They cannot be the same!

If you like tetration or unfamiliar functions, check out my notes on differentiating tetrations: here and here . You may find them interesting :) Also, let me know if you liked them so that I can do more stuff. Anyway, onto the solution!

Notations:

• $W(\cdot)$ denotes the product log (aka Lambert W Function)
• $\text{ssrt}(\cdot)$ denotes the super square root which is discussed in this solution

$x^x$ and $y^y$ can be written as ${^{2}x}$ and ${^{2}y}$ . We can take the super square root of these two and we can see that $x = \text{ssrt}(y^y) = \boxed{e^{W(y\ln y)}} = \boxed{\frac{y\ln y}{W(y\ln y)}}$ But hold up, what is a "super square root"? Don't worry, I will show you:

Let $x^x = k$ where $k$ is some number. We can take the natural log on both sides here to get $x\ln x = \ln k$ . From here, we can use the fact that $x = e^{\ln x}$ : $x\ln x = \ln k \implies \ln(x)e^{\ln(x)} = \ln k$ Since the coefficient and exponent of $e$ are the same here, we can use the product log: $\begin{aligned} \blue{\ln(x)}e^{\blue{\ln(x)}} = \ln k \implies \blue{\ln x} &= W(\ln k) \\ \implies x &= \boxed{e^{W(\ln k)}} = \boxed{\frac{\ln k}{W(\ln k)}}\end{aligned}$ Since we are solving for $x$ and $x$ is in the form $x^x$ or ${^{2}x}$ , we can say that we are taking the "super square root" of $x^x$ or ${^{2}x}$ , hence it is equal to what is shown above.

So finally, since we know that $y = 0.225$ , we can plug $0.225$ in and we see that $x=\green{\boxed{e^{W(0.225\ln(0.225))}}} = \green{\boxed{\frac{0.225\ln(0.225)}{W(0.225\ln(0.225))}}} \approx \green{\boxed{0.532349726242}}$