Simplify $x^\frac{1}{(\log_a x)}$ Simplify $x^\frac{1}{(\log_a x)}$
The solution in my textbook is the following:

Since $\log_a (x^\frac{1}{(\log_a x)}) = \frac{1}{\log_a x}$ $\log_a x
 = 1$,
therefore $x^\frac{1}{(\log_a x)} = a^1 = a.$

I understand the law used in the first line is $\log_a (x^y) = y \log_a x$. I also understand that $\log_a x$ multiplied by it's reciprocal equals to 1. However, I do not understand why it makes sense to take the $\log_a$ of our expression in the first place and neither do I understand how the conclusion in the first line helps us arrive at the conclusion in the second line that $x^\frac{1}{(\log_a x)} = a^1 = a.$
 A: First, the expression has sense only for $x > 0$.
So for $x > 0$, the number $y= x^{\frac{1}{\log_a(x)}}$ exists and is $> 0$, so you can compute its $\log_a$. The computation shows that $\log_a(y)=1$.
Now, do you know a lot of numbers whose $\log_a$ are equal to $1$ ?
A: For the second question, note that is the definition of logarithm:$$\log_a (x)$$ is that number $y$ such that $a^y = x$. Since you have shown that $\log_a(x^{1/\log_a(x)}) = 1$, this means that $a$ raised to the power of the RHS equals the term you are taking the $\log_a$ of. In other words, this means $$x^{1/\log_a(x)} = a^1 = a.$$

The first question is actually very good. Nothing is specified about the values of $a$ and $x$, which is a bit problematic. Here is how I would analyse when it makes sense to apply $\log_a$ to the initial expression:
Since we have a term of the form $\log_a(x)$ in $x^{1/\log_a(x)}$, it means that $a \in (0,\infty)$ and $a \neq 1$. Moreover $x > 0$ because that is the domain of $\log_a$. Since $\log_a(x)$ appears in the denominator of a fraction, it cannot equal zero, so $x \neq 1$ as well.
Secondly, since we have an expression of the form "$x$ raised to an exponent", we must have $x > 0$, but that does not give any new information. 
Lastly, to take $\log_a$ of an expression, that expression must have positive value. And, since $x > 0$, $x$ raised to any exponent is positive, so it makes sense to talk about
$$
\log_a(x^{1/\log_a(x)})
$$
whenever $a,x \in (0,\infty) \setminus \{ 1 \}$.

It is good practice to check, as done above, that the objects you are dealing with are well-defined.
A: Here's the proof with the steps arranged a little more logically, maybe this makes it clearer.
$x^\frac{1}{(\log_a x)} = a^{log_a(x^\frac{1}{(\log_a x)})}$
$=a^{\frac{1}{\log_a x}\log_a x}$
$=a^1$
$=a$
Here the first step works because $a^{log_a(z)}=z$, and the reason to do it is that it gets the answer - it's not an obvious thing to do but it works.
A: The argument uses the fact that, for $p,q>0$, we have
$$
p=q\qquad\text{if and only if}\qquad\log_ap=\log_bq.\tag{*}
$$
Since $x^{\frac{1}{\log_ax}}$ is quite a complicated expression as regards to the exponent, it makes sense to try and exploit $\log_a(x^k)=k\log_ax$. Indeed
$$
\log_a\bigl(x^{\frac{1}{\log_ax}}\bigr)=\frac{1}{\log_ax}\log_ax=1
$$
(for $x\ne1$, of course, or the given expression would not be defined to begin with).
Since $1=\log_aa$ by definition, we are in the situation
$$
\log_a\bigl(x^{\frac{1}{\log_ax}}\bigr)=\log_aa
$$
and therefore we can conclude
$$
x^{\frac{1}{\log_ax}}=a
$$
by the property (*).
A: By definition we have that
$$y=\log_a x \iff a^y=x \iff a=x^\frac1y$$
therefore
$$x^\frac{1}{(\log_a x)}=x^\frac1y =a$$
A: The second line simply uses the conclusion that we got to in the first line and the definition of logarithms in relation to exponents to conclude that $x^{\frac{1}{\log_a x}} = a$.
$x^{\frac{1}{\log_a x}}$
$\log_a\big({x^{\frac{1}{\log_a x}}}\big) \implies \frac{1}{\log_a x}(\log_a x) \implies 1$
$\log_a\big({x^{\frac{1}{\log_a x}}}\big) = 1$
$\log_a b = c \implies a^c = b$
$a^1 = {x^{\frac{1}{\log_a x}}}$
$x^{\frac{1}{\log_a x}} = a$
Of course, keep in mind that $x\neq1$ and $x>0$.
