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So, I asked a question earlier that basically asked "prove that $log(xy)=log(x)+log(y)$ for decimal/fraction logarithms", but I had not explained what I meant correctly. I meant: how can one prove that the law that $a^b*a^c= a^{b+c}$ holds for all real number exponents? For integers, as I said in my last post, it's rather easy. Just expand them and you'll notice that it holds true. Example:

$10^3*10^3 = (10*10*10)*(10*10*10) = 10^6$

However, this cannot be done with fractional or decimal exponents, at least in the form they're in.

Example: $10^{0.4} * 10^{0.9} = 10^{0.4+0.9}$. This holds true, but why does it hold true?

Thank you.

Blauheim44
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    Assuming the usual definition of $m^{th}$ root for $a \gt 0$, you can prove the equality for rational $b,c$ by finding an integer $m$ so that $mb, mc$ are both integers, then $\left(a^ba^c\right)^m=\left(a^b\right)^m\left(a^c\right)^m=a^{mb}a^{mc} \color{red}{=}a^{mb+mc}=\left(a^{b+c}\right)^m$ where the red $\color{red}{=}$ follows from the integer case. Then $\left(a^ba^c\right)^m=\left(a^{b+c}\right)^m \implies a^ba^c=a^{b+c}$ by taking the $m^{th}$ root. For irrational exponents, you must first define $a^b, a^c$, usually by continuity, then the equality follows from the rational case. – dxiv Nov 18 '21 at 07:23
  • @dxiv This should be an answer. Or a link to a duplicate. – Milten Nov 18 '21 at 07:34
  • @dxiv Thank you so much! This is perfect. So by adding the exponent m to turn them into integers, then using the same philosophy with integers, and then taking away that exponent, you can see that it also works for fractions and rational numbers at large! You are a real life saver; I've been ruminating on this for the past day now. But just one question: how come $a^{mb+mc} = (a^{b+c})^m$? – Blauheim44 Nov 18 '21 at 07:44
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    @Blauheim44 I left that as a comment rather than an answer because you gave little indication of the context of the question, or your background for that matter. Feel free to fill-in the missing blanks and write a self-answer. This would also be the best way to make sure you got everything right and covered. – dxiv Nov 18 '21 at 07:51
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    @Blauheim44 $a^{mb + mc} = a^{m(b + c)}$, and using the exponential law $(p^q)^n = p^{qn}$, we can put $a^{m(b + c)} = (a^{b + c})^m$ (going reverse). – Dstarred Nov 18 '21 at 07:56
  • @UnexpectedConfusion Ah, alright. Thank you. – Blauheim44 Nov 18 '21 at 08:02
  • @Milten My comment was an outline rather than a complete proof, and I left it more as an invitation for the OP (or anyone) to develop it into a full answer. There have been similar questions asked before e.g. 1 2 3, though not exact duplicates. The possible answers depend on the definitions and premises, and also the amount of calculus the OP is willing (or allowed) to use, which the question does not make clear. – dxiv Nov 18 '21 at 18:52

1 Answers1

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If $a > 0$, then taking $\log_a$ of both sides gives:

$\log_a(a^b * a^c) = \log_a(a^{b+c})$

$\log_a(a^b) + \log_a(a^c) = {b+c}$

$b + c = {b+c}$

If you are still not convinced, then try evaluating for real numbers.

Evaluate $10^{0.4}$ and $10^{0.9}$ on a software and multiply them together. Then check if the output is equal to $10^{0.4+0.9}$.

Dstarred
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  • Take a look at dxiv’s comment. “It’s true because it’s true” is rarely (I might dare say never) how maths works. – Milten Nov 18 '21 at 07:32
  • @Milten yeah and i have proved it by taking log of both sides... I will remove the heading though, I realise its a bit misleading – Dstarred Nov 18 '21 at 07:36
  • Okay I see your point. But the OP starts out by saying that the question came from wanting to prove the $\log$ identity, so that’s gonna be circular… – Milten Nov 18 '21 at 07:40