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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 15,130

L # 1

Show that

Character is who you are when no one is looking.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

So:

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

IPBLE: Increasing Performance By Lowering Expectations.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 15,130

Character is who you are when no one is looking.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 15,130

L # 2

If

show that xyz=1.

Character is who you are when no one is looking.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

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.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 15,130

**Well done, krassi_holmz!**

Character is who you are when no one is looking.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 15,130

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.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

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)*

IPBLE: Increasing Performance By Lowering Expectations.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 15,130

**Very well done, krassi_holmz! **

Character is who you are when no one is looking.

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

**L # 4**

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 15,130

Character is who you are when no one is looking.

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

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)*

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 15,130

No, I didn't

I remember

and

.

Character is who you are when no one is looking.

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

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)*

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**JaneFairfax****Member**- Registered: 2007-02-23
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**keveenjones****Member**- Registered: 2010-04-18
- Posts: 2

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)

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**rzaidan****Member**- Registered: 2009-08-13
- Posts: 59

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

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**rzaidan****Member**- Registered: 2009-08-13
- Posts: 59

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

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 15,130

Gentleman,

Thanks for the proofs.

Regards.

Character is who you are when no one is looking.

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**jonnyj99****Member**- Registered: 2010-08-16
- Posts: 4

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}. \,

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**reconsideryouranswer****Member**- Registered: 2011-05-11
- Posts: 172

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)*

Signature line:

I wish a had a more interesting signature line.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 15,130

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.

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