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**Jai Ganesh****Administrator**- Registered: 2005-06-28
- Posts: 47,766

When there are two numbers x and y, such that both x,y ≥1,

does it follow that y^x is always greater than x^y if x is greater than y?

No.

This is true only if y is greater than a certain CRITICAL Value.

Many years back, I tried to find this critical value of y for certain values of x.

Value of x Approximate value of y

10 1.3712886

100 1.04955

1000 1.0069805

10,000 1.000922309

100,000 1.00011514925

1,000,000 1.0000138158

10,000,000 1.00000161283

100,000,000 1.0000001843

1,000,000,000 1.0000000208

Illustration:- y^100 can be greater than 100^y only if the value of y ≥1.04955

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**Jai Ganesh****Administrator**- Registered: 2005-06-28
- Posts: 47,766

Mathsy, I think you missed this post!

With all the available technology, you could have well improved upon those digits!

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Occasionally, the box goes away before I can read all the new posts. Usually when there have been lots of new posts and it takes me a long time to read them all. I think that's what happened here.

There's a strong pattern emerging there, though.

Do you think it's possible to rearrange x^y=y^x to find y in terms of x?

If you could do that, you could find the critical value for any value of x.

I've got as far as the yth root of y = the xth root of x, but now I'm stuck.

Why did the vector cross the road?

It wanted to be normal.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,713

Hello, and welcome to the forum kylekatarn !

I will let Ganesh reply to this, but just thought I would say hi.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**Jai Ganesh****Administrator**- Registered: 2005-06-28
- Posts: 47,766

kylekatarn wrote:

it seems that yCritical "slowly"(?) approaches the limit of 1 as x approaches +oo

looking forward to hear comments on this topic!

Yes, you are correct! ycritical approaches 1, but is certainly greater than 1, as x approaches + ∞

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**Jai Ganesh****Administrator**- Registered: 2005-06-28
- Posts: 47,766

mathsyperson wrote:

I've got as far as the yth root of y = the xth root of x, but now I'm stuck.

Interestingly, if yth root of y = xth root of x,

it does not automatically follow that x=y

For example, if x=4 and y=2,

then this is true!

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**NIH****Member**- Registered: 2005-06-14
- Posts: 33

It's not possible to express y explicitly in terms of x using any of the standard elementary functions. However, there is a formal solution using something called the Lambert W function. This function can be evaluated using mathematical packages such as Maple and Mathematica, but no calculator currently has a button for it. See the references below for details.

http://mathworld.wolfram.com/LambertW-Function.html

http://www.americanscientist.org/template/AssetDetail/assetid/40804;_voi8-8bIm

http://www.orcca.on.ca/LambertW/

Another approach would be to use the Newton Raphson method. If a is an approximation to a root of

f(x) = 10^x - x^10 = 0, then a - f(a)/f'(a) will be a better approximation.

In this case, we have f'(x) = ln(10) * 10^x - 10x^9.

For example, if a = 1.4 is an approximate solution, then

1.4 - (10^1.4 - 1.4^10)/(ln(10) * 10^1.4 - 10*1.4^9) ~= 1.3744 is a better approximation.

This converges quite rapidly. The next two convergents are, to 10 decimal places, 1.3713296532 and

1.3712885814.

http://www.sosmath.com/calculus/diff/der07/der07.html

Finally, here's a page which will solve, for example, the equation 10^x = x^10, giving several answers in terms of the Lambert W function (aka the ProductLog function), along with the numeric values.

http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=equations&s2=solve&s3=basic

2 + 2 = 5, for large values of 2.

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