Two solutions are possible i.e. 2 and 4
I claim whoever has answered as 4,they must also be right.

It can be written as

(Sqrt 2 )^x = x
=> 2^(x/2) = x
=> x = 2 or 4

so both 2 and 4 are right

=> [Sqrt(2)]^x=x
=>Sqrt(2)=(x)^(1/x); //powering both side with (1/x)
=>x=2;//2^(1/2)=Sqrt(2).

Given,
[Sqrt(2)]^[Sqrt(2)]^[Sqrt(2)]^[Sqrt(2)] ........... infinite = x
then we can write,
[Sqrt(2)]^x=x
so, x=2
so option 10)2 is the right answer

[Sqrt(2)]^[Sqrt(2)]^[Sqrt(2)]^[Sqrt(2)] ........... infinite = x

or [Sqrt(2)]^x = x

or [Sqrt(2)] = 2^(1/2) = x^(1/x)

so, x = 2

The square root of 2 is the only real number other than 1 whose infinite is equal to its square.

sqrt=1/2
so 2 and 1/2 will cancellwd out last is 2^1=2

x=[sqrt(2)]^[sqrt(2)]........infinite
or, x=[sqrt(2)]^x
or, x=(2)^(x/2)
taking LOG on both side we get
or, log(x)=(x/2)*log(2)
or, 2*log(x)=x*log(2)
or, x^2=2^x
By using option we get x=2
But Here x=4 also satisfies the equation

option 10 is correct by solving

according to the sequence
x = sqrt(2)^x
so
x^(2/x) = 2
so
x = 2
so option 10
That's it.

sqrt(2)^x=x
x^2=2^x
=> x=2

2^x/2=x
x=2

By "infinite tetration", I'm going to assume you mean the limit of the sequence
A to the base 0 = 1
An + 1 = sqrt(2)^an
The following is meant to be a proof sketch:

1. An is an increasing sequence as long as
a < 2 this is true because
f(x) = sqrt(2)^x > x
for the interval
0 < x < 2

2. The sequence is bounded above namely if then
An < 2 then
An+1 = sqrt(2)^infinite < sqrt(2)^2 = 2
By induction, this is always true; moreover it also means that the sequence is always increasing.

3.The limitof L the sequence must satisfy,
limt An+1 = limt 2^an/2
i.e.
L = 2^L/2
(by continuity of exponentiation). Only
L = 2 and L = 4
andwork here - it's easy to verify that these are solutions, and writing
g(x) = 2^x/2 - x
and differentiating we find that (g) decreases on (0, 3.05) and increases on
(3.05 , infinite)

(note - it's not exactly 3.05) - thus we can have at most two zeros ofg(x) , which means only two possible values for L
(since our equation for
L is equivalent to)

g(L) = 0
Putting it all together:
An is bounded above by 2 and strictly increasing, and thus must converge (namely, to the least upper bound of the sequence). The only values it can converge to are 2 and 4; since we know every term in the sequence is less than 2 , it must converge to 2.

[[Sqrt(2)]^[Sqrt(2)]^[Sqrt(2)]^[Sqrt(2)] ........... infinite = x
xlog[Sqrt(2)=logx
log[Sqrt(2)=1/xlogx
1/2log2=1/xlogx
x=2 ans

[Sqrt(2)]^x=x;
[Sqrt(2)]=(x)^(1/x);
=2

[Sqrt(2)]^[Sqrt(2)]^[Sqrt(2)]^[Sqrt(2)] ........... infinite = x

or x=[Sqrt(2)]^x
or x^1/x=2^1/2
from above equation we can say that x=2

sqrt(2)^sqrt(2)^sqrt(2).....infinite=x
let sqrt(2)^(sqrt(2)^sqrt(2)^sqrt(2)....)=x
sqrt(2)^x=x
multiplying 1/x on both powers,
sqrt(2)=x^1/x
=>x=2
ans is 2.

[sqrt 2] ^ [sqrt 2] ^ [sqrt 2] ........ infinity =x
hence, [sqrt 2] ^ x =x
[2] ^ (x/2) = x
2= x ^ (2/x)
on solving we get x= 2

[Sqrt(2)]^[Sqrt(2)]^[Sqrt(2)]^[Sqrt(2)] ........... infinite = x

[Sqrt(2)]^x = x

By hit and trial, x=2 and x=4 satisfy

[Sqrt(2)]^[Sqrt(2)]^[Sqrt(2)]^[Sqrt(2)] ........... infinite = x

or [Sqrt(2)]^x = x

or [Sqrt(2)] = 2^(1/2) = x^(1/x)

so, x = 2..........option 10 is the answer

by elimination of invariant
2^x=x
=>x=2

sqrt(2) is roughly equal to 1.414.
So, our problem simplifies to 1.414^1.414...........
If we take finite steps say 2,3,4 etc we see that the value converges to 2.
So, x=2

this the answer.
2^1/2*x=x
by squaring both side
2^1/2=x^1/x
so x=2.

Answer is 10

Sqrt(2)]^[Sqrt(2)]^[Sqrt(2)]^[Sqrt(2)] ........... infinite = x
THUS, sqrt(2)^x=x
2^(x/2)=2,
which is possible at x=2
which is option 10

a/c to Q sqrt(2)^x=x
or x=2 satisfy this...

[Sqrt(2)]^[Sqrt(2)]^[Sqrt(2)]^[Sqrt(2)] ........... infinite = x then sqrt2=x ...(i) squiring both sides we get x=2