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From: Shin Dongjun on 8 Aug 2010 13:39 Prove (log3)^2 > (log4)^3. (Don't use an approximation like log3 = 0.4771) log is common logarithms that is, base of log is 10.
From: Raymond Manzoni on 8 Aug 2010 14:42 Shin Dongjun a �crit : > Prove (log3)^2 > (log4)^3. > > (Don't use an approximation like log3 = 0.4771) > > log is common logarithms that is, base of log is 10. You could, for example, study f(x)= (x-1) ln(ln(x)) for x > e ... Hoping this helped, Raymond
From: Rob Johnson on 9 Aug 2010 01:05 In article <9f8fba1b-df73-43af-b139-8cac16e201b9(a)z34g2000pro.googlegroups.com>, Shin Dongjun <acamath(a)gmail.com> wrote: >Prove (log3)^2 > (log4)^3. > >(Don't use an approximation like log3 = 0.4771) > >log is common logarithms that is, base of log is 10. The following can be verified by brute force exponentiation log(3)/log(4) > 7/9 <=> 3^9 > 4^7 49/81 > log(4)/log(10) <=> 10^49 > 4^81 Thus, ( log(3)/log(4) )^2 > log(4)/log(10) Rob Johnson <rob(a)trash.whim.org> take out the trash before replying to view any ASCII art, display article in a monospaced font
From: Rob Johnson on 9 Aug 2010 10:35 In article <4c5efa99$0$10185$ba4acef3(a)reader.news.orange.fr>, Raymond Manzoni <raymman(a)free.fr> wrote: >Shin Dongjun a �crit : >> Prove (log3)^2 > (log4)^3. >> >> (Don't use an approximation like log3 = 0.4771) >> >> log is common logarithms that is, base of log is 10. > > > You could, for example, study f(x)= (x-1) ln(ln(x)) for x > e ... This works well for proving the other direction for natural logs: ln(3)^2 < ln(4)^3 [1] The derivative of (x-1) ln(ln(x)) = ln(ln(x)) + (x-1) 1/(x ln(x)), which is positive for x > e. Thus, (x-1) ln(ln(x)) is monotonically increasing for x > e, which implies [1]. However, since the OP is about common logs, we get (x-1) log(log(x)) = (x-1) (ln(ln(x)) - ln(ln(10)))/ln(10) whose derivative is ln(ln(x)) - ln(ln(10)) + (x-1)/(x ln(x)) ---------------------------------------- ln(10) which is negative for x in [3,3.7517684] and positive for x in [3.7517684,4]. Thus, (x-1) log(log(x)) is decreasing for about 3/4 of the interval in question and increasing for about 1/4. However, trying to use this in a proof seems difficult. Rob Johnson <rob(a)trash.whim.org> take out the trash before replying to view any ASCII art, display article in a monospaced font
From: Raymond Manzoni on 9 Aug 2010 12:59
Rob Johnson a �crit : > In article <4c5efa99$0$10185$ba4acef3(a)reader.news.orange.fr>, > Raymond Manzoni <raymman(a)free.fr> wrote: >> Shin Dongjun a �crit : >>> Prove (log3)^2 > (log4)^3. >>> >>> (Don't use an approximation like log3 = 0.4771) >>> >>> log is common logarithms that is, base of log is 10. >> >> You could, for example, study f(x)= (x-1) ln(ln(x)) for x > e ... > > This works well for proving the other direction for natural logs: > > ln(3)^2 < ln(4)^3 [1] > > The derivative of (x-1) ln(ln(x)) = ln(ln(x)) + (x-1) 1/(x ln(x)), > which is positive for x > e. Thus, (x-1) ln(ln(x)) is monotonically > increasing for x > e, which implies [1]. > > However, since the OP is about common logs, we get > > (x-1) log(log(x)) > > = (x-1) (ln(ln(x)) - ln(ln(10)))/ln(10) > > whose derivative is > > ln(ln(x)) - ln(ln(10)) + (x-1)/(x ln(x)) > ---------------------------------------- > ln(10) > > which is negative for x in [3,3.7517684] and positive for x in > [3.7517684,4]. Thus, (x-1) log(log(x)) is decreasing for about 3/4 > of the interval in question and increasing for about 1/4. However, > trying to use this in a proof seems difficult. > > Rob Johnson <rob(a)trash.whim.org> You are right Rob I noticed my mistake too (from your other answer...) and the fact that the derivative became 0 in (3,4) as you indicated (invalidating my attempt at least directly). In fact I was searching a correct way to handle this and got some results 'in line' with your response : 1) Let's prove directly that (log(3)/log(2))^2 > 8 log(2) : Tables of powers : 2^n : 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048 3^n : 3, 9, 27, 81, 243, 729, 2187 so that 3^7 > 2^11 (2187>2048, 9>8 is too weak...) and (log(3)/log(2))^2 > (11/7)^2 On the other side 10 log(2)= log(1024)= log(1000*(1+24/1000)) = 3 + log(1+24/1000) < 3 + 24/(1000 ln(10)) (the series for log(1+x) is alternate) so (using ln(10)>2 and 24<25) 10 log(2) < 3 + 1/80 so that 8 log(2) < 241/100 < (11/7)^2 < (log(3)/log(2))^2 (laborious ...) 2) Starting with 3^12 > 2^19 is more direct : (log(3)/log(2))^2 > (19/12)^2 > (360/12^2 = 5/2) 2^16 < 10^5 so that 8 log(2) < 5/2 (to find 19/12 I used continued fractions of log(3)/log(2)) .... Other tricks could be tried (use 3^4 > 80 to get log(3)^2 > (1 + 3 log(2))^2/16 that has to be greater than 8 log(2)^3 and so on...) but I was overall dissatisfied by all these answers... (kind of cheating since we replace 'real values approximations' by large integers computations). Perhaps too that I didn't get the point of the problem since... I still prefer my initial answer even if of no value here! :-) Thanks for the correction anyway, Raymond |