Statement about the natural logarithm Is it true that $\sqrt{2ln(x)}=ln(x)$? I would have thought so since you could take the powers up?
 A: It's only true for $x=1$ and for $x=e^2$.
Infact:
$$\sqrt{2ln(x)}=ln(x)\implies2 \ln x=\ln^2 x \\\implies\ln^2x-2\ln x=0\implies \ln x(\ln x-2)=0 \\\implies\ln x=0 \quad \lor \quad \ln x =2 \implies x=1 \quad \lor \quad x=e^2$$
A: It is not equal for all $x$.
Simply because  $\sqrt{2lnx}=\sqrt{ln(x^2)}\neq\sqrt{(lnx)^2}$
A: You seem to be confused about the order of operations.
First, to show that $\sqrt{2\ln(x)} \not\equiv \ln x$, put $x=\mathrm e$ into your equation you get $\sqrt{2\ln \mathrm e} = \ln \mathrm e$. Given that $\ln \mathrm e = 1$, this means $\sqrt 2 = 1$.
If $x$ and $y$ are positive real numbers then $x\ln y \equiv \ln\left(x^y\right)$. That means $2\ln x = \ln\left( x^2 \right)$.
I understand why you think that $\sqrt {\ln\left( x^2 \right)} = \ln x$, but the order of operations is important.
In the case of $\sqrt {\ln\left( x^2 \right)}$ we start with $x$, we square to get $x^2$, we log to get $\ln(x^2)$, then we square root to get  $\sqrt {\ln\left( x^2 \right)}$. The square root operation does not cancel out the squaring operation because they didn't happen one after another. (Even if they did they might not cancel: $(\sqrt x)^2 \equiv |x|$, and not $x$.)
Just like $+1$, $\times 2$, $-1$ doesn't simplify to $\times 2$.
The first gives $x \mapsto x+1 \mapsto 2(x+1) \mapsto 2(x+1)-1 \equiv 2x+1$.
