Re: Opening
- From: Tina_Hall@xxxxxxxxxxx (Tina Hall)
- Date: Fri, 5 Oct 2007 23:29:00 GMT+1
Tim S <Tim@xxxxxxxxxxxxxxxxxxxxxxxx> wrote:
Tina Hall wrote:
Tim S <Tim@xxxxxxxxxxxxxxxxxxxxxxxx> wrote:
This follows from the fact that log(n^y) = y log(n), which is kind
of part of the definition of a logarithm.
I have no idea what that means.
OK, I don't think I'll talk about it any more, then. I was hoping a
bit of explanation might be more help than just telling you what keys
to press, but I guess sometimes it just doesn't work out.
Well, it would help if I understood what you intended as an explanation.
:)
(If you told me what keys to press, I'd forget, as I wouldn't understand
the idea behind it.)
Why only 'generally' "if n^y=x then x^(1/y)=n"?
Actually, when I said "generally" I meant "always". Now you've made
me think about it, I realise it isn't true if y = 0 (since then 1/y
doesn't exist).
Ok.
What exacty is done to get the y from one side of the equation to
the other, still up there as it is? [*]
You raise both sides to the power 1/y
Makes sense.
or (describing the same thing a different way) you take the yth root
of both sides.
Also makes sense, but the other looks better for my personal
understanding.
Thanks.
The yth root cancels out the yth power.
Simple fractions, no matter whether it's 'up there' or where numbers
usually are. Understood. :)
(Though I don't know what the abbreviation n^1/y could possibly look
like spelled out.)
And, no need to answer if too complicated, how would it look to get
the y on one side all by itself? Without that logarithm stuff?
(Unless you want to explain that.)
You can't do it without the logarithm stuff ... :-(
Ah, well. (I can't even begin to imagine what the logarithm stuff has to
do with exponentials.)
(The whole exponential stuff is still a mystery to me. I know how to
execute n^y, but to me that just looks like an abbreviation of a
long calculation, the y not really being part of the n*n*n*n.... If
you enter y as part of a genuine calculation, I don't know what's
going on anymore. That you get y zeros if n is 10 is coincidence,
not true with all the other possible numbers.
It's not a coincidence. It's because the number system we use is
based on powers of 10.
Ok, it's a coincidence because we happen to use a decimal system. :)
(We could use powers of some different number -- that's what binary
and hexadecimal are about.
You get y zeros if n is 16? I'd have to calculate that to check, but
proof doesn't mean understanding.
But it occurs to me that you get y zeros if n is 2 (binary). 2^3 = 8, 8
binary is 1000, three zeros. (But still, proof doesn't mean
understanding what's going on.)
I like binary. I think a worldwide overhaul, calculating in binary and
displaying in hexadecimal, would make things easier in the long run
(when those systems are familiar, instead of decimal as it is now with
the other two looking funny or even alien to a lot of people).
The ancient Babylonian used powers of 60 for some technical
calculations, e.g. in astronomy, which is where we get 60 seconds in a
minute and 60 minutes in an hour from.)
Interesting.
If it is in some weird way, I'd like to know how.)
[*] What I mean is, if it were:
a / b = x
You would do (*b) to get:
a = x * b
Basically, what the logarithms do is take something about powers and
roots, and turn it into something about multiplication and division.
So you take
n^y = x
and you take the logarithm of both sides, and get
The problem with this is 'the logarithm'. What is that?
Without that, your calculation looks like a formula; bits of it are
unknown things that have meanings to those people that know what it
means, but none to me.
I am not familiar with the mathematical procedure. It may be as simple
as understanding what a '*' means in a calculation, but unlike
multiplication, I don't know what the process is that is depicted that
way...
y * log(n) = log(x).
.... so, with the above, I don't know how you get there, and the
calculation itself says nothing to me.
And then you can solve for y, by dividing both sides by log(n).
That, again, would be easy if I knew what came before. :)
This in turn follows from the fact that logarithms are designed to
turn multiplication into addition, so if we have
No idea how (or that) they do that.
a * b = c
and we take the logarithm of both sides, we get
log(a) + log(b) = log(c).
?
So if we have
a^n
which is just
a * a * a * ... * a
That's the bit I know.
(where there are n copies of a, and the dots are because couldn't be
bothered to write out n copies, where n is an arbitary number :-) )
Of course. :)
then we take the logarithm and get
log(a) + log(a) + log(a) + ... + log(a)
?
(where there are n copies of log(a) )
which is just
n * log(a).
??
Tim
Having a bad feeling this discussion is just going to get deeper and
deeper ...
You're free to jump the boat and swim for your life. :) (Snipping
everything and inserting a <splash!> would tell me that you did, without
any negative feelings or impressions involved.)
I'd like to know what you're talking about with the logarithm stuff [*],
but I'm happy with the "if n^y = x, then x^(1/y) = n", and knowing why.
[*] As I mentioned before, recently to another poster, too, I'd like to
know more about maths, so anything added to what I know is welcome; I
like maths, I wish I knew more. It's not a must, though.
--
Tina
WIP: Space: 5365 words
WISuspension: Seasons & Elements trilogy | Magic Earth series
Posted to Usenet newsgroup rec.arts.sf.composition.
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