Can we get the log of two sides of an inequation without modifying the symbol? This might sound retarded for your standards. But the higher x gets, the higher $\log_{10}(x)$ gets, right? Does the same principle apply for $\log_{2}(x)$ ? And by that matter if I have the following expression: $$n! \leq n^n$$ Does $\leq $ hold? $$\log_{2}(n!) \leq \log_{2}(n^n)$$

Yeah the logarithm is a strict monotone increasing function, so if $$a\leq b \implies \log(a) \leq \log(b)$$ This is true for the natural logarithm and hence for all logarithm with base greater 1.

The derivative of the logarithm is $\frac{1}{x}$ and hence for $x>0$ always positive. So in fact if $a,b$ are strictly greater zero you get $$ a\leq b \iff \log (a) \leq \log(b)$$

To see it is monotone you just can use logarithm laws and the fact that $\log(x)>0$ whenever $x>1$. So you have $$\log(a)=\log\left(b\cdot \frac{a}{b}\right)=\log(b)+\log\left(\frac{a}{b}\right)$$ so if $\frac{a}{b}>1$ you add something positive else you add something negative.