# Logarithm property needs an explanation

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.

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

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

• I'm sorry but I didn't understand that, how does that exhibit why the property works? – Willie3838 Sep 10 '16 at 20:35
• Replacing $a, b, c$ by $10, a, x$ respectively gives you $\log_{10}(a)\cdot\log_a(x) = \log_{10}(x)$. – Anon Sep 10 '16 at 20:58
• 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 – Willie3838 Sep 10 '16 at 22:09

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!

• 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 – Willie3838 Sep 10 '16 at 22:09

One may notice that $$\log_a(x)=\frac{\log x}{\log a}=\frac{\frac{\log x}{\log 10}}{\frac{\log a}{\log 10}}=\frac{\log_{10} x}{\log_{10} a}.$$

• 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. – Willie3838 Sep 10 '16 at 20:52
• @Willie3838 I think multiplying the initial quotient by $\frac{\frac1{\log 10}}{\frac1{\log 10}}=1$ as I did shows why it works. – Olivier Oloa Sep 10 '16 at 20:56
• 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? – Willie3838 Sep 10 '16 at 21:05
• The first equation says $\log_a(x) = \log_e(x)/\log_e(a)$. Could you please explain why that is true? – Anon Sep 10 '16 at 21:20
• 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. – egreg Sep 10 '16 at 21:24
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5377193689346313, "perplexity": 385.37952454449396}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570987833766.94/warc/CC-MAIN-20191023122219-20191023145719-00349.warc.gz"}