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I was asked to prove the third law of logarithm where

$\log A^y = y · \log A$

My "proof" is that:

$\log A^y = y · \log A$

If $\log A^y = \log 10^{(\log A)·y}$

That means that

$\log 10^{(\log A)·y} = y · \log A$

$\log 10$ cancels out leaving $\log A · y = y · \log A$

I am by no means an expert at maths and proofs thus I wonder if my "proof" is correct. I have seen two other ways of proving this law but I am unsure of whether this one is right too.

Jair Taylor
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Hale
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  • Hi Emily. It is unclear to me what you are doing. First I wonder, what kind of number is $y$? If you only care about positive integers $y$, the proof can be easier. Otherwise, you might check this out: https://math.stackexchange.com/questions/1340616 – 311411 May 19 '21 at 19:59
  • Hi @311411 . Excuse me for being unclear, this field of mathematics is new to me. I think $y$ is considered to be a positive integer. The question does not specify. Also, thank you for the link – Hale May 19 '21 at 20:10
  • That's what I thought, alright. Let's start modestly, and choose $y = 2$. Also, I will write \log, more standard way. So do you agree we want to prove $\log (A^2),=,2\log(A)$? – 311411 May 19 '21 at 20:20
  • by the way, I am achieving the math formatting by enclosing \log (A^2),=,2\log(A) inside dollar signs. – 311411 May 19 '21 at 20:20
  • Thanks, yes I agree. – Hale May 19 '21 at 20:32
  • Also should I use the format that you proposed or just enclose an expression using dollar signs? – Hale May 19 '21 at 20:34
  • okay, so you know that $\log(AB),=,\log(A)+\log(B)$. If $A=B$, then we get $\log(AA),=,\log(A)+\log(A)=2\log(A)$. Can you now show it for $y=3$? – 311411 May 20 '21 at 01:56
  • Oh yes! If y = 3 then A must be multiplied with itself 3 times so we get that: $log(AAA) = log(A) + log(A) + log(A) = 3log(A)$. We could generalise this so $log(A^y) = log(y) + log(y) + log(y) + ... + log(y) = ylog(A)$ A must be multiplied with itself y times which we could write as log(A) + log(A) y times until we achieve our desired result. – Hale May 20 '21 at 08:27
  • Good, you see the pattern. Now, to really do it for all integers 1,2,3,4,5,6,7,.. forever you need the principle of Induction: https://en.wikipedia.org/wiki/Mathematical_induction – 311411 May 20 '21 at 11:55
  • Hi Emily, the basic idea of this proof works. However it's a bit hard to follow and the order of the logic is unclear. You can re-arrange things to make more clear. It's easiest to understand when you write is as a series of re-writes: $\log A^y = \log\left( (10^{\log A})^y)\right) = \ldots = y \log A$ (you fill in the $\ldots$ and justify each step.) – Jair Taylor May 20 '21 at 17:53
  • Hi Jair. What I tried to show above in my post was that by using the expression $x = 10^{(logx)}$ I could substitute $A$ for $10^{(logA)}$. I don't know if it is right but I agree with you! – Hale May 20 '21 at 18:00
  • @Emily Yup $x = 10^{\log x}$ is true for any $x > 0$ so it is fine to substitute $A$ for $10^{\log A}$. Cheers – Jair Taylor May 20 '21 at 18:04
  • @JairTaylor So it's right? Thank you and 311411 very much for your help!! – Hale May 20 '21 at 18:07

1 Answers1

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Remember also that $A^y$ means the $y$th product if $A$ with itself, so

$$\ln A^y = \ln(A \times A \times A \ldots \times A)$$

Where the number of $A$s in the product in the last step is $y$. Then by properties of logs

$$\ln(AB) = \ln A + \ln B$$

One has $y$ copies of $\ln A$ or $y \ln A$.