Why is $\sqrt[n]{a}$ equal to $a^{1/n}$ only when $n\ge 2$? My teacher gave us the following logarithmic equation to solve for $x$.

$\log 4$ + $(1+1/(2x))\log 3$ = $\log(\sqrt[x]{3} + 27)$

As you can see it involves the term $\sqrt[x]{3}$, which later in the solution was converted into $3^{1/x}$. The values of $x$ obtained were $1/4$ and $1/2$. However my teacher said these two values were to be rejected, citing this &quot;rule&quot;; $\sqrt[n]{a}$ is equal to $a^{1/n}$ only when $n\ge 2$. Why is this so? My teacher didn't provide any reasoning for this and told us just to remember this rule and apply it accordingly.
$\sqrt[n]{x}$ is typically treated as equivalent to $x^{\frac{1}{n}}$, as seen in symbolic solvers such as WolframAlpha.
However, there is a lot of convention needed to be taken into account. Typically, the radical is used when dealing with $n\in\mathbb{N}$. As dxiv pointed out, the case when $n=1$, i.e; $\sqrt[1]{x}$ is redundant, so we're left with the cases when $n\geq2$.
Personally, I use them interchangeably, so will your calculator. We can presume $\sqrt[\sqrt{2}]{x}$, for example, should be treated not so differently to $x^{\frac{1}{\sqrt{2}}}$. Obviously, the latter notation is preferred in this case.
I know some mathematicians differentiate the use in complex analysis, for instance. It's convention to use the radical only when referring to a principal root, and $\frac{1}{n}$ when dealing with other branches.
Should you be docked marks for using them interchangeably? Probably not.
Will it affect calculation? If in your calculation separate branches need to be taken into account, or the more likely scenario, the condition that $n\in\mathbb{N}$ is implied, as seems to be the case in the eyes of your teacher.
Because we define $y=\sqrt[n]x$ as the inverse function of $y=x^n$ in the suitable interval just for $n\in\mathbb N$ and not for other numbers.