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You are not logged in. #2 20060310 01:41:40
Re: LogarithmsSo: Last edited by krassi_holmz (20060310 01:44:53) IPBLE: Increasing Performance By Lowering Expectations. #4 20060310 02:06:00
Re: LogarithmsL # 2 show that xyz=1. Character is who you are when no one is looking. #5 20060310 06:19:45
Re: LogarithmsLet and Now, consider that from xyz=1 follows: But x'=log x; y'=log y' z'=log z, so we must prove that: x'+y'+z'=0. Rewriting in terms of x' gives: q.E.d. IPBLE: Increasing Performance By Lowering Expectations. #6 20060310 15:17:14
Re: LogarithmsWell done, krassi_holmz! Character is who you are when no one is looking. #7 20060311 15:48:02
Re: LogarithmsL # 3 Character is who you are when no one is looking. #8 20060311 19:03:38
Re: Logarithmsloga=2logx+3logy Last edited by krassi_holmz (20060311 19:06:29) IPBLE: Increasing Performance By Lowering Expectations. #9 20060311 19:23:36
Re: LogarithmsVery well done, krassi_holmz! Character is who you are when no one is looking. #10 20090105 10:58:47#12 20090105 22:37:30
Re: LogarithmsYou are not supposed to use a calculator or log tables for L # 4. Try again! Last edited by JaneFairfax (20090105 22:40:20) #13 20090106 00:36:33
Re: LogarithmsNo, I didn't and . Character is who you are when no one is looking. #14 20090106 20:57:49
Re: LogarithmsYou still used a calculator / log table in the past to get those figures (or someone else did and showed them to you). I say again: no calculators or log tables to be used (directly or indirectly) at all!! Last edited by JaneFairfax (20090106 23:30:04) #15 20090207 22:31:40
#16 20100419 17:06:39
Re: Logarithmslog a = 2log x + 3log y #17 20100419 19:04:41
Re: LogarithmsHi ganesh #18 20100419 19:14:13
Re: LogarithmsHi ganesh #19 20100419 23:02:18
Re: LogarithmsGentleman, Character is who you are when no one is looking. #20 20100817 14:15:11
Re: Logarithmslog_2(16) = \log_2 \left ( \frac{64}{4} \right ) = \log_2(64)  \log_2(4) = 6  2 = 4, \, #21 20110528 17:59:31
Re: Logarithms
I don't want a method that will rely on defining certain functions, taking derivatives, Change of base: Each side is positive, and multiplying by the positive denominator keeps whatever direction of the alleged inequality the same direction: On the righthand side, the first factor is equal to a positive number less than 1, while the second factor is equal to a positive number greater than 1. These facts are by inspection combined with the nature of exponents/logarithms. Because of (log A)B = B(log A) = log(A^B), I may turn this into: I need to show that Then Then 1 (on the lefthand side) will be greater than the value on the righthand side, and the truth of the original inequality will be established. I want to show Raise a base of 3 to each side: Each side is positive, and I can square each side:  Then I want to show that when 2 is raised to a number equal to (or less than) 1.5, then it is less than 3. Each side is positive, and I can square each side: Last edited by reconsideryouranswer (20110528 18:05:01) Signature line: I wish a had a more interesting signature line. #22 20110528 21:57:00
Re: LogarithmsHi reconsideryouranswer, Character is who you are when no one is looking. 