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## #1 2006-03-09 22:54:27

ganesh
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### Logarithms

L # 1

Show that

Character is who you are when no one is looking.

## #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.

## #3 2006-03-10 02:01:49

ganesh
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### Re: Logarithms

Character is who you are when no one is looking.

## #4 2006-03-10 02:06:00

ganesh
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### Re: Logarithms

L # 2

If

show that xyz=1.

Character is who you are when no one is looking.

## #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

JaneFairfax
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### Re: Logarithms

L # 4

Q: Who wrote the novels Mrs Dalloway and To the Lighthouse?

## #11 2009-01-05 15:13:57

ganesh
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### Re: Logarithms

Character is who you are when no one is looking.

## #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)

Q: Who wrote the novels Mrs Dalloway and To the Lighthouse?

## #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)

Q: Who wrote the novels Mrs Dalloway and To the Lighthouse?

## #15 2009-02-07 22:31:40

JaneFairfax
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### Re: Logarithms

Q: Who wrote the novels Mrs Dalloway and To the Lighthouse?

## #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

## #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

## #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

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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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