VERY VERY VERY poor penman ship :( My apologies to all :(

    • one year ago
  82. Limitless
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    Whoa whoa whoa. One moment. Is this \(\log(x-15)^4\) or is it meant to be \(\log[(x-15)^4]\)?

    • one year ago
  83. FoolForMath
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    I think both are same.

    • one year ago
  84. Limitless
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    That implies \(x=-15\) is a solution.

    • one year ago
  85. FoolForMath
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    No. that's the tricky part. Try substituting x=-15 in the equation.

    • one year ago
  86. Limitless
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    So, probably not. But this is cool, so here you go: \(\log(2x)=\frac{1}{4}\log((x-15)^4)\) \(4\log(2x)=\log((x-15)^4)\) \(\log((2x)^4)=\log((x-15)^4)\) \(2x=x-15\) \(x=-15\)

    • one year ago
  87. asnaseer
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    \[\log(2x)=\frac{1}{4}\log(x-15)^4\]therefore:\[\log(2x)^4=\log(x-15)^4\]\[(2x)^4=(x-15)^4\]\[2x=\pm(x-15)\]

    • one year ago
  88. FoolForMath
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    Yes yes but as I said x=-15 is not a valid solution. Okay I am going to going to dig a hole and hide my face on it. I will see you guys later.

    • one year ago
  89. asnaseer
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    therefore only ONE real solution, x=5

    • one year ago
  90. FoolForMath
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    That's right asnaseer.!

    • one year ago
  91. Limitless
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    Asnaseer has slain me. :p

    • one year ago
  92. Limitless
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    I forgot, once again, the \(\pm\).

    • one year ago
  93. asnaseer
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    :) we all have our days - in my case - seconds, of fame :)

    • one year ago
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