To the power of what?

Many of you are reading this problem as such: $(((2^{2})^{2})^{2})^{0}$ If this were what the problem looked like, then you would be right to use the property of exponents that says $(x^{a})^{b}=x^{ab}$ and say this simplifies to $2^{0}=1$ . But that would be wrong, because you see no parenthesis in the original problem. If you see no parenthesis in a problem like this, then you should assume it looks something more like this: $\Huge{2^{(2^{(2^{(2^{0})})})}}$ In this case you would simply use the order of operations and work from the inner to outer parenthesis, and that simplifies to $2^{4}=\color{#D61F06}{16}\space\space\space\Box$

To read more, see: how are exponent towers evaluated?

When solving towers of exponents, simply do from the top.

$\Large 2^{2^{2^{2^{0}}}} = 2^{2^{2^{1}}}=2^{4}= \boxed{16}$

$\Large { 2 }^{ { 2 }^{ { 2 }^{ { 2 }^{ 0 } } } }={ 2 }^{ { 2 }^{ { 2 }^{ { 1 } } } }={ 2 }^{ { 2 }^{ { 2 } } }={ 2 }^{ 4 }=16$

Exponents must be resolved starting from the highest level:

= 2^(2^(2^(2^0)))

= 2^(2^(2^1))

= 2^(2^2)

= 2^4

= 16

If you start solving from the top, so, 2 to the power 0 is 1. Then 2 to the power of 1 is 2. Then you can simply solve the problem.

$\large { 2 }^{ { 2 }^{ { 2 }^{ { 2 }^{ 0 } } } }={ 2 }^{ { 2 }^{ { 2 }^{ 1 } } }\\ { 2 }^{ { 2 }^{ { 2 }^{ { 2 }^{ 0 } } } }={ 2 }^{ { 2 }^{ 2 } }\\ { 2 }^{ { 2 }^{ { 2 }^{ { 2 }^{ 0 } } } }={ 2 }^{ 4 }\\ { 2 }^{ { 2 }^{ { 2 }^{ { 2 }^{ 0 } } } }=16$

Evaluate from top to bottom, not bottom to top ! $\begin{aligned} 2^{2^{2^{2^0}}} &= 2^{2^{2^1}} \\ &= 2^{2^2} \\ &= 2^4 \\ &= \boxed{16} \end{aligned}$

Therefore, the answer is 16 .

2 to the 0 power, like anything to the 0 power, is 1. Thus this problem reduces to 2 to the power of (2 to the power of 2) = 2 to the power of 4 = 16

The indices are always in the form of a^b = c. It means the base (a) is at the bottom of its form. The rest (b) remains as the exponent (index). Determine the exponent by calculating the exponentials. You must get c = 16, which is the result.

We can calculate from up to the down.. i.e. 2^0 =1, 2^1=2, 2^2= 4, 2^4=16....

Here, 2^0=1 and 2^2=4 ; so 2^4 = 16

In my views: Right now assume there is two solution of this problem 1 and 16. But we know it can't. So, we will check which one is correct. Assume given problem is x=((((2)^2)^2)^2)^0. Now we will take log of base 2 like this, log(x)=(((2)^2)^2)^0*log2 (it's been difficult to write proper expression of mathematics through mobile so sorry for that and here base of each log is 2 so log2=1).now log2=1,hence log(x)=(((2)^2)^2)^0. Similarly take log again like this, log(log(x))=((2)^2)^0. Next is log(log(log(x)))=2^0. Next log(log(log(log(x))))=0. Now select x=1 and check whether log is existing or not, so log(1)=0, but log(0)=???? It will not exist so, here 1 can't be the answer. Now check for 16, due to base of log is 2, so we will write 16=2^4, take log, log(16)=4, log(log(16))=2, log(log(log(16)))=1, and log(log(log(log(16))))=0. Hence 16 will be answer not 1.

Thanks Suman

For all maths calculations you must follow BODMAS/BIDMAS before using any other rules. There are no brackets which means that indices are dealt with first. This results in starting at the top of the tower and working your way down which means that the answer is indeed 16.

Many people simplify the expression as follows :

$\LARGE {2^{2^{2^{2^0}}} = (2^{2})^{2})^{2})^{0}}.$

According to this it's right to use the property $(a^m)^n = a^{mn}$

This results in $2^0,$ which is $1.$

In the problem brackets are not given , hence considering it like that is not right.

While simplifying towers of exponents like this one , we should start evaluating from the top and not from bottom.

$\LARGE {2^{2^{2^{2^0}}}} = 2^{(2^2)} = 2^4 = \boxed{16}$

So, this problem is confusing a lot of people.

You must break this down because the way it is written the rule (x^y)^z = x^(y*z) does not apply because it is not written that way.

So to simplify this you must start with the highest powers.

So the correct way to solve 2^2^2^2^0is: 2^0 = 1 Then 2^1 = 2 then 2^2 = 4 then 2^4 = 16

You must reduce the tower to a single level starting at the top and working your way down.

The answer is 16. Not 1. Google tetration and you will see that that is what this problem is. This is not a base raised to 2 exponents (as it would be (2^3)^2).

This is an exponent being raised to another exponent. So in order to simplify the base you have to know what the number it's being raised to it. BUT in order to know that number, you have to raise the exponent to its' exponent, and so on and so forth.

The fact that the exponents are not all on the same level should be a big clue that this is not simply multiplying exponents.

if you hover the mouse over the expression it reveals {2 ^ {2 ^ {2 ^ {2 ^ 0}}}}. The inmost operators are resolved first, so = {2 ^ {2 ^ {2 ^ 1}}} = {2 ^ {2 ^ 2}} = 2 ^ 4 = 16

First, solve 2^0. That will give you 1. Then, solve 2^(2^0). That will give you 2^1=2. Now, solve 2^[2^(2^0)]. That will give you 4. Finally, solve 2^{2^[2^(2^0)]}. That will give 2^4=16.

Rule is solve from top. So Answer is 16

$2^{0}$ = 1 therefore $2^{2}$ $^{2}$ $^{2}$ $^{0}$ = $2^{2}$ $^{2}$ $^{1}$ and $2^{1}$ = 2 so $2^{2}$ $^{2}$ $^{1}$ = $2^{2}$ $^{2}$ = $2^{4}$ = 16

2^2^2^2^0=2^2^2^1=2^2^2=2^4=16 (you can check this with a calculator too for those that don't believe this to be correct and the answer to be 1). you start with 2^0=1 then move down to 2^1=2 then move down again to 2^2 leaving you with 2^4=16. BODMAS.

2^2^2^2^0= 2^2^2^1= 16, so answer is 16 smile emoticon

My analysis here is purely by inspection and logic. One can mentally raise 2 to the 6th power which makes it 16. Then , since the last power is 0, you can discard it , stop right there and you don't do any operation anymore. Therefore, the answer is your last answer which is 16.

this is a perplexing question. but the thing to remember in this question is the rule of exponents. just come from top to bottom u will get the answer ((2)^(((2)^(((2)^(((2)^(0))))))))=((2)^(((2)^(((2)^(1))))))=((2)^(((2)^(2))))=((2)^(4))=16