Math Stumper

[MrCodeDude]

e^x^infinity (infinite x's) = 10

What does x =?

 

[Juice Box]

ln e

 

[MrCodeDude]

That just means e = 10?

 

[dopcombo]

e^x^x^x.... =10

x^x^x^x.... = ln 10

x^(x^x^x...) = ln 10

x^(ln 10) = ln 10

x = (ln 10)^(1/ln 10)

?

 

[Brutuskend]

My dog's butt smells like ass.

 

[upsciLLion]

Originally posted by: Brutuskend
My dog's butt smells like ass.

Why have you been smelling your dogs butt? :shocked:

 

[MrCodeDude]

Originally posted by: Brutuskend
My dog's butt smells like ass.

FANTASTIC!

 

[MrCodeDude]

Originally posted by: dopcombo
e^x^x^x.... =10

x^x^x^x.... = ln 10

x^(x^x^x...) = ln 10

x^(ln 10) = ln 10

x = (ln 10)^(1/ln 10)

?

Thank you, I'll tell you if you were correct tomorrow.

 

[TuxDave]

x=ln(10)^(1/ln(10))

GODDAMIT... too slow...

 

[OOBradm]

Originally posted by: Brutuskend
My dog's butt smells like ass.

Teh Winnar!

 

[chuckywang]

this is just one of those recursive problems.....just have to be careful that it actually converges. not a problem for this problem.

 

[Goosemaster]

if x=0, then the answer would be 1.

Looks like it is taylor polynomials...🙁...x^x.......should approach ln10

 

[chuckywang]

Originally posted by: Goosemaster
if x=0, then the answer would be 1.

Looks like it is taylor polynomials...🙁...x^x.......should approach ln10

you sure about that?
if x = 0, then 0^0^0^ .... = y, which implies that 0^y = y which has no solution.

 

[Goosemaster]

whoops....I forgot about e..I meant that e^0=1 , so with laylor polynomails the ploynomaial functions would approach 10...not ln10.

Poor bastard..I hated these type of problems....what was it again: x^0 + (x-n)^N/n!....

 

[chuckywang]

Originally posted by: Goosemaster
whoops....I forgot about e..I meant that e^0=1 , so with laylor polynomails the ploynomaial functions would approach 10...not ln10.

Poor bastard..I hated these type of problems....what was it again: x^0 + (x-n)^N/n!....

e^x = SUM(x^n/n!,n,0,infinity). I'm not sure what you mean when you say the taylor polynomials would appraoch 10.

 

[dopcombo]

Originally posted by: MrCodeDude

Originally posted by: dopcombo
e^x^x^x.... =10

x^x^x^x.... = ln 10

x^(x^x^x...) = ln 10

x^(ln 10) = ln 10

x = (ln 10)^(1/ln 10)

?

Thank you, I'll tell you if you were correct tomorrow.

MrCodeDude, Eheh, were TuxDave and I right? Or was it some other way of converging it?