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:

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


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

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


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


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:

$log_ax^n = n \cdot log_ax$


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

According to the previously proven property

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

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


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.

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

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.


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