JaneFairfax, here is a basic proof of L4:

For all real a > 1, y = a^x is a strictly increasing function.

log(base 2)3 versus log(base 3)5

2*log(base 2)3 versus 2*log(base 3)5

log(base 2)9 versus log(base 3)25

2^3 = 8 < 9

2^(> 3) = 9

3^3 = 27 > 25

3^(< 3) = 25

So, the left-hand side is greater than the right-hand side, because

Its logarithm is a larger number.

I saw this post today and saw the probs on log. Well, they are not bad, they are good. But you can also try these problems here by me (Credit: to a book):

http://www.mathisfunforum.com/viewtopic … 93#p399193

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

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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:

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log_2(\sqrt[3]4) = \frac {1}{3} \log_2 (4) = \frac {2}{3}. \,

]]>Thanks for the proofs.

Regards.

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]]>

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]]>

b = log x log y

log a + 3 b = 5log x

loga - 2b = 3logy + 2logy = 5logy

logx / logy = (loga+3b) / (loga-2b)

]]>I remember

and

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