Is it true that $\frac{\ln(a)}2=\ln(\sqrt{a})$ for $a>0$? In particular, is $\frac{\ln(2)}{2}=\ln(\sqrt2)$? I believe the following two identities are correct. For some reason, they look wrong to me. Are they?
$$ \frac{ \ln \left( 2 \right) } { 2 } = \ln( \sqrt{2} ) $$
$$ \frac{ \ln \left( a \right) } { 2 } = \ln( \sqrt{a} ) $$
The second one being valid for all $a > 0$.
 A: They are both correct. To prove them, use the logarithm property $\ln\left(a^b\right)=b\ln(a)$, for $a\gt0$.

This can be rewritten as
$$b\ln(a)=\ln\left(a^b\right),\;\;\;\text{for }a\gt0$$
$\frac{\ln(a)}{2}$ can be written as $\frac12\ln(a)$, and $a^{(1/2)}\equiv\sqrt a$.
You can finish it from here.
A: These are in fact correct. Notice it comes from the fact that
$${e^{\ln(\sqrt{a})}=a^{\frac{1}{2}}=(e^{\ln(a)})^{\frac{1}{2}}=e^{\frac{\ln(a)}{2}}}$$
Now, since ${e^{x}}$ is bijective (and hence injective) on ${\mathbb{R}}$, then
$${e^x=e^y \Leftrightarrow x=y}$$
And so finally
$${\ln(\sqrt{a}) = \frac{\ln(a)}{2}}$$
A: Yes, this is a logarithm property.
For all $a \geq 0$ and for all $c > 0$, this property holds:
$$ \log_c(a^b) = b\log_c(a) $$
This property is known as the “logarithm power rule”.
Your question is about the specific case where $b = \dfrac12$. You can see that it’s true by rewriting $\ln(\sqrt{a})$ and then using the logarithm property, like this:
$$\ln(\sqrt{a}) = \ln\left(a^{1/2}\right) = \frac{\ln a}{2}$$
The proof of the rule is as follows:
$$ a = c^{\log_c(a)}  \tag*{Exponentiation as inverse of $\log$} $$
$$ a^b = \left(c^{\log_c(a)}\right)^b  \tag*{Each side to the power of $b$}$$
$$ a^b = c^{b\log_c(a)}  \tag*{Power rule of exponentiation}$$
$$ \log_c\left(a^b\right) = \log_c\left(c^{b\log_c(a)}\right)  \tag*{$\log_c$ of both sides} $$
$$ \boxed{\log_c\left(a^b\right) = b\log_c(a)}  \tag*{$\log$ as inverse of exponentiation} $$
A: $\ln x = \log_e x$. Multiply $\frac{\ln(a)}{2} = \ln(\sqrt{a})$ by $2$ to get $\ln(a) = 2\ln(\sqrt{a})$, and by the log rule $x\log_a b = \log_a b^x$, we get $\ln(a) = \ln(\sqrt{a}^2)$, which is obviously true.
-FruDe
A: If you restrict $\log$ function to real numbers, then it is defined for all arguments $ >0$. Also, square root of a positive number is positive. Hence, $\log \sqrt{a}$ exists as long as $a>0$
