## satellite73 2 years ago $\frac{4^{\left(\frac{1}{\log_4(\frac{3}{4})}\right)}}{3^{\left(\frac{1}{\log_3(\frac{3}{4})}\right)}}$

1. ParthKohli

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

2. satellite73

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

3. satellite73

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

4. ParthKohli

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

5. satellite73

having the damndest time typesetting as well

6. FoolForMath

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

7. FoolForMath

Is this right?

8. satellite73

yes that is it

9. satellite73

@experimentX you lost me there

10. experimentX

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

11. satellite73

yes

12. FoolForMath

13. satellite73

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

14. FoolForMath

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

15. satellite73

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

16. experimentX
17. satellite73

not times, to the power of

18. satellite73
19. satellite73

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

20. experimentX

Oh .. my mistake

21. satellite73

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

22. FoolForMath
23. experimentX

i wish i could answer ... lol

24. FoolForMath

typesetting is a mess :/

25. satellite73

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

26. satellite73

?

27. experimentX

it seems that i can wrap latex with \$ here too .. i thought $only works here 28. satellite73 doesn't work for me \frac{a}{b} 29. satellite73 for in line i use $$30. FoolForMath Work : \frac{a}{b} 31. satellite73 ok how about changing the base to \(\frac{4}{3}$$ maybe that would work 32. FoolForMath  one ;) 33. Zarkon 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

34. satellite73

12 to some power?

35. Zarkon

$12^{\log_{12}(\cdot)}$

36. Zarkon

the dot it the original problem

37. Zarkon

*is the..

38. satellite73

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?

39. Zarkon

I also did it for $a\times b$

40. satellite73

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

41. Zarkon

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

42. satellite73

and why? this of course gives it right away

43. Zarkon

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

44. FoolAroundMath

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

45. experimentX

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

46. experimentX

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

47. satellite73

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

48. satellite73

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

49. maheshmeghwal9

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