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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
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

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ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Wasnt that log4^3/4 and log_3^3/4 asked in a question before?
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
it was ambiguous before apparently it is \(\frac{1}{12}\)
 2 years ago

satellite73 Group TitleBest 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
 2 years ago

ParthKohli Group TitleBest 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
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
having the damndest time typesetting as well
 2 years ago

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

FoolForMath Group TitleBest ResponseYou've already chosen the best response.1
Is this right?
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
yes that is it
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
@experimentX you lost me there
 2 years ago

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

FoolForMath Group TitleBest ResponseYou've already chosen the best response.1
1/12 is the right answer.
 2 years ago

satellite73 Group TitleBest 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 \[\huge \frac{4^{\left(\frac{\ln(4)}{\ln(\frac{3}{4})}\right)}}{3^{\left(\frac{\ln(3)}{\ln(\frac{3}{4})}\right)}}\]
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.1
I am gonna M.SE it .. if you don't have any objection?
 2 years ago

satellite73 Group TitleBest 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
 2 years ago

experimentX Group TitleBest 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
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
not times, to the power of
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
somehow we are supposed to arrive at \(\frac{1}{4\times 3}\)
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
Oh .. my mistake
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
thing maybe rewrite exponents as \(\frac{\log_4(4)}{\log_4(\frac{3}{4})}\)
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.1
http://math.stackexchange.com/questions/161685/
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
i wish i could answer ... lol
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.1
typesetting is a mess :/
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
oh maybe we can change the base to \(\frac{4}{3}\)!!
 2 years ago

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

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
doesn't work for me $\frac{a}{b}$
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
for in line i use \(
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.1
Work : $$ \frac{a}{b}$$
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
ok how about changing the base to \(\frac{4}{3}\) maybe that would work
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.1
$$ one ;)
 2 years ago

Zarkon Group TitleBest 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
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
12 to some power?
 2 years ago

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

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

satellite73 Group TitleBest 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?
 2 years ago

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

satellite73 Group TitleBest 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
 2 years ago

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

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
and why? this of course gives it right away
 2 years ago

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

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

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

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

satellite73 Group TitleBest 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 \[\exp^{\ln(ab)}=\frac{1}{ab}\]
 2 years ago

satellite73 Group TitleBest 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
 2 years ago

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.0
Hmm............................ gr8* :D
 2 years ago
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