Reducing logarithms and exponentials

Question

I know that...
log(m·n) = log(m) + log(n)
log(m/n) = log(m) – log(n)
log(m·n) = n · log(m)
loga(x) = logb x / logb a
log(m^n) = n·log(m)
$a^n · a^m = a^{n+m}$
$a^n · b^n = (a·b)^n$
$(a^b)^c=a^{b·c}$
$e^{log(x)} = x$ if $x>0$
$log(e^x) = x$ if $x>0$
And many other.

But what about this one?
$e^{1/log(x)}$

How can I reduce that equation or other similar?

When I write log() I mean neperian logarithm, sometimes written ln()

Answer 1

$$\huge e^{1/\log(x)}=(x^{\log_x(e)})^{\log_x(e)}=x^{[\log_x(e)]^2}$$

$$\frac{\log(e)}{\log(x)}=\log_x(e)\qquad e=x^{\log_x(e)}$$