Financier wrote:

The operation x#n for all positive integers greater than 1 is defined in the following manner:

x#n = x to the power of x#(n-1)

If x#1 = x,

which of the following expressions has the greatest value?

A. (3#2)#2

B. 3#(1#3)

C. (2#3)#2

D. 2#(2#3)

E. (2#2)#3

Couple of things before solving:If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus:

\(a^m^n=a^{(m^n)}\) and not \((a^m)^n\), which on the other hand equals to \(a^{mn}\).

So:

\((a^m)^n=a^{mn}\);

\(a^m^n=a^{(m^n)}\) and not \((a^m)^n\).

Back to the original question:Let's replace # by @ as # looks like the symbol "not equal to" and it might confuse someone.

Given: \(x@n=x^{(x@(n-1))}\) and \(x@1=x\):

______________________________________

\(x@2=x^{(x@1)}=x^x\), as \(x@1=x\);

\(x@3=x^{(x@2)}=x^{(x^x)}=x^{x^x}\);

\(x@4=x^{(x@3)}=x^{(x^{x^x})}=x^{x^{x^x}}\);

...

Basically n in x@n represents the # of stacked x-es.

A. \((3@2)@2=(3^3)@2=(27)@2=27^{27}=3^{81}\)

B. \(3@(1@3)=3@(1^{1^1})=3@1=3\)

C. \((2@3)@2=(2^{2^2})@2=16@2=16^{16}=2^{64}\)

D. \(2@(2@3)=2@(2^{2^2})=2@16\) this will be huge number 2^2^2^2^2^2^2^2^2^2^2^2^2^2^2^2

=2^2^2^2^2^2^2^2^2^2^2^2^2^2^4

=2^2^2^2^2^2^2^2^2^2^2^2^2^16

=.... ;

E. \((2@2)@3=(2^2)@3=4@3=4^{4^4}=4^{256}=2^{512}\)

Option D (2^2^2^2^2^2^2^2^2^2^2^2^2^2^2^2) will be much bigger number than numbers from other answer choices.

Answer: D.

Sorry to bump this topic but I don't see what you did with x@3. Could you write out the complete calculation?