Proving log ⁡ x 2 = 2 log ⁡ xHow does log ⁡ x 2...

Apply the exponential to get:
$e^{\log (x^2)}=x^2={\left(e^{\log (x)}\right)}^2=e^{\log (x)}{e}^{\log (x)}=e^{2\log (x)}$

Assuming you have the following understaning of what is a logarithm:
$x={\log }_{b}a⟺a=b^x$
So, let we have the following values:
$x={\log }_{b}a$ and $y={\log }_{b}c$
Which means that:
$a=b^x$ and $y={\log }_{b}c$
If we multiply $a$ and $c$:
$a\cdot c=b^x\cdot b^y=b^{x+y}$
Hence, by our definition of logarithm:
$x+y={\log }_b(a\cdot c)={\log }_{b}a+{\log }_{b}c$ (do you remember that $x$ and $y$ are the logarithms of $a$ and $b$?)
So, we have prooved that:
${\log }_b(a\cdot c)={\log }_{b}a+{\log }_{b}c$
Now just use that formula in your particular case to get:
${\log }_b(x^2)={\log }_b(x\cdot x)=\log (x)+\log (x)=2\log (x)$