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**Inequality**

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**great_math****Member**- Registered: 2008-10-07
- Posts: 5

Prove that

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**All_Is_Number****Member**- Registered: 2006-07-10
- Posts: 258

Apply the change of base formula:

From there, both expressions can be easily reduced to numerical values.

*Last edited by All_Is_Number (2008-11-04 05:18:13)*

*You can shear a sheep many times but skin him only once.*

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

What do you mean by "reduced to numerical values"? You are NOT supposed to use a calculator in this problem!!

Here is my solution.

Note that the function

is strictly increasing for .Hence, for all

, .Now define

Then

.Since

for all , we have by above.Thus

is strictly decreasing for .In particular,

.But

.*Last edited by JaneFairfax (2008-11-07 08:54:58)*

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**All_Is_Number****Member**- Registered: 2006-07-10
- Posts: 258

JaneFairfax wrote:

You are NOT supposed to use a calculator in this problem!!

Have you ever given any consideration to leaving your arrogance and anger at the login screen? The OP did not list any such restrictions on the problem. My method is perfectly legitimate. If you have a different way, fine. I have no doubt I could also find other ways to do it. However, I posted what, thus far, appears to be the easiest way to go about it.

*You can shear a sheep many times but skin him only once.*

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

Look. If the problem were so easy as to be solvable by mechanically pushing keys on a gadget, there would have no need for **great_math** to post it at all. If you had taken a look at the problems that **great_math** usually posts here, you should have realized that **great_math**s problems are usually more sophisticated than that.

If your mathematical ability is limited to pushing keys on your calculator, I suggest you take a good look around at what constitutes a good mathematical proof. You might learn something.

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**All_Is_Number****Member**- Registered: 2006-07-10
- Posts: 258

JaneFairfax wrote:

Look. If the problem were so easy as to be solvable by mechanically pushing keys on a gadget, there would have no need for

great_mathto post it at all. If you had taken a look at the problems thatgreat_mathusually posts here, you should have realized thatgreat_maths problems are usually more sophisticated than that.If your mathematical ability is limited to pushing keys on your calculator, I suggest you take a good look around at what constitutes a good mathematical proof. You might learn something.

Perhaps you should re-read the first sentence of my reply to you.

Be careful not to confuse recognition of the usefulness of a calculator with reliance on a calculator.

*You can shear a sheep many times but skin him only once.*

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

Do you like my non-calculator solution, though? Do you think its a brilliant solution?

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**bitus****Member**- Registered: 2008-10-12
- Posts: 20

Here's another suggestion; First we prove: if a,b,c,d positives, b<a<d<c, a-b>c-d, then a/b>c/d. Let a=b+k(k>0), c=d+m(m>0). Since a-b>c-d, b+k-b>d+m-d so k>m or k/m>1. Since b/d<1, k/m>b/d or k/b>m/d or a/b>c/d. Now; log(7)10=log10/log7 & log(11)13=log13/log11. Let log10=a, log7=b, log13=c & log11=d. Then, 10^a=10, 10^b=7, 10^c=13 & 10^d=11 or 10^(a-b)=10/7 & 10^(c-d)=13/11. You don't need a calculator to see that 10/7>13/11. So, 10^(a-b)> 10^(c-d) so, a-b>c-d (1). Since 0<log7<log10<log11<log13, 0<b<a<d<c (2). From (1) & (2) ; a/b>c/d or log10/log7>log13/log11 or log(7)10>log(11)13.

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