mathematics - A truly amazing way of making every possible positive integer

This is a follow-up to "A truly amazing way of making the number 2016":

For every positive integer $n$, find a mathematical expression that yields the value $n$ while obeying the following rules:

• Each of the digits $1,2,3,4,5,6,7,8,9$ is used exactly once


• Decimal points are allowed


• You may use brackets "(" and ")" to structure your expression, and to make it well-defined



• The only allowed mathematical operations are addition (+), subtraction (-), multiplication (*), division (/); the minus sign may also be used as the sign of a negative number.


• The only allowed mathematical functions are square-roots and logarithms. Logarithms must be written in the form $\log[b](x)$ to denote the base-$b$ logarithm of number $x$

Note that in particular the following is not allowed:

• Juxtaposition of digits (as juxtaposing 1 and 3 to get "31")


• using the digit 0, or using non-decimal digits


• other mathematical operations and functions (cube-roots, exponentiation, factorials, absolute values, trigonometric functions, etc)



• integration, differentiation, limits, matrices, and determinants


• rounding up, rounding down, rounding to the nearest integer

Answer

We can use:

$$\log[2]\left(\frac{1}{\log[3]\left(\underbrace{\sqrt{\sqrt{ \ldots \sqrt{5-\sqrt{4}}}}}_\text{square root repeated $n$ times}\right)}\right) + 6 - 7 - 8 + 9$$

This works because:

$a^{\overbrace{b \cdot b \ldots b}^\text{$n$ times}} = a^{b^n}$, and since $\sqrt{a} = a^{\frac12}$ we have that $\sqrt{\sqrt{ \ldots \sqrt{5 - \sqrt{4}}}} = 3^{\left(\frac12\right)^n} = 3^{\frac1{2^n}}$
Now, $\text{log}[a](a^c) = c$, so $\text{log}[3]\left(3^{\frac1{2^n}}\right) = \frac1{2^n}$
Finally, by the same token: \begin{equation}\text{log}[2]\left(\frac1{\frac1{2^n}}\right) = \text{log}[2]\left(2^n\right) = n\end{equation} The rest is simply adding the unneeded numbers in a zero sum.

If you want a version with the numbers in order for 1 to 9, we can do a small manipulation:

$$-1\cdot\log[2]\left(\log[3]\left(\underbrace{\sqrt{\sqrt{ \ldots \sqrt{-\sqrt{4}+5}}}}_\text{square root repeated $n$ times}\right)\right) + 6 - 7 - 8 + 9$$