Proving $\log x^2 = 2\log x$ How does $\log x^2 = 2\log x$?
Can you do a proof please. I know that this is true but I don't know why.
 A: Apply the exponential to get:
$$\large e^{\log(x^2)} = x^2 = \left(e^{\log(x)}\right)^2 = e^{\log(x)} e^{\log(x)} = e^{2\log(x)}$$
A: Assuming you have the following understaning of what is a logarithm:
$x = \log_{b} a \iff a = b^x$
So, let we have the following values:
$x =  \log_{b} a$ and $y =  \log_{b} c$
Which means that:
$a = b^x$ and $c = b^y$
If we multiply $a$ and $c$:
$a\cdot c = b^x \cdot b^y = b^{x+y}$
Hence, by our definition of logarithm:
$x+y = \log_{b}(a\cdot c)=\log_{b}a+\log_bc$ (do you remember that $x$ and $y$ are the logarithms of $a$ and $b$?)
So, we have prooved that:
$\log_{b}(a\cdot c)=\log_{b}a+\log_bc$
Now just use that formula in your particular case to get:
$\log_{b}(x^2)=\log_{b}(x\cdot x) = \log(x) + \log(x)=2\log(x)$
A: $$\log (x^2) = \log (x\cdot x)$$
Using this logarithmic rule, $$\log_n (ab)=\log_na+\log_nb$$The above can be rewritten as:
$$\log (x\cdot x)=\log x +\log x=2\log x$$

More generally, $$\log_a x^n = n\log_a x$$
Let $$m = \log_a x$$
Using the logarithmic definition: $$\log_ax=N\implies a^N=x$$
The above becomes:
$$x = a^m$$
Raise both sides to the power of $n$: 
$$x^n = ( a^m )^n = a^{m\cdot n}$$
Convert back to logs, 
$$\log_ax^n = \log_aa^{m\cdot n} \implies \log_a x^n = m \cdot n \space (\log_aa^x=x, \space \text{inverse log law})$$
But $m = \log_a x$ so,
$$\log_a x^n = n \log_a x$$ 
A: By definition, $\log(x^{2})$ is the unique number $y$ such that $e^{y}=x^{2}$
$$x^{2}=e^{\log(x^{2})}$$
But also,
$$x^{2}=(x)^{2}=(e^{\log(x)})^{2}=e^{2\log(x)}$$
A: Let's say that $\log x $ is a number, call it $n$. By definition, that means that $10^n = x$. So then what if we square $x$? It follows that $x^2 = \left( 10^n \right)^2 = 10^{2n}$.  So then, what's $\log 10^{2n}$? That is, to what power do we raise $10$ to get $10^{2n}$?  Obviously, we raise it to the power of $2n$; therefore, $\log 10^{2n} = 2n$. But $10^{2n} = x^2$, and so $\log 10^{2n} = \log x^2$. And $n = \log x$, and so $2n = 2\log x$. Therefore, $\log x^2 = 2\log x.$
