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

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

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

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

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

so

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

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

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

x^(x^x^x^x...) must be x^2

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...

This was wrong. Melody's approach is correct.

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 ?

Thanks Alan :)