How can $\frac12 \log(x) = \log(\sqrt{x})$? How can $\frac{\log (x)}{2}= \log \left(\sqrt{x}\right) \\$? 

How would I come to this conclusion?
I assume that you know $\log(ab)=\log a+\log b$. Then
$$
\log x=\log(\sqrt{x}\cdot\sqrt{x})=\log\sqrt{x}+\log\sqrt{x}=2\log\sqrt{x}
$$
and so
$$
\log\sqrt{x}=\frac{1}{2}\log x.
$$
Anyway, it's just a special case of the general law
$$
\log(x^k)=k\log x
$$
(for $x&gt;0$ and any $k$).
Firstly you should know that this is a direct result of the Logarithms of Powers Rule:
$$\Large\log(\sqrt{x}) = \log(x^{1/2}) = \frac{1}{2} \log(x)$$

Here's my attempt at a proof nonetheless,

To Prove: $ \frac{1}{2} \log(x) = \log(\sqrt{x})$

Proof: Let us assume that the base of the logarithm is $b$.

We know that $x = b^{\log x} \implies \sqrt{x} =  b^{\log \sqrt{x}}$
$$
\Rightarrow x = b^{2\log \sqrt{x}} \\ 
\Rightarrow b^{\log x} = b^{2\log \sqrt{x}} \\ 
\Rightarrow \log(x) = 2\log(\sqrt{x}) \\
\Rightarrow \frac{1}{2} \log(x) = \log(\sqrt{x})
$$



Aliter: 
$$\large\frac{1}{2} \log(x) = \log(\sqrt{x}) \Leftrightarrow \log(x) = 2\log(\sqrt{x}) \Leftrightarrow x = b^{2\log(\sqrt{x})}$$

$$
\large\text{RHS} = b^{\log(\sqrt{x}) + \log(\sqrt{x})}
=  b^{\log(\sqrt{x})} \times b^{\log(\sqrt{x})}
= \sqrt{x} \times \sqrt{x} = x = \text{LHS}$$