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satellite73
 4 years ago
\[\frac{4^{\left(\frac{1}{\log_4(\frac{3}{4})}\right)}}{3^{\left(\frac{1}{\log_3(\frac{3}{4})}\right)}}\]
satellite73
 4 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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ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0Wasnt that log4^3/4 and log_3^3/4 asked in a question before?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it was ambiguous before apparently it is \(\frac{1}{12}\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0tried changing the base, and also rewriting the logs as a difference, but it is not coming snappy at all

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0So we just have to find \(\Large {4^{1 } \over 3^{1 \over 12}}\)? I'm new with logs so mercy

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0having the damndest time typesetting as well

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0For my own reason: \[ \huge \frac{4^{\left(\frac{1}{\log_4(\frac{3}{4})}\right)}}{3^{\left(\frac{1}{\log_3(\frac{3}{4})}\right)}} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@experimentX you lost me there

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1i got \[ \frac 4 3 \log_4 3\] is 1/12 supposed to be answer??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.01/12 is the right answer.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\huge \frac{4^{\left(\frac{1}{\log_4(\frac{3}{4})}\right)}}{3^{\left(\frac{1}{\log_3(\frac{3}{4})}\right)}}\] yes it is i do not see the gimmick change base? rewrite as \[\huge \frac{4^{\left(\frac{\ln(4)}{\ln(\frac{3}{4})}\right)}}{3^{\left(\frac{\ln(3)}{\ln(\frac{3}{4})}\right)}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I am gonna M.SE it .. if you don't have any objection?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i do not know what that means, but i have no objection this was a previous problem so go nuts

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1lol ... 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0not times, to the power of

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0somehow we are supposed to arrive at \(\frac{1}{4\times 3}\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thing maybe rewrite exponents as \(\frac{\log_4(4)}{\log_4(\frac{3}{4})}\)

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1i wish i could answer ... lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0typesetting is a mess :/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh maybe we can change the base to \(\frac{4}{3}\)!!

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1it seems that i can wrap latex with $ here too .. i thought \[ only works here

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0doesn't work for me $\frac{a}{b}$

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Work : $$ \frac{a}{b}$$

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok how about changing the base to \(\frac{4}{3}\) maybe that would work

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.2I 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

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.2\[12^{\log_{12}(\cdot)}\]

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.2the dot it the original problem

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you 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?

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.2I also did it for \[a\times b\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0christ no wonder you are not going to write it here, i can barely write it on paper

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.2\[\huge \frac{a^{\left(\frac{1}{\log_a(\frac{b}{a})}\right)}}{b^{\left(\frac{1}{\log_b(\frac{b}{a})}\right)}}=\frac{1}{a\times b}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and why? this of course gives it right away

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.2i derived it...seemed a natural thing to do...combine into one exponent

FoolAroundMath
 4 years ago
Best ResponseYou've already chosen the best response.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) \[\log_{b/a}y = \frac{1}{\log_{a}(b/a)}\log_{b/a}a  \frac{1}{\log_{b}(b/a)}\log_{b/a}b \] \[\log_{b/a}y =\frac{loga.loga}{\log(b/a)\log(b/a)}  \frac{logb.logb}{\log(b/a).\log(b/a)}\] \[\log_{b/a}y = \frac{(logalogb)(loga+logb)}{\log(b/a).\log(b/a)}\] \[\log_{b/a}y = \frac{\log(ab)}{\log(b/a)}\] \[\log_{b/a}y = \log_{b/a}(\frac{1}{ab})\] \[=> y = \frac{1}{ab}\]

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1$$ \huge a = b^{\log_ba} $$ $$ \huge \frac{\log_ba}{\log_a b  1 }  \frac{1}{1  \log_ba} = \frac{(\log_ba)^2}{1\log_ba}  \frac{1}{1\log_ba} \\ \frac{(1 + \log_ba)(1  \log_ba)}{(1  \log_ba)} = (\log_b + \log_a) = \log_b(ab) $$

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1$$ \huge b^{\log_b(ab)} = 1/ab$$

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok 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 \[\exp^{\ln(ab)}=\frac{1}{ab}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0using the definition \(a^x=e^{x\ln(a)}\) i have to say other ways were snappier though

maheshmeghwal9
 4 years ago
Best ResponseYou've already chosen the best response.0Hmm............................ gr8* :D
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