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So here's the property: (Excuse the spacing, I'm just trying to make it clearer)

$\log_a{x} = \log_{10} {x} / \log_{10} {a}$

What I don't understand is WHY it's correct, can someone please give me some proof/explanation? Thank you so much for your help.

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Consider this rule: $\log_a(b)\cdot c = \log_a(b^c)$. Replace $c$ by $\log_b(c)$, and you get

$$\log_a(b) \cdot \log_b(c) = \log_a(b^{\log_b(c)}) = \log_a(c).$$

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  • $\begingroup$ I'm sorry but I didn't understand that, how does that exhibit why the property works? $\endgroup$ – Willie3838 Sep 10 '16 at 20:35
  • $\begingroup$ Replacing $a, b, c$ by $10, a, x$ respectively gives you $\log_{10}(a)\cdot\log_a(x) = \log_{10}(x)$. $\endgroup$ – Anon Sep 10 '16 at 20:58
  • $\begingroup$ I appreciate your answer, I am still unsure whether or not to accept this due to my inexperienceness with logarithms, I am tackling all these answers to see if they help me solve my problem $\endgroup$ – Willie3838 Sep 10 '16 at 22:09
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Here’s an explanation that’s perhaps a bit more consistent with the way that the logarithms are first introduced in school:

$N=\log_ax$ means $a^N=x$.
$M=\log_{10}a$ means $10^M=a$.
Thus $x=\left(10^M\right)^N=10^{MN}$.
Consequently, $\log_{10}x=MN=\log_{10}a\cdot\log_ax$.

And there you are!

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  • $\begingroup$ I appreciate your answer, I am still unsure whether or not to accept this due to my inexperienceness with logarithms, I am tackling all these answers to see if they help me solve my problem $\endgroup$ – Willie3838 Sep 10 '16 at 22:09
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One may notice that $$ \log_a(x)=\frac{\log x}{\log a}=\frac{\frac{\log x}{\log 10}}{\frac{\log a}{\log 10}}=\frac{\log_{10} x}{\log_{10} a}. $$

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  • $\begingroup$ This may seem dumb, but I'm really getting into a habit of questioning math for my greater knowledge so... what makes that equivalent? It doesn't tell me why it works. $\endgroup$ – Willie3838 Sep 10 '16 at 20:52
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    $\begingroup$ @Willie3838 I think multiplying the initial quotient by $\frac{\frac1{\log 10}}{\frac1{\log 10}}=1$ as I did shows why it works. $\endgroup$ – Olivier Oloa Sep 10 '16 at 20:56
  • $\begingroup$ Sorry, I can't wrap my head around the context, excuse me if I may seem rude but how does multiplying the logrithm by 1 show why it works? $\endgroup$ – Willie3838 Sep 10 '16 at 21:05
  • $\begingroup$ The first equation says $\log_a(x) = \log_e(x)/\log_e(a)$. Could you please explain why that is true? $\endgroup$ – Anon Sep 10 '16 at 21:20
  • $\begingroup$ Since the proof of $\log_a x=\frac{\log_e x}{\log_e a}$ is the same as the proof of $\log_a x=\frac{\log_{10} x}{\log_{10} a}$, I think this explains nothing at all. $\endgroup$ – egreg Sep 10 '16 at 21:24

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