Parentheses in Power Tower

Algebra Level 2

Suppose we start with a number 2 2 . Trivially, no matter the number of parentheses, the result is still 2 2 .

Squaring that, we have 2 2 2^2 . Then, the results are the same for any choice of parentheses.

For three 2 2 's, wherever the parentheses are set, the result is 16 16 . For instance, ( 2 2 ) 2 = 2 ( 2 2 ) (2^2)^2 = 2^{(2^2)} .

What about the power tower of at least four 2 2 's? Would the result still be the same for any choice of parentheses and for all number of 2 2 's?

Note: Order of operations holds.

Yes No

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1 solution

( 2 2 ) ( 2 2 ) = 4 4 = 2 8 = 256 \Large (2^{2})^{(2^{2})} = 4^{4} = 2^{8} = 256 , but

2 ( 2 ( 2 2 ) ) = 2 ( 2 4 ) = 2 16 = 65536 \Large 2^{(2^{(2^{2})})} = 2^{(2^{4})} = 2^{16} = 65536 .

So the answer is No \boxed{\text{No}} .

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