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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 $a > 0$.

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    $\begingroup$ It would be more helpful if you told us what your reason for thinking the identities are wrong is. To check the identities, note that $x = y$ iff $e^x = e^y$. $\endgroup$ – Rob Arthan Jul 18 '20 at 21:12
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    $\begingroup$ No they aren't. $c\ln x$ is always equal to $\ln (x^c)$ where $c$ is a constant. $\endgroup$ – Devansh Kamra Jul 18 '20 at 21:13
  • $\begingroup$ I guess they look wrong to me because we are dividing by $2$ rather than multiplying by $2$. Of course dividing by $2$ is the same as multiplying by $\frac{1}{2}$. I am glad to hear that they are right. $\endgroup$ – Bob Jul 18 '20 at 21:16
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    $\begingroup$ This is simply the identity $\ln a^b = b\ln a$ and the definition $\sqrt a = a^{\frac 12}$. It's a bit nice to know when we were defining things we took care that they made sense and were consistant and we weren't just pulling things out of muck. $\endgroup$ – fleablood Jul 18 '20 at 21:26
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    $\begingroup$ @Bob: you may find it helpful to think of $\sqrt{2}$ as $2^{\frac{1}{2}}$ and $\frac{1}{2}$ as $2^{-1}$ so that the division is confined to the exponents. $\endgroup$ – Rob Arthan Jul 18 '20 at 21:26
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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.

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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

$${\ln(\sqrt{a}) = \frac{\ln(a)}{2}}$$

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    $\begingroup$ Your answer is among the clearest. $\endgroup$ – Sebastiano Jul 18 '20 at 22:18
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    $\begingroup$ @Sebastiano Thank you! :) $\endgroup$ – Riemann'sPointyNose Jul 18 '20 at 22:25
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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:

$$ a = c^{\log_c(a)} \tag*{Exponentiation as inverse of $\log$} $$ $$ a^b = \left(c^{\log_c(a)}\right)^b \tag*{Each side to the power of $b$}$$ $$ a^b = c^{b\log_c(a)} \tag*{Power rule of exponentiation}$$ $$ \log_c\left(a^b\right) = \log_c\left(c^{b\log_c(a)}\right) \tag*{$\log_c$ of both sides} $$ $$ \boxed{\log_c\left(a^b\right) = b\log_c(a)} \tag*{$\log$ as inverse of exponentiation} $$

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$\ln x = \log_e x$. 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

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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 $a>0$

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