Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

satellite73 Group Title

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

  • 2 years ago
  • 2 years ago

  • This Question is Closed
  1. ParthKohli Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • 2 years ago
  2. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    it was ambiguous before apparently it is \(\frac{1}{12}\)

    • 2 years ago
  3. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    tried changing the base, and also rewriting the logs as a difference, but it is not coming snappy at all

    • 2 years ago
  4. ParthKohli Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • 2 years ago
  5. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    having the damndest time typesetting as well

    • 2 years ago
  6. FoolForMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    For my own reason: \[ \huge \frac{4^{\left(\frac{1}{\log_4(\frac{3}{4})}\right)}}{3^{\left(\frac{1}{\log_3(\frac{3}{4})}\right)}} \]

    • 2 years ago
  7. FoolForMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Is this right?

    • 2 years ago
  8. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    yes that is it

    • 2 years ago
  9. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    @experimentX you lost me there

    • 2 years ago
  10. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    i got \[ \frac 4 3 \log_4 3\] is 1/12 supposed to be answer??

    • 2 years ago
  11. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    yes

    • 2 years ago
  12. FoolForMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    1/12 is the right answer.

    • 2 years ago
  13. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

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

    • 2 years ago
  14. FoolForMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    I am gonna M.SE it .. if you don't have any objection?

    • 2 years ago
  15. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    i do not know what that means, but i have no objection this was a previous problem so go nuts

    • 2 years ago
  16. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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

    • 2 years ago
  17. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    not times, to the power of

    • 2 years ago
  18. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    somehow we are supposed to arrive at \(\frac{1}{4\times 3}\)

    • 2 years ago
  19. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Oh .. my mistake

    • 2 years ago
  20. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    thing maybe rewrite exponents as \(\frac{\log_4(4)}{\log_4(\frac{3}{4})}\)

    • 2 years ago
  21. FoolForMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    http://math.stackexchange.com/questions/161685/

    • 2 years ago
  22. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    i wish i could answer ... lol

    • 2 years ago
  23. FoolForMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    typesetting is a mess :/

    • 2 years ago
  24. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    oh maybe we can change the base to \(\frac{4}{3}\)!!

    • 2 years ago
  25. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    ?

    • 2 years ago
  26. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    it seems that i can wrap latex with $ here too .. i thought \[ only works here

    • 2 years ago
  27. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    doesn't work for me $\frac{a}{b}$

    • 2 years ago
  28. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    for in line i use \(

    • 2 years ago
  29. FoolForMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Work : $$ \frac{a}{b}$$

    • 2 years ago
  30. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    ok how about changing the base to \(\frac{4}{3}\) maybe that would work

    • 2 years ago
  31. FoolForMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    $$ one ;)

    • 2 years ago
  32. Zarkon Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    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

    • 2 years ago
  33. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    12 to some power?

    • 2 years ago
  34. Zarkon Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    \[12^{\log_{12}(\cdot)}\]

    • 2 years ago
  35. Zarkon Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    the dot it the original problem

    • 2 years ago
  36. Zarkon Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    *is the..

    • 2 years ago
  37. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    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?

    • 2 years ago
  38. Zarkon Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    I also did it for \[a\times b\]

    • 2 years ago
  39. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    christ no wonder you are not going to write it here, i can barely write it on paper

    • 2 years ago
  40. Zarkon Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    \[\huge \frac{a^{\left(\frac{1}{\log_a(\frac{b}{a})}\right)}}{b^{\left(\frac{1}{\log_b(\frac{b}{a})}\right)}}=\frac{1}{a\times b}\]

    • 2 years ago
  41. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    and why? this of course gives it right away

    • 2 years ago
  42. Zarkon Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    i derived it...seemed a natural thing to do...combine into one exponent

    • 2 years ago
  43. FoolAroundMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  44. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    $$ \huge a = b^{\log_ba} $$ $$ \huge \frac{\log_ba}{\log_a b - 1 } - \frac{1}{1 - \log_ba} = \frac{(\log_ba)^2}{1-\log_ba} - \frac{1}{1-\log_ba} \\ \frac{-(1 + \log_ba)(1 - \log_ba)}{(1 - \log_ba)} = -(\log_b + \log_a) = -\log_b(ab) $$

    • 2 years ago
  45. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    $$ \huge b^{-\log_b(ab)} = 1/ab$$

    • 2 years ago
  46. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    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 \[\exp^{-\ln(ab)}=\frac{1}{ab}\]

    • 2 years ago
  47. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    using the definition \(a^x=e^{x\ln(a)}\) i have to say other ways were snappier though

    • 2 years ago
  48. maheshmeghwal9 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Hmm............................ gr8* :D

    • 2 years ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.