Is
$$y^{\frac{1}{\infty}} = 1 \quad ? $$
If that's correct. $\log_1 (1)$ will be solved.
For $y\in(0,\infty)$, $\lim_{x\to\infty} y^{1/x}=1$. For $y=0$, $\lim_{x\to\infty}y^{1/x}=0$.
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For $y\in(0,\infty)$, $\lim_{x\to\infty} y^{1/x}=1$. For $y=0$, $\lim_{x\to\infty}y^{1/x}=0$.