proving log properties

[richardt]

Greetings:

I have seen the proofs of log(uv) = log(u) + log(v), etc., in several texts and each reverts to substitution (as I am confident you are familiar). Why can we not prove the case above (for base, b) by simply saying, b^[log(u) + log(v)] = b^log(u) * b^log(v) = uv. Hence log(uv) = log(u) + log(v). ??

By substitution, the same property is typically proved as follows: Let w = log u and z = log v. Then b^w = u and b^z = v and so log(uv) = etc.

It just seems to me that the whole substitution routine is unnecessary.

What do you think?

Thanks

Rich

 

[Ishuda]

Formally you would also have to mention (prove) that y=log(x), x \(\displaystyle \epsilon\) R⁺ [= (0, ∞) ] is a one to one mapping onto R [= (-∞, ∞) ], that is the mapping is bijective. Or a similar proof for x = b^y, y \(\displaystyle \epsilon\) R onto R⁺.

 

[pka]

I have seen the proofs of log(uv) = log(u) + log(v), etc., in several texts and each reverts to substitution (as I am confident you are familiar). Why can we not prove the case above (for base, b) by simply saying, b^[log(u) + log(v)] = b^log(u) * b^log(v) = uv. Hence log(uv) = log(u) + log(v). ?

It seems to me that you are confusing history of and the development of the teaching of logarithms. Logarithms have long history which lacks a well defined development. On the other hand over the last fifty years, there has developed a consistence on how the subject should be presented. It is save to attribute to Leonard Gillman basic ideas of this approach: see his Calculus text from 1973.

The basic idea is not to assume what a logarithm is but rather to find a function with which we can derive all the historical properties. What follows is an outline of such a development.
Define a function \(\displaystyle L:\mathbb{R}^+\to\mathbb{R}\) by \(\displaystyle x\mapsto\int\limits_1^x {\frac{1}{t}dt} ~.\)

Now note that \(\displaystyle L(1)=0\). Prove that \(\displaystyle L(uv)=L(u)+L(v)\). From which all the standard properties of logarithms easily follow.

For inverse simply define \(\displaystyle e\) as the number such that \(\displaystyle L(e)=1\).

In summation: \(\displaystyle \log(x)=L(x)\).