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

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ParthKohliBest ResponseYou've already chosen the best response.0
Wasnt that log4^3/4 and log_3^3/4 asked in a question before?
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
it was ambiguous before apparently it is \(\frac{1}{12}\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
tried changing the base, and also rewriting the logs as a difference, but it is not coming snappy at all
 one year ago

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

satellite73Best ResponseYou've already chosen the best response.2
having the damndest time typesetting as well
 one year ago

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

satellite73Best ResponseYou've already chosen the best response.2
@experimentX you lost me there
 one year ago

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

FoolForMathBest ResponseYou've already chosen the best response.1
1/12 is the right answer.
 one year ago

satellite73Best ResponseYou've already chosen the best response.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 \[\huge \frac{4^{\left(\frac{\ln(4)}{\ln(\frac{3}{4})}\right)}}{3^{\left(\frac{\ln(3)}{\ln(\frac{3}{4})}\right)}}\]
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.1
I am gonna M.SE it .. if you don't have any objection?
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
i do not know what that means, but i have no objection this was a previous problem so go nuts
 one year ago

experimentXBest ResponseYou've already chosen the best response.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
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
not times, to the power of
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
somehow we are supposed to arrive at \(\frac{1}{4\times 3}\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
thing maybe rewrite exponents as \(\frac{\log_4(4)}{\log_4(\frac{3}{4})}\)
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.1
http://math.stackexchange.com/questions/161685/
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
i wish i could answer ... lol
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.1
typesetting is a mess :/
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
oh maybe we can change the base to \(\frac{4}{3}\)!!
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
it seems that i can wrap latex with $ here too .. i thought \[ only works here
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
doesn't work for me $\frac{a}{b}$
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
for in line i use \(
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.1
Work : $$ \frac{a}{b}$$
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
ok how about changing the base to \(\frac{4}{3}\) maybe that would work
 one year ago

ZarkonBest ResponseYou've already chosen the best response.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
 one year ago

ZarkonBest ResponseYou've already chosen the best response.2
\[12^{\log_{12}(\cdot)}\]
 one year ago

ZarkonBest ResponseYou've already chosen the best response.2
the dot it the original problem
 one year ago

satellite73Best ResponseYou've already chosen the best response.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?
 one year ago

ZarkonBest ResponseYou've already chosen the best response.2
I also did it for \[a\times b\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
christ no wonder you are not going to write it here, i can barely write it on paper
 one year ago

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

satellite73Best ResponseYou've already chosen the best response.2
and why? this of course gives it right away
 one year ago

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

FoolAroundMathBest 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) \[\log_{b/a}y = \frac{1}{\log_{a}(b/a)}\log_{b/a}a  \frac{1}{\log_{b}(b/a)}\log_{b/a}b \] \[\log_{b/a}y =\frac{loga.loga}{\log(b/a)\log(b/a)}  \frac{logb.logb}{\log(b/a).\log(b/a)}\] \[\log_{b/a}y = \frac{(logalogb)(loga+logb)}{\log(b/a).\log(b/a)}\] \[\log_{b/a}y = \frac{\log(ab)}{\log(b/a)}\] \[\log_{b/a}y = \log_{b/a}(\frac{1}{ab})\] \[=> y = \frac{1}{ab}\]
 one year ago

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

experimentXBest ResponseYou've already chosen the best response.1
$$ \huge b^{\log_b(ab)} = 1/ab$$
 one year ago

satellite73Best ResponseYou've already chosen the best response.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 \[\exp^{\ln(ab)}=\frac{1}{ab}\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
using the definition \(a^x=e^{x\ln(a)}\) i have to say other ways were snappier though
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.0
Hmm............................ gr8* :D
 one year ago
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