Intutively, why does $x^{\frac{1}{\ln x}} = e$?

Question

For any $x \gt 0$, we have this identity:

$$x^{\frac{1}{\ln x}} = e\text.$$

You can see this by using the fact that $x = e^{\ln x}$.

I'm wondering if there's a good intuitive explanation for this one, given that $x^{\frac{1}{k}}$ is the operation that inverts raising $x$ to the $k$th power and $\ln x$ is the inverse of the exponential function. Is there some compelling intuitive or geometric argument that makes this identity more obvious than algebraic rearrangement?

And this is because $\frac{1}{\log_{a} x} =\log_x{a}$. More generally: $$\frac{\log_a b}{\log_a c} = \log_c b.$$ — Thomas Andrews, Feb 24 '16 at 05:56
$x^{1/\ln(x)}=e$ when $x>0$. For $x\leq0$, you have $x^{1/\ln(x)}$ is undefined. — 2'5 9'2, Feb 24 '16 at 06:04
I don't really understand what sort of argument you have in mind. Do you have any intuitions about $e$, or about the natural log as distinct from logarithms with respect to other bases? — Qiaochu Yuan, Feb 24 '16 at 06:33
@Thomas Andrews That second observation is beautiful. You should concert that to an answer. — templatetypedef, Feb 24 '16 at 07:09

score 4 · Accepted Answer · answered Feb 24 '16 at 05:59

The expression $\log x$ describes a number that is an exponent: specifically, it is the exponent when, to which $e$ is raised, yields $x$. When we speak of $1/\log x$, this is in some sense an inverse exponent: for example, if I write $$a^b = c,$$ then $$c^{1/b} = a.$$ So $1/b$ is the exponent to which $c$ is raised to "recover" $a$. Similarly, then, as $\log x$ is the exponent to which $e$ is raised to yield $x$, we must then have $1/\log x$ as the exponent to which $x$ must be raised to yield $e$.

Of course, all of the above is nothing more than a more prosaic restatement of the algebraic rules that you already wrote in your question.

score -1 · Answer 2 · answered Feb 24 '16 at 05:56

-1

$$x^{\displaystyle\left(\frac 1{\ln x}\right)}=e\iff \left(x^{\displaystyle\left(\frac 1{\ln x}\right)}\right)^{\ln x}=e^{\ln x}\iff x^1=x^{\ln e}=x^1\equiv{\bf T}$$

answered Feb 24 '16 at 05:56

RE60K

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Intutively, why does $x^{\frac{1}{\ln x}} = e$?

2 Answers2