+0

Help!!

0
189
13
+146

$${x}^{{x}^{{x}^{...}}} = 2$$. Solve for x.

Sep 2, 2018

#9
+27775
+2

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

.

Sep 7, 2018

#1
+100778
+3

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

.
Sep 2, 2018
#2
+146
+2

How does $${x}^{{x}^{{x}^{{x}^{...}}}} = x^2$$?

Mathgenius  Sep 2, 2018
#3
+100778
+2

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

so

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

Melody  Sep 2, 2018
#4
+146
+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
+100778
+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
+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...

Sep 3, 2018
#7
+5038
+1

This was wrong.  Melody's approach is correct.

.
Sep 7, 2018
edited by Rom  Sep 9, 2018
#8
+100778
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
+27775
+2

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

.

Alan  Sep 7, 2018
#11
+100778
0

Thanks Alan :)

Melody  Sep 8, 2018