Simplify $x^\frac{1}{(\log_a x)}$

Question

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

Answer 1

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 $1$ ?

Answer 2

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 $$x^{1/\log_a(x)} = a^1 = a.$$

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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 $a,x \in (0,\infty) \setminus \{ 1 \}$.

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It is good practice to check, as done above, that the objects you are dealing with are well-defined.

Answer 3

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.

Answer 4

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 (*).

Answer 5

By definition we have that

$$y=\log_a x \iff a^y=x \iff a=x^\frac1y$$

therefore

$$x^\frac{1}{(\log_a x)}=x^\frac1y =a$$

Answer 6

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 $x>0$.