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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 \(\frac{1}{12}\)

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

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.

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

not times, to the power of

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 ;)

12 to some power?

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

the dot it the original problem

*is the..

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

and why? this of course gives it right away

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

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

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

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