\({x}^{{x}^{{x}^{...}}} = 2\). 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 :)