# Simple logarithm properties proof

I was reading a school algebra book about logarithm function (on $$\mathbb{R}^+$$). There were several properties without proof. So I decided to prove 2 of them myself.

The first property:

$$log_a(x\cdot y) = log_ax + log_ay$$

## Proof

By definition of logarithm we know, that: $$a^{log_ax} = x$$ (axiom 1)

So $$x\cdot y = a^{log_ax} \cdot a^{log_ay} = a^{log_ax\;+\;log_ay}$$

Therefore

$$log_a(x\cdot y) = log_a(a^{log_ax} \cdot a^{log_ay}) = log_a(a^{log_ax\;+\;log_ay}) = log_ax\;+\;log_ay$$

### question

I am not sure about this equality: $$log_a(a^{log_ax\;+\;log_ay}) = log_ax\;+\;log_ay$$, because we don't have an axiom, that $$\forall x\;log_aa^x = x$$, despite the fact, that we use it here. It must be a conclusion from the definition of logarithm:

$$log_ax$$ is a power to which we must take a to get x. If we want to know, to which power we must take a to get $$a^x$$ --- obviously, it's x.

This trivial axiom was missing in the book. Shouldn't we explicitly denote it?

The second property also aroused a question:

$$log_ax^n = n \cdot log_ax$$

## Proof

$$log_ax^n = log_a\prod_{i=1}^{n} x$$

According to the previously proven property

$$log_a(x\cdot y) = log_ax + log_ay$$

Therefore $$log_a\prod_{i=1}^{n} x = \sum_{i=1}^{n}log_ax = n \cdot log_ax$$

### question

Equality $$x^n = \prod_{i=1}^{n} x$$ is true only for $$n \in \mathbb{Z}^+$$, so formally speaking our proof is valid only for $$n \in \mathbb{Z}^+$$.

While throughout the book the equality $$log_ax^n = n \cdot log_ax$$ is considered to be true $$\forall n \in \mathbb{R}$$.

I don't understand, how to prove it.

• If you have $a^{\log_a x} = x$, then taking $\log_a$ of both sides gives $\log_a(a^{\log_a x}) = \log_a x$. – copper.hat Nov 25 '13 at 22:20
• @copper.hat Is it a contradiction? Please, give a more detailed explanation. – user4035 Nov 25 '13 at 22:23
• No contradiction, if you let $y = \log_a x$, you get $\log_a(a^y) = y$. – copper.hat Nov 25 '13 at 23:05

## 1 Answer

To the first question, as you stated, it is a consequence of the definition of the logarithm, so there's no need for an axiom for it.

For the second, realize that: $$a^{n\cdot\log_a x}=(a^{\log_a x})^n=x^n=a^{\log_a x^n}\Longrightarrow n\cdot\log_a x=\log_a x^n, \forall n\in \mathbb{R},$$ since the property used is valid for real $n$ and the exponential function is injective.