Math Is Fun Forum

great_math

Member

Registered: 2008-10-07

Posts: 5

Inequality

Prove that

All_Is_Number

Member

Registered: 2006-07-10

Posts: 258

Re: Inequality

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)

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

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Inequality

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)

All_Is_Number

Member

Registered: 2006-07-10

Posts: 258

Re: Inequality

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.

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

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Inequality

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.

All_Is_Number

Member

Registered: 2006-07-10

Posts: 258

Re: Inequality

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.

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.

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

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Inequality

Do you like my non-calculator solution, though? Do you think it’s a brilliant solution?

bitus

Member

Registered: 2008-10-12

Posts: 20

Re: Inequality

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.