If $x^{x^{x+1}}=\sqrt{2}$, then evaluate $x^{x^{p}}$, where $p = 2x^{x+1}+x+1$

Question

I can't figure out how to give a proper form to this expression to use the root of two.

If $$x^{x^{x+1}}=\sqrt{2}$$ find the value of $W$ if $$W=x^{x^{p}} \quad\text{where}\; p = 2x^{x+1}+x+1$$

EDIT: This is an algebraic manipulation problem with the exponents. There was an error in the previous version (see the Edit History) that I have corrected.

Answer 1

Doing it numerically I get $W=2.7564025221095 $. Is this what you are after or do you need to represent this with $\sqrt 2$?

Answer 2

$x^{x^x} = \sqrt{2} \implies x^x = log_x\sqrt2$

Now:

$W = x^{x^{2x^{x+1}+x+1}} = x^{\left(x^{2x^{x+1}}x^xx\right)} = x^A$

Let the exponent be A for the time being:

$A = x^{2x^{x+1}}x^xx = \left(x^{x^{x+1}}\right)^2x^xx = \left(x^{x^xx}\right)^2x^xx$

Since $x^x = log_x\sqrt2$, we have $A = \left(x^{xlog_x\sqrt{2}}\right)^2xlog_x\sqrt{2} = (\sqrt{2}^x)^2xlog_x\sqrt{2}$

Let $\sqrt{2}^x = B \implies A = B^2log_x(B) = log_x(B^{B^2})$

Remember $W = x^A = B^{B^2} = (\sqrt{2}^x)^{2^x}$

What to do from here?

Answer 3

Assuming $x^{x^{x+1}} = \sqrt 2$, \begin{align*} x^{x^{2x^{x+1} + x + 1}} &= x^{x^{2x^{x+1}} \cdot x^{x+1}} \\ &= x^{\left(x^{x^{x+1}}\right)^{2} \cdot x^{x+1}} \\ &= x^{\left(\sqrt{2}\right)^{2} \cdot x^{x+1}} \\ &= x^{2 \cdot x^{x+1}} \\ &= \left(x^{x^{x+1}}\right)^{2} \\ &= \left(\sqrt{2}\right)^{2} \\ &= 2 \end{align*}

Answer 4

To avoid 3-story exponents, let $y = x^{x+1}$

Then it is given that $x^y = \sqrt{2}$ or $x^{2y} = 2$

$$p = 2y + (x+1)$$

$$x^p = x^{2y} x^{x+1} = 2y$$

$$x^{x^{p}} = x^{2y} = 2$$