Is it true that $\frac{\ln(a)}2=\ln(\sqrt{a})$ for $a>0$? In particular, is $\frac{\ln(2)}{2}=\ln(\sqrt2)$?

Question

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 $a > 0$.

Answer 1

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.

Answer 2

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

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

Answer 3

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:

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

Answer 4

$\ln x = \log_e x$. 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

Answer 5

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 $a>0$