Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
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

3 Answers 3

up vote 2 down vote accepted

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

share|improve this answer

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

share|improve this answer

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

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

share|improve this answer
3  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.