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satellite73

  • 2 years ago

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

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

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

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

  4. ParthKohli
    • 2 years ago
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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

  5. satellite73
    • 2 years ago
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    having the damndest time typesetting as well

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

  7. FoolForMath
    • 2 years ago
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    Is this right?

  8. satellite73
    • 2 years ago
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    yes that is it

  9. satellite73
    • 2 years ago
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    @experimentX you lost me there

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

  11. satellite73
    • 2 years ago
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    yes

  12. FoolForMath
    • 2 years ago
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    1/12 is the right answer.

  13. satellite73
    • 2 years ago
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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 \[\huge \frac{4^{\left(\frac{\ln(4)}{\ln(\frac{3}{4})}\right)}}{3^{\left(\frac{\ln(3)}{\ln(\frac{3}{4})}\right)}}\]

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

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

  16. experimentX
    • 2 years ago
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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

  17. satellite73
    • 2 years ago
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    not times, to the power of

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

  19. experimentX
    • 2 years ago
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    Oh .. my mistake

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

  21. FoolForMath
    • 2 years ago
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    http://math.stackexchange.com/questions/161685/

  22. experimentX
    • 2 years ago
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    i wish i could answer ... lol

  23. FoolForMath
    • 2 years ago
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    typesetting is a mess :/

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

  25. satellite73
    • 2 years ago
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    ?

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

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

  28. satellite73
    • 2 years ago
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    for in line i use \(

  29. FoolForMath
    • 2 years ago
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    Work : $$ \frac{a}{b}$$

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

  31. FoolForMath
    • 2 years ago
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    $$ one ;)

  32. Zarkon
    • 2 years ago
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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

  33. satellite73
    • 2 years ago
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    12 to some power?

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

  35. Zarkon
    • 2 years ago
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    the dot it the original problem

  36. Zarkon
    • 2 years ago
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    *is the..

  37. satellite73
    • 2 years ago
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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?

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

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

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

  41. satellite73
    • 2 years ago
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    and why? this of course gives it right away

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

  43. FoolAroundMath
    • 2 years ago
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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) \[\log_{b/a}y = \frac{1}{\log_{a}(b/a)}\log_{b/a}a - \frac{1}{\log_{b}(b/a)}\log_{b/a}b \] \[\log_{b/a}y =\frac{loga.loga}{\log(b/a)\log(b/a)} - \frac{logb.logb}{\log(b/a).\log(b/a)}\] \[\log_{b/a}y = \frac{(loga-logb)(loga+logb)}{\log(b/a).\log(b/a)}\] \[\log_{b/a}y = \frac{-\log(ab)}{\log(b/a)}\] \[\log_{b/a}y = \log_{b/a}(\frac{1}{ab})\] \[=> y = \frac{1}{ab}\]

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

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

  46. satellite73
    • 2 years ago
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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 \[\exp^{-\ln(ab)}=\frac{1}{ab}\]

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

  48. maheshmeghwal9
    • 2 years ago
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    Hmm............................ gr8* :D

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