# basic rules logarithm of exponential

I am looking for proof of the basic rules of logarithm. I can prove all basic rules except this $$\log_ab^y=y\log_ab$$ how to get this rule using definition of logarithm.

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What definition? There are several equivalent definitions. Most either have it as the inverse exponential function or $\int_1^x\frac{1}{x}dx$ – CBenni Feb 18 '13 at 15:29
@CBenni: I expect it's just this definition: "$z = \log_a b$ if $a^z = b$". – Tara B Feb 18 '13 at 15:32
@TaraB well that might be the definition, but I know a lot of professors choose another definition for various (or no) reasons. – CBenni Feb 18 '13 at 15:38
Yes, I know, I just meant that I expect that in this case what I said is the relevant definition (I was the one who 'upvoted' your comment). – Tara B Feb 18 '13 at 15:41

If $\log_ab=m\iff a^m=b$

So, $\log_ab^y=n\iff a^n=b^y=(a^m)^y=a^{my}\implies n=my$ if $a\ne0,1$

So, $\log_ab^y=n=my=y\log_ab$

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Just check that $$a^{y \log_{a}(b)} = a^{\log_{a}(b) y} = \left(a^{\log_{a}(b)}\right)^{y} = b^{y}.$$

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Assuming you know, or have proven that $\log_a(xy) = \log_a x + \log_z y$ We use this to prove:

$$\log_ab^y= \log_a \underbrace{(b \cdot b \cdot ...\cdot b)}_{\large y \; times}$$ $$\iff \log_a b^y = \underbrace{\log_a b + \log_a b + \cdots + \log_z b}_{\large y \;\;times}$$ $$\iff \log_ab^y = y\log_z b$$

Else: by definition only

$$a^{y \log_{a}(b)}= a^{\log_{a}(b) y}=\left(a^{\log_{a}(b)}\right)^{y}=b^{y}.$$

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what if $y\notin \mathbb{N}$? – CBenni Feb 18 '13 at 15:36
@CBenni That's why I included the "else"...by definition... – amWhy Feb 18 '13 at 15:53