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

Mathematics
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Wasnt that log4^3/4 and log_3^3/4 asked in a question before?
it was ambiguous before apparently it is \(\frac{1}{12}\)
tried changing the base, and also rewriting the logs as a difference, but it is not coming snappy at all

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Other answers:

So we just have to find \(\Large {4^{1 } \over 3^{1 \over 12}}\)? I'm new with logs so mercy
having the damndest time typesetting as well
For my own reason: \[ \huge \frac{4^{\left(\frac{1}{\log_4(\frac{3}{4})}\right)}}{3^{\left(\frac{1}{\log_3(\frac{3}{4})}\right)}} \]
Is this right?
yes that is it
@experimentX you lost me there
i got \[ \frac 4 3 \log_4 3\] is 1/12 supposed to be answer??
yes
1/12 is the right answer.
\[\huge \frac{4^{\left(\frac{1}{\log_4(\frac{3}{4})}\right)}}{3^{\left(\frac{1}{\log_3(\frac{3}{4})}\right)}}\] yes it is i do not see the gimmick change base? rewrite as \[\huge \frac{4^{\left(\frac{\ln(4)}{\ln(\frac{3}{4})}\right)}}{3^{\left(\frac{\ln(3)}{\ln(\frac{3}{4})}\right)}}\]
I am gonna M.SE it .. if you don't have any objection?
i do not know what that means, but i have no objection this was a previous problem so go nuts
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
not times, to the power of
http://www.wolframalpha.com/input/?i=%284^%281+%2Flog+base+4+%283%2F4%29%29%2F%283^%281+%2Flog+base+3+%283%2F4%29%29%29
somehow we are supposed to arrive at \(\frac{1}{4\times 3}\)
Oh .. my mistake
thing maybe rewrite exponents as \(\frac{\log_4(4)}{\log_4(\frac{3}{4})}\)
http://math.stackexchange.com/questions/161685/
i wish i could answer ... lol
typesetting is a mess :/
oh maybe we can change the base to \(\frac{4}{3}\)!!
?
it seems that i can wrap latex with $ here too .. i thought \[ only works here
doesn't work for me $\frac{a}{b}$
for in line i use \(
Work : $$ \frac{a}{b}$$
ok how about changing the base to \(\frac{4}{3}\) maybe that would work
$$ one ;)
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
12 to some power?
\[12^{\log_{12}(\cdot)}\]
the dot it the original problem
*is the..
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?
I also did it for \[a\times b\]
christ no wonder you are not going to write it here, i can barely write it on paper
\[\huge \frac{a^{\left(\frac{1}{\log_a(\frac{b}{a})}\right)}}{b^{\left(\frac{1}{\log_b(\frac{b}{a})}\right)}}=\frac{1}{a\times b}\]
and why? this of course gives it right away
i derived it...seemed a natural thing to do...combine into one exponent
\[y = \frac{a^{\frac{1}{\log_{a}(\frac{b}{a})}}}{b^{\frac{1}{\log_{3}(\frac{b}{a})}}}\] take log on both sides with base (b/a) \[\log_{b/a}y = \frac{1}{\log_{a}(b/a)}\log_{b/a}a - \frac{1}{\log_{b}(b/a)}\log_{b/a}b \] \[\log_{b/a}y =\frac{loga.loga}{\log(b/a)\log(b/a)} - \frac{logb.logb}{\log(b/a).\log(b/a)}\] \[\log_{b/a}y = \frac{(loga-logb)(loga+logb)}{\log(b/a).\log(b/a)}\] \[\log_{b/a}y = \frac{-\log(ab)}{\log(b/a)}\] \[\log_{b/a}y = \log_{b/a}(\frac{1}{ab})\] \[=> y = \frac{1}{ab}\]
$$ \huge a = b^{\log_ba} $$ $$ \huge \frac{\log_ba}{\log_a b - 1 } - \frac{1}{1 - \log_ba} = \frac{(\log_ba)^2}{1-\log_ba} - \frac{1}{1-\log_ba} \\ \frac{-(1 + \log_ba)(1 - \log_ba)}{(1 - \log_ba)} = -(\log_b + \log_a) = -\log_b(ab) $$
$$ \huge b^{-\log_b(ab)} = 1/ab$$
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 \[\exp^{-\ln(ab)}=\frac{1}{ab}\]
using the definition \(a^x=e^{x\ln(a)}\) i have to say other ways were snappier though
Hmm............................ gr8* :D

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