$log_1(x)$ existance

Question

I was looking at the equation $2^x=29$ and I saw that it was the same thing as asking $2=\sqrt[x]{29}$ So $$ 1^x=27$$ but is also $$1=\sqrt[x]{27}$$ I also saw that $\sqrt[x]{27}$ goes to 1 as gets bigger. Which can be expressed as $log_1(27)=lim_{x \to \infty} \sqrt[x]{27}$So does that mean that $log_1(x)$ could exist?

Answer 1

$\sqrt[x]{27}=27^{1/x}$

And

$\lim_{x\to\infty} \frac{1}{x}=0$

So

$\lim_{x\to\infty} 27^{1/x} = 1$

Therefore:

$\nexists x\in \mathbb{R}$ s.t. $\sqrt[x]{27} =1$

Note this follows for any positive real $\gt 1.$