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Does the following equation is true for all $ a,b\in{\mathbb R}$?

$$\log_a b = \log_\sqrt a \sqrt b$$

I have tried to proove this, and I didnt find any contradiction. Is it true?

EDIT

Thanks guys for the answers. I've just found a different proof without using $\ln$, for the record, here it is :

$$\log_\sqrt{a} \sqrt{b} = \frac{ \log_a \sqrt{b}}{\log_a \sqrt{a}} = \frac{\frac{1}{2} \log_a b}{\frac{1}{2} \log_a a} = \frac{\log_a b}{1} = \log_a b$$

Second EDIT

Is it true to generelize this rule, like this?

$$\log_{a^n} {b^n} = \frac{ \log_a {b^n}}{\log_a {a^n}} = \frac{n \log_a b}{n \log_a a} = \frac{n\log_a b}{n} = \log_a b$$

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  • $\begingroup$ It's true for positive $a,b$ and $a\neq1$, which is when that's defined. $\endgroup$ – Macavity Nov 24 '14 at 18:41
  • $\begingroup$ "I find it very usefull in cases when you can't change the bases or don't want to." But you are changing bases. The base is $a$ on the left and $\sqrt{a}$ on the right, and most of the time $a\not=\sqrt a$. Can you give an example where you find it useful? I'm genuinely curious. I've never needed anything other than the "conventional" rules. $\endgroup$ – Randy E Nov 24 '14 at 18:54
  • $\begingroup$ In general, $\log_{\large~a^c}b^c=\log_ab.~$ Here, $c=\frac12$. $\endgroup$ – Lucian Nov 25 '14 at 14:44
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It's true for all positive $b$, $a \neq 1$ as

$$\log_\sqrt{a} \sqrt{b} = \frac{ \ln \sqrt{b}}{\ln \sqrt{a}} = \frac{\frac{1}{2} \ln b}{\frac{1}{2} \ln a} = \frac{\ln b}{\ln a} = \log_a b$$

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