I encountered a question in which I had to find $x$ and the question had $3^{1/x}$ and I got $x=1/2$ and $1/4$ as its solution. But my textbook said that "$x$th root of $3$" is valid for $x\geq 2$ and $x$ should be a natural number.
But the question did not directly mention $x$th root 3 and instead "$3^{1/x}$". Do they actually mean the same? Do they have same restrictions on $x$?
PS: The actual question was "Solve for real values of $x$: log $4$ + ($1$+ $1/2x$) log $3$= log ($3^{1/x}$ + $27$)"
Best Answer
In general if $b > 0; b\ne 1$ then $b^x; x \in \mathbb R$ has a rather complicated definition based on calculus which I won't actually go into detail what it is but it is all based on the fundamental relation $b^{x+y} = b^xb^y$.
But there are some basic concepts that fall into generalizations.
$b^0 = 1$ (it has to if $b^x = b^{0+x} = b^0b^x$)
And if $n\in \mathbb N$ then $b^n = \underbrace{b\cdot ..... \cdot b}$.
And if $x = \frac 1n; n\in \mathbb N$ then, yes, $b^{\frac 1n}$ will indeed have to be then $n$th root of $b$. They are the same thing.
(It has to be..... If $b^{x+y} = b^xb^y$ then $(b^{\frac 1n})^n = \underbrace{b^{\frac 1n } \cdot ...\cdot b^{\frac 1n}} = b^{\frac 1n + \frac 1n + ....+\frac 1n} = b^1 = b$.)
So IF we assume $x$ is a natural number then $3^{\frac 1x}$ will be the $x$th root of $3$.
But you haven't actually told us what the question was or why we should assume $x$ is a natural number.
If $x$ is a real number we do have $(b^{\frac 1x})^x = b^{\frac xx} = b^1 = b$. So whatever $b^{\frac 1x}$ is, if we raise it to the $x$ power we will get $b$. We can casually refer to that as "the $x$th root of $b$" but if $x$ is not a natural number that might no be clear what that means. After all $3^{\frac 1{\pi}} = 3^{0.318....} \approx 1.4186....$ and $(3^{\frac 1{\pi}})^\pi =(1.4186.....)^{3.1415....} = 3$. So we could so $3^{\frac 1\pi}$ is the $\pi$th root of $3$ but... that's not usually what we mean by "root". Which usually means a root is a solution to a polynomial. And as a polynomial only has natural numbers as powers as solution to $x^\pi = 3$ isn't usually called a "root".... at least, I wouldn't call it that.
But... FWIW..... The number $3^{\frac 1x}$ will always be a solution to "$w^x = 3$ solve for $w$".