What is value of $2^{2^{2^{2^0}}}$?

Question

I did Google search. I gone through PEDMAS rule. I think answer would be 16.

So can anyone explain this ?

$$\huge 2^{2^{2^{2^0}}} = \text{?}$$

Answer 1

$2^0 = 1$. So $2^{2^0} = 2^1 = 2$ (the order of operations is that you consider each power acting on the number directly beneath it). Similarly, $2^{2^{2^0}} = 2^2 = 4$, and so $2^{2^{2^{2^0}}} = 2^4 = 16$, as you suspected.

If you want to apply order of operations differently, you have to use parentheses. For example, if you wanted to find $2^3$, raised to the 5th power, you would write $\left(2^3\right)^5$, whereas $2^{3^5}$ is read as 2 raised to the $3^5$-th power.

Answer 2

The usual rule is to read $a^{b^c}$ without parentheses as $a^{(b^c)}$ because $(a^b)^c=a^{bc}$ and the latter is a simpler expression. So we come down from the top and $ 2^{2^{2^{2^0}}} =2^{2^{2^1}}=2^{2^2}=2^4=16$ supporting your claim.

Answer 3

PEMDAS does not really specify what to do here. Generally you are told that if the operation has the same priority, then you should do the operation "left to right". So for instance: $5-3-2 = (5-3)-2 = 0$ not $5-(3-2) = 4$.

Most people would read this as exactly how you did, which is that it's $16$.

The problem is that PEMDAS doesn't address the style that you wrote it in. If I simply assume that a superscript means exponentiation then there are a couple of ways to interpret your expression. To get away from this style, instead of using a superscript, let's use a symbol for exponentiation: $\text{^}$ such that we understand that $a\text{^}b = a^b$.

So if we do that, then your expression (at least appears) to become:

$$ 2\text{^}2\text{^}2\text{^}2\text{^}0 $$

Now, if we use the "left to right rule" this becomes:

$$ ((((2\text{^}2)\text{^}2)\text{^}2)\text{^}0) $$

Which clearly equals $1$ (since it's eventually some number raised to the $0$ power).

This problem is actually incredibly clear when you try and write your expression in latex. You must write: 2^{2^{2^{2^0}}}...do you see the implied parenthesis. On the other hand the expression 2^2^2^2^0 gives an error: $2^2^2^2^0$ because the expression doesn't make any sense--if you were to just manually write it, it would like: $2^{2220}$ which is clearly not what was intended.

So PEMDAS has no bearing here just as PEMDAS has no bearing in the following expression:

$$ \frac{5-3}{2} $$

This expression, we all know, equals $1$. But if the fraction symbol means divide we would use PEMDAS to read this as $5-3/2 = 5-1.5 = 3.5 \neq 1$. But it's because PEMDAS has no bearing on this notation--this notation has an implied parentheses and actually means $(5-3)/2$ just as the notation $2^{2^0}$ has an implied parentheses: $2\text{^}\left(2\text{^}0\right)$.

So this question isn't actually related at all to PEMDAS, rather it's related to how we interpret certain notations.

Answer 4

$2^{(2^{(2^{(2^0)})})} =2^{(2^{(2^1)})}=2^{(2^2)}=2^4=16$

But $((((2^{2})^2)^2)^0)=(((4^2)^2)^0)=((16^2)^0)=256^0=1$

The basically $(a^b)^c \ne a^{(b^c)}$. So which of these two different terms is meant by $a^{b^c} $?

The usual accepted value is $a^{(b^c)} $. The choose is arbitrary but as $(a^b)^c =a^{bc} $ it seems $a^{(b^c)}$ "needs it more".

So.... I agree with you. It should be 16.

Answer 5

If you mean this $2^{(2^{(2^{(2^0)})})}$ then it is tetration solution.

$2^{(2^{(2^{(2^0)})})} =^42^0=^32=16$

Otherwise $2^{2×2×2×0}=2^0=1$

Answer 6

My Answer is: It SHOULD equal 16.

but if your super-script sizing is accurately portrayed, then my answer is 2 read on to find out why...

The problem:

Your notation makes it ambiguous because of the SIZING of the super-script exponents

the 3rd & 4th 2, and the top 0 are all the same size, giving us only ambiguity.

The SHOULD ANSWER (16):

so now the assumption has to be made that you intended them to get smaller at each level going up.

in which case it only has the 1 answer and would not be ambiguous. I disagree with the chosen answer; yet provide similar detail.

if all 4 exponents do get smaller and hence are true super-script exponents, then each super-script digit is on the one preceding, and so on backwards down through the notation, equaling 16.

To show the =16 scenario with brackets the size is no longer important because the brackets control the order. $$2^{(2^{(2^{(2^0)})})}$$

Executed in this order...

$2^0 = 1$

$2^1 = 2$

$2^2 = 4$

$2^4 = 16$

For the is NOT answer equals 1 (I do not agree with)

people discuss previously (which I do not believe is an option) answer equals 1; to be written without brackets it would be confusing, the super-script SIZE at each level would be the same, and hence to write it cleanly would require Brackets, otherwise it would look like this $$2^{2\,2\,2\,0} $$ but this looks wrong (maybe each level does need to lift a little but they all should be the same size), and here with brackets: $$(((2^2)^2)^2)^0$$ Notice how the 2nd, 3rd, 4th and 5th levels are the same size super-script

In this scenario, which I believe is not an option:

executed in this order:

$2^2 = 4$

$4^2 = 16$

$16^2 = 256$

$256^0 = 1$

Nope not how the super-script works.

Devil's advocate: (the only other option)

If your sizing was purposeful (and the top three Super-scripts are the same size), then someone could conclude that the 4th is superScript around the bottom 2nd and 3rd, and the 5th is superscript around 2nd-4th

Meaning the first 2 is raised to the power of 1 in which case THE ANSWER = 2

I demonstrate here

$$2^{((2^2)^2)^0}$$

Check Sizes of my Super-script here, they pretty much match the question: I used this MathJax: $$2^{((2^2)^2)^0}$$

Hence executed in this specific order:

$2^2 = 4$

$4^2 = 16$

$16^0 = 1$

$2^1 = 2$

So my main point is your Exponent (super-script) sizing is the problem and hence it cannot be explained, without more information.