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[Core] log 唔識做

引用:
原帖由 ehbb 於 2020-3-19 09:22 AM 發表
i) log_9 p = log(p)/log9
ii) log_12 q = log(q)/log12
iii) log_16 (p+q) = log(p+q)/log16 = log(p)+log(q)/log16
既然大家相等
即log(p)/log9 = log(q)/log12 = log(p)+log(q)/log16
咁樣做唔做到?
(iii) is incorrect. The relation is "log (pq) = log p + log q". Can't just expand log (p+q) into log p + log q.
Sidenote, "(log p)/(log q) = log (p/q)" is not correct either.

The question looks very weird. The only lead I have is log9 + log16 = 2log12 (Go proof it!). So far no manipulation seems useful.
   

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[隱藏]
引用:
原帖由 ehbb 於 2020-3-19 09:22 AM 發表
i) log_9 p = log(p)/log9
ii) log_12 q = log(q)/log12
iii) log_16 (p+q) = log(p+q)/log16 = log(p)+log(q)/log16
既然大家相等
即log(p)/log9 = log(q)/log12 = log(p)+log(q)/log16
咁樣做唔做到?
(iii) is incorrect. The relation is "log (pq) = log p + log q". Can't just expand log (p+q) into log p + log q.
Sidenote, "(log p)/(log q) = log (p/q)" is not correct either.

The question looks very weird. The only lead I have is log9 + log16 = 2log12 (Go proof it!). So far no manipulation seems useful.

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The solution in #2 is incorrect.
log (p+q) =/= log p + log q; log(pq) = log p + log q is the correct relation.

So this question is very weird, but the major breakthrough for me is, surprisingly, going back to the definition of logarithms.
(embed the image to the solution in just a sec)

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The solution in #2 is incorrect, can't quote for some reason.
Remember only log(pq) = log p + log q. Can't do anything with log(p+q)

So this question is quite difficult. The major "aha" moment for me is to consider the definition of logarithms.
Also surprise! It's the golden ratio (or its buddy, I can't remember).
Sorry for the huge image lmao. Message me for any queries. Good luck!



[ 本帖最後由 風之男 於 2020-3-21 02:47 AM 編輯 ]
本帖最近評分記錄
  • ehbb 論壇積分 +3 敢於指正, 用心回應, 詳盡回覆 2020-3-21 10:07 AM

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