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\({x}^{{x}^{{x}^{...}}} = 2\). Solve for x.

Mathgenius  Sep 2, 2018

Best Answer 

 #9
avatar+27130 
+2

Certainly there is such a value of x!  Melody's original reply is correct.

.

Alan  Sep 7, 2018
 #1
avatar+93872 
+3

\({x}^{{x}^{{x}^{...}}} = 2 \qquad\\ so\\ x^2=2\\ so\\ x=+\sqrt{2}\)

Melody  Sep 2, 2018
 #2
avatar+82 
+2

How does \({x}^{{x}^{{x}^{{x}^{...}}}} = x^2\)?

Mathgenius  Sep 2, 2018
 #3
avatar+93872 
+2

because x^x^x^x .... =2

so

x^(x^x^x^x..) = x^2

Melody  Sep 2, 2018
 #4
avatar+82 
+1

I'm sorry, I am confused on x^(x^x^x^x..) = x^2. How do you decide this?

Mathgenius  Sep 2, 2018
 #5
avatar+93872 
+1

It says in the question that x^x^x^x.... =2 

 

so

 

x^(x^x^x^x...)  must be  x^2

Melody  Sep 3, 2018
edited by Melody  Sep 3, 2018
 #6
avatar+20 
0

Actually you are not aksed about the limited equation.

You are asking that x^x^x^x^x so on = 2

then the multiplication is done through out the equation by power rule : 

2^2^2^2...

RockHouzy  Sep 3, 2018
 #7
avatar+2765 
+1

 

This was wrong.  Melody's approach is correct.

Rom  Sep 7, 2018
edited by Rom  Sep 9, 2018
 #8
avatar+93872 
0

Thanks Rom,

Would it be fair to say that the fallacy in my answer is that i assumed such an x existed in the first place ?

Melody  Sep 7, 2018
 #9
avatar+27130 
+2
Best Answer

Certainly there is such a value of x!  Melody's original reply is correct.

.

Alan  Sep 7, 2018
 #11
avatar+93872 
0

Thanks Alan :)

Melody  Sep 8, 2018

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