# 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 $e^x = e^y$. – Rob Arthan Jul 18 '20 at 21:12
• No they aren't. $c\ln x$ is always equal to $\ln (x^c)$ where $c$ is a constant. – Devansh Kamra Jul 18 '20 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 '20 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. – fleablood Jul 18 '20 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. – Rob Arthan Jul 18 '20 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. – Sebastiano Jul 18 '20 at 22:18
• @Sebastiano Thank you! :) – Riemann'sPointyNose Jul 18 '20 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 $$a>0$$