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satellite73

\[\frac{4^{\left(\frac{1}{\log_4(\frac{3}{4})}\right)}}{3^{\left(\frac{1}{\log_3(\frac{3}{4})}\right)}}\]

  • one year ago
  • one year ago

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  1. ParthKohli
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    Wasnt that log4^3/4 and log_3^3/4 asked in a question before?

    • one year ago
  2. satellite73
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    it was ambiguous before apparently it is \(\frac{1}{12}\)

    • one year ago
  3. satellite73
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    tried changing the base, and also rewriting the logs as a difference, but it is not coming snappy at all

    • one year ago
  4. ParthKohli
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    So we just have to find \(\Large {4^{1 } \over 3^{1 \over 12}}\)? I'm new with logs so mercy

    • one year ago
  5. satellite73
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    having the damndest time typesetting as well

    • one year ago
  6. FoolForMath
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    For my own reason: \[ \huge \frac{4^{\left(\frac{1}{\log_4(\frac{3}{4})}\right)}}{3^{\left(\frac{1}{\log_3(\frac{3}{4})}\right)}} \]

    • one year ago
  7. FoolForMath
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    Is this right?

    • one year ago
  8. satellite73
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    yes that is it

    • one year ago
  9. satellite73
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    @experimentX you lost me there

    • one year ago
  10. experimentX
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    i got \[ \frac 4 3 \log_4 3\] is 1/12 supposed to be answer??

    • one year ago
  11. satellite73
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    yes

    • one year ago
  12. FoolForMath
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    1/12 is the right answer.

    • one year ago
  13. satellite73
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    \[\huge \frac{4^{\left(\frac{1}{\log_4(\frac{3}{4})}\right)}}{3^{\left(\frac{1}{\log_3(\frac{3}{4})}\right)}}\] yes it is i do not see the gimmick change base? rewrite as \[\huge \frac{4^{\left(\frac{\ln(4)}{\ln(\frac{3}{4})}\right)}}{3^{\left(\frac{\ln(3)}{\ln(\frac{3}{4})}\right)}}\]

    • one year ago
  14. FoolForMath
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    I am gonna M.SE it .. if you don't have any objection?

    • one year ago
  15. satellite73
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    i do not know what that means, but i have no objection this was a previous problem so go nuts

    • one year ago
  16. experimentX
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    lol ... something went wrong http://www.wolframalpha.com/input/?i=4%2F3+*+%281+%2Flog+base+4+%283%2F4%29%29%2F%281+%2Flog+base+3+%283%2F4%29%29

    • one year ago
  17. satellite73
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    not times, to the power of

    • one year ago
  18. satellite73
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    somehow we are supposed to arrive at \(\frac{1}{4\times 3}\)

    • one year ago
  19. experimentX
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    Oh .. my mistake

    • one year ago
  20. satellite73
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    thing maybe rewrite exponents as \(\frac{\log_4(4)}{\log_4(\frac{3}{4})}\)

    • one year ago
  21. FoolForMath
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    http://math.stackexchange.com/questions/161685/

    • one year ago
  22. experimentX
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    i wish i could answer ... lol

    • one year ago
  23. FoolForMath
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    typesetting is a mess :/

    • one year ago
  24. satellite73
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    oh maybe we can change the base to \(\frac{4}{3}\)!!

    • one year ago
  25. satellite73
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    ?

    • one year ago
  26. experimentX
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    it seems that i can wrap latex with $ here too .. i thought \[ only works here

    • one year ago
  27. satellite73
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    doesn't work for me $\frac{a}{b}$

    • one year ago
  28. satellite73
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    for in line i use \(

    • one year ago
  29. FoolForMath
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    Work : $$ \frac{a}{b}$$

    • one year ago
  30. satellite73
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    ok how about changing the base to \(\frac{4}{3}\) maybe that would work

    • one year ago
  31. FoolForMath
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    $$ one ;)

    • one year ago
  32. Zarkon
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    I changed it to 12 to some power...then converted the exponent to natural logs...simplified and got -1 ie \[12^{-1}\] though I'm lazy and am not going to type that much

    • one year ago
  33. satellite73
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    12 to some power?

    • one year ago
  34. Zarkon
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    \[12^{\log_{12}(\cdot)}\]

    • one year ago
  35. Zarkon
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    the dot it the original problem

    • one year ago
  36. Zarkon
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    *is the..

    • one year ago
  37. satellite73
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    you mean you converted \(\frac{4}{3}\) ... oh ok but that gimmick works if you know the answer i guess, otherwise where would the 12 come from?

    • one year ago
  38. Zarkon
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    I also did it for \[a\times b\]

    • one year ago
  39. satellite73
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    christ no wonder you are not going to write it here, i can barely write it on paper

    • one year ago
  40. Zarkon
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    \[\huge \frac{a^{\left(\frac{1}{\log_a(\frac{b}{a})}\right)}}{b^{\left(\frac{1}{\log_b(\frac{b}{a})}\right)}}=\frac{1}{a\times b}\]

    • one year ago
  41. satellite73
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    and why? this of course gives it right away

    • one year ago
  42. Zarkon
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    i derived it...seemed a natural thing to do...combine into one exponent

    • one year ago
  43. FoolAroundMath
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    \[y = \frac{a^{\frac{1}{\log_{a}(\frac{b}{a})}}}{b^{\frac{1}{\log_{3}(\frac{b}{a})}}}\] take log on both sides with base (b/a) \[\log_{b/a}y = \frac{1}{\log_{a}(b/a)}\log_{b/a}a - \frac{1}{\log_{b}(b/a)}\log_{b/a}b \] \[\log_{b/a}y =\frac{loga.loga}{\log(b/a)\log(b/a)} - \frac{logb.logb}{\log(b/a).\log(b/a)}\] \[\log_{b/a}y = \frac{(loga-logb)(loga+logb)}{\log(b/a).\log(b/a)}\] \[\log_{b/a}y = \frac{-\log(ab)}{\log(b/a)}\] \[\log_{b/a}y = \log_{b/a}(\frac{1}{ab})\] \[=> y = \frac{1}{ab}\]

    • one year ago
  44. experimentX
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    $$ \huge a = b^{\log_ba} $$ $$ \huge \frac{\log_ba}{\log_a b - 1 } - \frac{1}{1 - \log_ba} = \frac{(\log_ba)^2}{1-\log_ba} - \frac{1}{1-\log_ba} \\ \frac{-(1 + \log_ba)(1 - \log_ba)}{(1 - \log_ba)} = -(\log_b + \log_a) = -\log_b(ab) $$

    • one year ago
  45. experimentX
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    $$ \huge b^{-\log_b(ab)} = 1/ab$$

    • one year ago
  46. satellite73
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    ok here is my effort \[\large \exp{\frac{\ln(a)}{\log_a(\frac{b}{a})}}=\exp{\frac{\ln^2(a)}{\ln(b)-\ln(a)}}\] and similarly \[\large \exp{\frac{\ln(b)}{\log_b(\frac{b}{a})}}=\exp\frac{\ln^2(b)}{\ln(b)-\ln(a)}\] subtract and get \[\frac{\ln^2(a)-\ln^2(b)}{\ln(b)-\ln(a)}=-(\ln(b)+\ln(a))=-\ln(ab)\]and finally \[\exp^{-\ln(ab)}=\frac{1}{ab}\]

    • one year ago
  47. satellite73
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    using the definition \(a^x=e^{x\ln(a)}\) i have to say other ways were snappier though

    • one year ago
  48. maheshmeghwal9
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    Hmm............................ gr8* :D

    • one year ago
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