Question

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

Answer 1

Wasnt that log4^3/4 and log_3^3/4 asked in a question before?

Answer 2

it was ambiguous before
apparently it is \(\frac{1}{12}\)

Answer 3

tried changing the base, and also rewriting the logs as a difference, but it is not coming snappy at all

Answer 4

So we just have to find \(\Large {4^{1 } \over 3^{1 \over 12}}\)? I'm new with logs so mercy

Answer 5

having the damndest time typesetting as well

Answer 6

For my own reason: \[ \huge \frac{4^{\left(\frac{1}{\log_4(\frac{3}{4})}\right)}}{3^{\left(\frac{1}{\log_3(\frac{3}{4})}\right)}} \]

Answer 7

Is this right?

Answer 8

yes that is it

Answer 9

@experimentX you lost me there

Answer 10

i got
\[ \frac 4 3 \log_4 3\]
is 1/12 supposed to be answer??

Answer 11

yes

Answer 12

1/12 is the right answer.

Answer 13

\[\huge \frac{4^{\left(\frac{1}{\log_4(\frac{3}{4})}\right)}}{3^{\left(\frac{1}{\log_3(\frac{3}{4})}\right)}}\]
yes it is
i do not see the gimmick
change base?
rewrite as \[\huge \frac{4^{\left(\frac{\ln(4)}{\ln(\frac{3}{4})}\right)}}{3^{\left(\frac{\ln(3)}{\ln(\frac{3}{4})}\right)}}\]

Answer 14

I am gonna M.SE it .. if you don't have any objection?

Answer 15

i do not know what that means, but i have no objection
this was a previous problem so go nuts

Answer 16

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

Answer 17

not times, to the power of

Answer 18

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

Answer 19

somehow we are supposed to arrive at \(\frac{1}{4\times 3}\)

Answer 20

Oh .. my mistake

Answer 21

thing maybe rewrite exponents as \(\frac{\log_4(4)}{\log_4(\frac{3}{4})}\)

Answer 22

http://math.stackexchange.com/questions/161685/

Answer 23

i wish i could answer ... lol

Answer 24

typesetting is a mess :/

Answer 25

oh maybe we can change the base to \(\frac{4}{3}\)!!

Answer 26

?

Answer 27

it seems that i can wrap latex with $ here too .. i thought \[ only works here

Answer 28

doesn't work for me
$\frac{a}{b}$

Answer 29

for in line i use \(

Answer 30

Work : $$ \frac{a}{b}$$

Answer 31

ok how about changing the base to \(\frac{4}{3}\) maybe that would work

Answer 32

$$ one ;)

Answer 33

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

Answer 34

12 to some power?

Answer 35

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

Answer 36

the dot it the original problem

Answer 37

*is the..

Answer 38

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?

Answer 39

I also did it for \[a\times b\]

Answer 40

christ no wonder you are not going to write it here, i can barely write it on paper

Answer 41

\[\huge \frac{a^{\left(\frac{1}{\log_a(\frac{b}{a})}\right)}}{b^{\left(\frac{1}{\log_b(\frac{b}{a})}\right)}}=\frac{1}{a\times b}\]

Answer 42

and why? this of course gives it right away

Answer 43

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

Answer 44

\[y = \frac{a^{\frac{1}{\log_{a}(\frac{b}{a})}}}{b^{\frac{1}{\log_{3}(\frac{b}{a})}}}\]
take log on both sides with base (b/a)
\[\log_{b/a}y = \frac{1}{\log_{a}(b/a)}\log_{b/a}a - \frac{1}{\log_{b}(b/a)}\log_{b/a}b \]
\[\log_{b/a}y =\frac{loga.loga}{\log(b/a)\log(b/a)} - \frac{logb.logb}{\log(b/a).\log(b/a)}\]
\[\log_{b/a}y = \frac{(loga-logb)(loga+logb)}{\log(b/a).\log(b/a)}\]
\[\log_{b/a}y = \frac{-\log(ab)}{\log(b/a)}\]
\[\log_{b/a}y = \log_{b/a}(\frac{1}{ab})\]
\[=> y = \frac{1}{ab}\]

Answer 45

$$
\huge a = b^{\log_ba}
$$
$$
\huge \frac{\log_ba}{\log_a b - 1 } - \frac{1}{1 - \log_ba} = \frac{(\log_ba)^2}{1-\log_ba} - \frac{1}{1-\log_ba} \\
\frac{-(1 + \log_ba)(1 - \log_ba)}{(1 - \log_ba)} = -(\log_b + \log_a) = -\log_b(ab)
$$

Answer 46

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

Answer 47

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
\[\exp^{-\ln(ab)}=\frac{1}{ab}\]

Answer 48

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

Answer 49

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