SOLUTION: This question talks about how the sqrt of x is to the exponent of sqrt of x indefinitely. The answer to it is 2 but the goal is to find x. How would you solve this question? Ima

Algebra ->  Expressions-with-variables -> SOLUTION: This question talks about how the sqrt of x is to the exponent of sqrt of x indefinitely. The answer to it is 2 but the goal is to find x. How would you solve this question? Ima      Log On


   



Question 1134981: This question talks about how the sqrt of x is to the exponent of sqrt of x indefinitely. The answer to it is 2 but the goal is to find x. How would you solve this question?
Image : https://ibb.co/pdHnWmt''.
I don’t get how it’s 2. Please explain a method without involving graphing calculator.

Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Previously answered (question 1134770); the response is repeated here.

Given: sqrt(x)^sqrt(x)^sqrt(x)^sqrt(x)^sqrt(x)^sqrt(x)... = 2

Square both sides, remembering that %28a%5Eb%29%5E2+=+a%5E%282b%29

sqrt(x)^(2*sqrt(x)^sqrt(x)^sqrt(x)^sqrt(x)...) = 4

sqrt(x)^(2*2) = 4 [everything after the 2 in that equation is equal to what we started with; its value is 2]

sqrt(x)^4 = 4

x^2 = 4

x = 2

The result can be verified using excel, or a graphing calculator.

(1) Calculate the square root of 2;
(2) calculate the square root of 2 raised to that power;
repeat step (2) repeatedly

Answer by ikleyn(52787) About Me  (Show Source):
You can put this solution on YOUR website!
.


Given :  sqrt(x)^sqrt(x)^sqrt(x)^sqrt(x)^sqrt(x)^sqrt(x)... = 2.    (1)



Look into the many-floor degree index at the lowest  sqrt%28x%29.


This  many-floor degree index is nothing else as 2 (as it is given !).


So you can rewrite equation (1) in an equivalent form


    %28sqrt%28x%29%29%5E2 = 2.     (2)


Since  %28sqrt%28x%29%29%5E2 = x,   equation (2) is equivalent to


    x = 2,    which gives you the ANSWER.

Solved.