Logarithms

ganesh · 2006-03-08 23:54:27

L # 1

Show that

krassi_holmz · 2006-03-09 02:41:40

So:

Last edited by krassi_holmz (2006-03-09 02:44:53)

ganesh · 2006-03-09 03:01:49

ganesh · 2006-03-09 03:06:00

L # 2

If

show that xyz=1.

krassi_holmz · 2006-03-09 07:19:45

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.

ganesh · 2006-03-09 16:17:14

Well done, krassi_holmz!

ganesh · 2006-03-10 16:48:02

L # 3

If x²y³=a and log (x/y)=b, then what is the value of (logx)/(logy)?

krassi_holmz · 2006-03-10 20:03:38

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 (2006-03-10 20:06:29)

ganesh · 2006-03-10 20:23:36

Very well done, krassi_holmz!

JaneFairfax · 2009-01-04 11:58:47

L # 4

ganesh · 2009-01-04 16:13:57

JaneFairfax · 2009-01-04 23:37:30

You are not supposed to use a calculator or log tables for L # 4. Try again!

Last edited by JaneFairfax (2009-01-04 23:40:20)

ganesh · 2009-01-05 01:36:33

No, I didn't
I remember

and
.

JaneFairfax · 2009-01-05 21:57:49

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 00:30:04)

JaneFairfax · 2009-02-06 23:31:40

keveenjones · 2010-04-18 19:06:39

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)

rzaidan · 2010-04-18 21:04:41

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

rzaidan · 2010-04-18 21:14:13

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

ganesh · 2010-04-19 01:02:18

Gentleman,

Thanks for the proofs.
Regards.

jonnyj99 · 2010-08-16 16:15:11

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}. \,

reconsideryouranswer · 2011-05-27 19:59:31

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 (2011-05-27 20:05:01)

ganesh · 2011-05-27 23:57:00

Hi reconsideryouranswer,

This problem was posted by JaneFairfax. I think it would be appropriate she verify the solution.

#1 2006-03-08 23:54:27

Logarithms

#2 2006-03-09 02:41:40

Re: Logarithms

#3 2006-03-09 03:01:49

Re: Logarithms

#4 2006-03-09 03:06:00

Re: Logarithms

#5 2006-03-09 07:19:45

Re: Logarithms

#6 2006-03-09 16:17:14

Re: Logarithms

#7 2006-03-10 16:48:02

Re: Logarithms

#8 2006-03-10 20:03:38

Re: Logarithms

#9 2006-03-10 20:23:36

Re: Logarithms

#10 2009-01-04 11:58:47

Re: Logarithms

#11 2009-01-04 16:13:57

Re: Logarithms

#12 2009-01-04 23:37:30

Re: Logarithms

#13 2009-01-05 01:36:33

Re: Logarithms

#14 2009-01-05 21:57:49

Re: Logarithms

#15 2009-02-06 23:31:40

Re: Logarithms

#16 2010-04-18 19:06:39

Re: Logarithms

#17 2010-04-18 21:04:41

Re: Logarithms

#18 2010-04-18 21:14:13

Re: Logarithms

#19 2010-04-19 01:02:18

Re: Logarithms

#20 2010-08-16 16:15:11

Re: Logarithms

#21 2011-05-27 19:59:31

Re: Logarithms

#22 2011-05-27 23:57:00

Re: Logarithms

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