interesting dilemma (1 Viewer)

[AAEldar]

But since it goes on infinitely, the power of x must equal 7 (since the power is )

Hence

I can see what you're doing but I don't think it works. As funnytomato said, if you just use your calculator and store the seventh root of 7 in memory, then raise it to itself 7 times you'll get 7, then it becomes larger.

 

[RealiseNothing]

I can see what you're doing but I don't think it works. As funnytomato said, if you just use your calculator and store the seventh root of 7 in memory, then raise it to itself 7 times you'll get 7, then it becomes larger.

I'm pretty sure you can't use the substitution that x^x^x .... = 7 back into the original x^x^x.....

So that's why it works using that method.

 

[RealiseNothing]

It reminds me of this:

http://www.google.com.au/imgres?img...BNqisiAesv4jbDg&sqi=2&ved=0CCwQ9QEwAA&dur=236

 

[seanieg89]

Have to be more explicit about what the infinite power tower means. I think the standard meaning of such a notation is as follows:

Let S(n) be the sequence defined by:

S(1)=x, S(n+1)=x^S(n).

x^x^x^x^... is lim (n->inf) S(n) provided this limit exists.

Under this interpretation, known as tetration, 7^(1/7) is indeed a solution to x^x^x^...=7.