Are there positive solutions to $x^q- a = 0$ other than $\sqrt[q]{a}?$ I'm reading a book that says the following (translated):
"Before we proceed, let us remember that, given a real number $a > 0$ and a integer $q > 0$, the symbol $\sqrt[q]{a}$ represents a positive real number such that its $q$-power equals to $a$, that is to say it's the only positive solution of $x^q-a=0$".
My whole problem is to show that there is one, and only one, positive real root to $x^q-a=0$. In the case $x^2 - a = 0$ I already couldn't go further. Any ideas? Thanks in advance.
 A: Consider the graph of $y=x^q - a$. Then the gradient is $\frac{dy}{dx} = qx^{q-1}$. When $x>0$, since $q>0$, $\frac{dy}{dx} > 0$. In other words, the function is increasing for $x>0$. At $x=0$, $y=-a$ and so $y<0$ since $a>0$. So there must be a positive solution, since the function is negative at 0, and is increasing, continuous and clearly has no upper limit. There must be only one positive solution because the function is increasing.
A: Suppose there exists another value, $b$ so that $b^q -a =0,$ and $b\in \Bbb{R}_{>0}.$ We know that $b\ne \sqrt[q]{a},$ so, $b<\sqrt[q]{a}$ or $b>\sqrt[q]{a}.$
Let's suppose that $\sqrt[q]{a}>1,$ and proceed as follows:
First, suppose $b> \sqrt[q]{a},$ then $b^q > a,$ and so $b^q-a > 0,$ but this is not possible, as $b^q-a =0.$
Finally, suppose $b< \sqrt[q]{a},$ then $b^q <a,$ and so $b^q-a <0,$ again  this is not possible, as $b^q-a =0.$
Conclude that $\sqrt[q]{a}$ is the only positive solution.
Please attempt the argument where $\sqrt[q]{a}<1,$ using this as a template.
A: Suppose that we know there is a positive number $\sqrt[q]{a}$ such that $(\sqrt[q]{a})^q-a=0$, but we don't know yet if there is a different positive number $x$ such that $x^q-a=0$. Using the identity
$$
w^n-z^n=(w-z)(w^{n-1}+w^{n-2}z+\dots+z^{n-1})
$$
we see that the equation $x^q-a=0$ is equivalent to
$$
\left(x-\sqrt[q]{a}\right)\left(x^{q-1}+x^{q-2}(\sqrt[q]{a})+\dots+(\sqrt{a})^{q-1}\right)=0 \, .
$$
One of the bracketed terms on the LHS must be equal to zero for the above equality to hold. Since the second bracketed expression is positive, we must have $x-\sqrt[q]{a}=0$, and so $x=\sqrt[q]{a}$.
