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- Path: sparky!uunet!mcsun!uknet!pavo.csi.cam.ac.uk!camcus!gjm11
- From: gjm11@cus.cam.ac.uk (G.J. McCaughan)
- Newsgroups: sci.math
- Subject: Re: Retarded log question
- Message-ID: <1992Sep11.145135.20098@infodev.cam.ac.uk>
- Date: 11 Sep 92 14:51:35 GMT
- References: <1992Sep10.225134.21189@hellgate.utah.edu>
- Sender: news@infodev.cam.ac.uk (USENET news)
- Organization: U of Cambridge, England
- Lines: 40
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-
- In article <1992Sep10.225134.21189@hellgate.utah.edu>, tolman%asylum.cs.utah.edu@cs.utah.edu (Kenneth Tolman) writes:
-
- > If you are given a number, such as 2^X, how can you determine what
- > the same number would be with a different base, such as 10?
- >
- > For example, 2^100,000,000 = 10^ ??
-
- 2^100000000 = 10^x
- <=> log_10(2^100000000) = log_10(10^x) = x
-
- and now you need to know:
- log_b(x^y) = y.log_b(x) (1)
- which gives
-
- <=> 100000000.log_10(2) = x
-
- and now you need to know:
- log_b(x) = log_c(x)/log_c(b) for any b,c [proof in a moment]
- which gives
-
- <=> 100000000.log(2)/log(10) = x
-
- where "log" is to any base you want. This is your answer.
-
- Proof of the base-conversion thing above:
- it's the same as log_b(x).log_c(b) = log_c(x);
- well, x=b^log_b(x) so
- log_c(x) = log_c(b^log_b(x))
- = log_b(x).log_c(b) by (1).
-
- I might as well prove (1) too while I'm at it.
-
- log_b(x^y) is defined by b^log_b(x^y) = x^y
- but we know b^(y.log_b(x)) = (b^log_b(x))^y = x^y.
-
- Here I've used x^yz = (x^y)^z, which I am *not* going to prove.
-
- --
- Gareth McCaughan Dept. of Pure Mathematics & Mathematical Statistics,
- gjm11@cus.cam.ac.uk Cambridge University, England. [Research student]
-