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I was watching a video that proves the "Log of a power" rule.

  1. I'm just having trouble understanding the loga(a^x) = x rule - which he uses in the proof

  2. And I don't get why you can log both sides. I know whatever you do to one side of a equation you can do to the other - but I still think there's more to it than just that shallow understanding. As soon as I log something I am saying its a exponent - I am basically going from working with exponentiations to working with exponents - what are the steps behind this?

  3. Also if 10^x = 10^2 - the bases would be the same - so intuitively I would say that the only differences can possibly be in the exponents - so x = 2. But what is the actual way to prove this? Can you show me all the steps that get us to x = 2

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avkaapstad
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    Why you can log both sides? Because if $a=b$ then $f(a)=f(b)$ by the definition of a function, here $f:x\mapsto\log x$, given $a,b\in\operatorname{dom}f$. – Hakim Oct 25 '14 at 20:21
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    For $1$: intuitively, $\log_a(a^x)$ is what we're asking for in the question: to what power do I have to raise $a$ to get $a^x$, which is $x$, and therefore $\log_a(a^x)=x$. – rae306 Oct 25 '14 at 20:24
  • Rainier thanks - I understand am smacking myself on the head now – avkaapstad Oct 25 '14 at 20:44

2 Answers2

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Here use that the function $f(x)=\log_a(x)$ is well defined, that is $a=b \Rightarrow f(a) = f(b)$. Then $$a^{mn} = x^n \Rightarrow \log_a(a^{nm})= \log_a(x^n)$$

And the facts $\log_aa^{mn} = mn\log_aa$, $\log_a a = 1$ and $\log_ax^n = n\log_a x$.

Aaron Maroja
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As Hakim points out in the comments, doing an operation to both sides of an equation is logically valid because logs are well-defined functions.

If you still don't like the given proof, then here's an equivalent proof that does everything in a single chain of equalities but uses substitutions instead: \begin{align*} \log_a(x^n) &= \log_a((a^{\log_a x})^n) &\text{by substituting } x = a^{\log_a x}\\ &= \log_a(a^{n\log_a x}) &\text{since in general $(b^c)^d = b^{cd} = b^{dc}$}\\ &= n\log_a x &\text{since in general} \log_m (m^y) = y \end{align*}

Community
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Adriano
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