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?