# 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 $$a > 0$$.

• 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 $e^x = e^y$. Jul 18, 2020 at 21:12
• No they aren't. $c\ln x$ is always equal to $\ln (x^c)$ where $c$ is a constant. Jul 18, 2020 at 21:13
• 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.
– Bob
Jul 18, 2020 at 21:16
• 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. Jul 18, 2020 at 21:26
• @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. Jul 18, 2020 at 21:26

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.

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

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

• Your answer is among the clearest. Jul 18, 2020 at 22:18
• @Sebastiano Thank you! :) Jul 18, 2020 at 22:25

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:

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

$$\ln x = \log_e x$$. 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

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 $$a>0$$