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I found the following rule while reviewing logarithms: "Raising the logarithm of a number by its base equals the number.", i.e.,

$$ b^{\log_b (k)} =k.$$

Why is this true?

(Wording of rule credit: https://www.chilimath.com/lessons/advanced-algebra/expanding-logarithms/)

Update:

I found this definition of a logarithm: "a quantity representing the power to which a fixed number (the base) must be raised to produce a given number."

Gary
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user982216
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    How do you define logarithm? It could be true because that’s the definition – J. W. Tanner Oct 20 '21 at 01:14
  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Oct 20 '21 at 01:14
  • I saw the above rule about logarithms. I found proofs for the other rules, but not this one. How would one derive the above rule? – user982216 Oct 20 '21 at 02:48
  • Also, I really don't know (due to my limited knowledge) how I would give more details. The context I'm getting it from is in the URL provided. I would be happy to expand the question if I knew how. I guess I'm looking for a proof? – user982216 Oct 20 '21 at 02:50
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    If you were to take your definition of a logarithm, which is completely in words, and try to write it using an equation instead, that's essentially the equation you would write. – David K Oct 20 '21 at 03:03
  • @DavidK Thank you. That makes perfect sense. I feel silly for not seeing it earlier. I have a long way to go! – user982216 Oct 20 '21 at 03:07
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    An important thing to note, too, is that in some sense it's arbitrary which of the properties of the logarithm we take as a definition. What's important is the relationships between the properties (i.e. which properties imply which other properties). What you start out with as a definition, may later show up as a theorem in another context where you've chosen a different starting point. This is true about mathematical definitions in general. – user3716267 Oct 20 '21 at 03:10

2 Answers2

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$\log_b(k)$ is defined as the power that $b$ needs to be raised to produce $k$. Clearly, if $b$ is raised to this power, then the result will produce $k$. Hence, $b^{\log_b(k)}=k$.

Alan Abraham
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  • Thank you! This is exactly what I was looking for. I'm going to see if I can work this out on paper. If no other answers come in, I'll be sure to mark this as accepted. – user982216 Oct 20 '21 at 03:05
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I think Alan A. gave a good answer. If the asker likes equation, I have by definition of logarithm,$$ \log_b (k)=dummy.$$ by definition of logarithm,$$ b^{dummy} =k$$ $$ b^{\log_b (k)} =k.$$

Kav
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