Intuitive understanding of logarithms

I know logarithms are supposed to be the inverse of exponential functions, and while this makes sense, it seems to me that a more intuitive and significant property is $$\log (ab) = \log(a)+\log(b)$$

So in this way, the logarithm is a fundamental relationship between addition and multiplication. Should logarithms in schools be taught this way? Should I think of them primarily in this way?

EDIT: This probably related to the fact that the only continuous functions $f$ that satisfy $f(x+y) = f(x)f(y)$ are exponential functions (there are apparently some super-weird non-continuous non-exponential functions that satisfy that multiplicativity but I have no idea what they are).

• Logarithms turn multiplication into addition (and vice-versa) just as exponentation does, do you think of exponentation primarily in this way? $a^b\times a^c=a^{b+c}$ Oct 9 '14 at 17:52
• Well $e^{a+b} = e^a e^b,$ right? I think it's best to think of logs as inverses of exponents, and if you want to think of it in terms of addition and multiplication, you might think in terms of abstract algebra - that the function $f(x)=log(x)$ defines an isomorphism between $R^*$ under multiplication, and $R$ under addition. More on this in an answer if you don't understand. Oct 9 '14 at 17:53
• @AlfredYerger More please :) what is $R^*$? I think I kind of know what an isomorphism is. Oct 9 '14 at 17:54
• @AlfredYerger the logarithmic function can be thought of as an isomorphism but from the positive real numbers under multiplication to the real numbers under addition Oct 9 '14 at 18:08
• @AlfredYerger \mathbb{R} renders as $\mathbb{R}$, you can use it with any letter Oct 9 '14 at 19:45

The property you posted is derived from the exponent property $$(a^n)(a^m)=a^{n+m}$$ therefore we can see that logarithms and exponents are essentially the same, however in a different notation. I think that makes more sense than them being a relationship between multiplication and addition.
If: $$10^n=A$$ Then: $$log(A)=n$$ And if: $$10^m=B$$ Then: $$logB=m$$
If we take $$10^c=(10^n)(10^m)=10^(m+n)$$ It follows to say that: $$c=n+m$$ therefore: $$log(AB)=logA+logB$$
• I don't think it's directly derived from $a^n \cdot a^m = a^{n+m}$ - as I said in a comment on the original question, the relationship between addition and multiplication that the $\log$ gives is quite different than what $\exp$ gives, since $\exp(ab) \neq \exp(a) + \exp(b)$ Oct 9 '14 at 18:06
• @TerrenceTown Since $\log$ is the inverse of $\exp$, the conversion between multiplication and addition goes in the other way. $\exp$ converts addition into multiplication, therefore its inverse converts multiplication into addition. Oct 9 '14 at 18:09