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**Makonorth****Member**- Registered: 2005-10-31
- Posts: 2

Someone showed me this problem today and I couldn't think of how to solve it:

x^x^x^x^x = 2

Anyone know how to solve it and if so can you explain how you get to your answer?

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

Is that (((x^x)^x)^x)^x or x^(x^(x^(x^x)))?

I'm just asking that to stall for time, because I have no idea how to start trying to solve it. Probably using logs.

Why did the vector cross the road?

It wanted to be normal.

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**kylekatarn****Member**- Registered: 2005-07-24
- Posts: 445

x^x^x^x^x=2

x ≈ 1.425385621 ≈ √2

I think the solution of x^x^x^...........=2 converges into √2.

If someone proved that it would be remarkable.

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**Makonorth****Member**- Registered: 2005-10-31
- Posts: 2

That's exactly how it was written.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,615

Well then, you should exponentiate in this order I believe:

x^(x^(x^(x^x)))

In Excel (by trial and error) I get: 1.432694...

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

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**kylekatarn****Member**- Registered: 2005-07-24
- Posts: 445

yes, its very interesting.

As you add up more "^x" terms, the solution get closer to SQRT(2).

Again, I would like to see a demonstration of this (or a counter example)

: )

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

Hmm. It's obvious that once it gets to 2 then it will stay there because (√2)² = 2, but that's not anywhere close to a proof.

Why did the vector cross the road?

It wanted to be normal.

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**kylekatarn****Member**- Registered: 2005-07-24
- Posts: 445

It's not a proof, but a start.

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