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#2 2006-03-10 01:41:40
- krassi_holmz
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Re: Logarithms
So:
Last edited by krassi_holmz (2006-03-10 01:44:53)
IPBLE: Increasing Performance By Lowering Expectations.
#4 2006-03-10 02:06:00
- ganesh
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Re: Logarithms
L # 2
If
show that xyz=1.
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#5 2006-03-10 06:19:45
- krassi_holmz
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Re: Logarithms
Let
log x = x'
log y = y'
log z = z'.
Then:
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 2006-03-10 15:17:14
- ganesh
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Re: Logarithms
Well done, krassi_holmz!
Character is who you are when no one is looking.
#7 2006-03-11 15:48:02
- ganesh
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Re: Logarithms
L # 3
If x²y³=a and log (x/y)=b, then what is the value of (logx)/(logy)?
Character is who you are when no one is looking.
#8 2006-03-11 19:03:38
- krassi_holmz
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Re: Logarithms
loga=2logx+3logy
b=logx-logy
loga+3b=5logx
loga-2b=3logy+2logy=5logy
logx/logy=(loga+3b)/(loga-2b).
Last edited by krassi_holmz (2006-03-11 19:06:29)
IPBLE: Increasing Performance By Lowering Expectations.
#9 2006-03-11 19:23:36
- ganesh
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Re: Logarithms
Very well done, krassi_holmz! 
Character is who you are when no one is looking.
#10 2009-01-05 10:58:47
#12 2009-01-05 22:37:30
- JaneFairfax
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Re: Logarithms
You are not supposed to use a calculator or log tables for L # 4. 
Try again!
Last edited by JaneFairfax (2009-01-05 22:40:20)
#13 2009-01-06 00:36:33
- ganesh
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Re: Logarithms
No, I didn't 
I remember
and
.
Character is who you are when no one is looking.
#14 2009-01-06 20:57:49
- JaneFairfax
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Re: Logarithms
You 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 (2009-01-06 23:30:04)
#15 2009-02-07 22:31:40
- JaneFairfax
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#16 2010-04-19 17:06:39
- keveenjones
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Re: Logarithms
log a = 2log x + 3log y
b = log x – log y
log a + 3 b = 5log x
loga - 2b = 3logy + 2logy = 5logy
logx / logy = (loga+3b) / (loga-2b)
#17 2010-04-19 19:04:41
- rzaidan
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Re: Logarithms
Hi ganesh
for L # 1
since log(a)= 1 / log(b), log(a)=1
b a a
we have
1/log(abc)+1/log(abc)+1/log(abc)=
a b c
log(a)+log(b)+log(c)= log(abc)=1
abc abc abc abc
Best Regards
Riad Zaidan
#18 2010-04-19 19:14:13
- rzaidan
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Re: Logarithms
Hi ganesh
for L # 2
I think that the following proof is easier:
Assume Log(x)/(b-c)=Log(y)/(c-a)=Log(z)/(a-b)=t
So Log(x)=t(b-c),Log(y)=t(c-a) , Log(z)=t(a-b) So Log(x)+Log(y)+Log(z)=tb-tc+tc-ta+ta-tb=0
So Log(xyz)=0 so xyz=1 Q.E.D
Best Regards
Riad Zaidan
#19 2010-04-19 23:02:18
- ganesh
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Re: Logarithms
Gentleman,
Thanks for the proofs.
Regards.
Character is who you are when no one is looking.
#20 2010-08-17 14:15:11
- jonnyj99
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Re: Logarithms
log_2(16) = \log_2 \left ( \frac{64}{4} \right ) = \log_2(64) - \log_2(4) = 6 - 2 = 4, \,
log_2(\sqrt[3]4) = \frac {1}{3} \log_2 (4) = \frac {2}{3}. \,
#21 2011-05-28 17:59:31
- reconsideryouranswer
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Re: Logarithms
JaneFairfax wrote:
L # 4
I don't want a method that will rely on defining certain functions, taking derivatives,
noting concavity, etc.
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 right-hand 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 left-hand side) will be greater than the value on the
right-hand 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 (2011-05-28 18:05:01)
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#22 2011-05-28 21:57:00
- ganesh
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Re: Logarithms
Hi reconsideryouranswer,
This problem was posted by JaneFairfax. I think it would be appropriate she verify the solution.
Character is who you are when no one is looking.