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Joined: 07 Nov 2007
Posts: 120

Posted: 28 Oct 2009 03:23:09 pm    Post subject:


As the topic title suggests... I was busy with the harmonic series, so I was wondering: If there's a formula where the input of 2^x outputs 0.5 x. What's the formula?
Now, my logic was: 2^x results in 0.5 x, x results in Ln(0.5x)/ln(2) = ln(0.5x)/ln(2) - 1.
Apparently though, that is not the answer. I am totally unfamiliar with formula's where the input is not simply x... So could somebody help me out with this maybe?
Btw, I already discovered that the answer is supposed to be 0.5 * (ln(x)/ln(2)), but I was wondering if there's a fast and logical way to find that?
Actually, now that I think about it, maybe the way I found it out eventually was as logical as it gets:
First I made a formula for a in 2^a = b, where b is your input, then I did 0.5*a...

Okay, sorry I made this thread, I solved my own problem already... -.-

Last edited by Guest on 28 Oct 2009 03:35:26 pm; edited 1 time in total
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Puzzleman 3000

Active Member

Joined: 02 Nov 2008
Posts: 604

Posted: 28 Oct 2009 03:50:17 pm    Post subject:

you can simplify your formula to f(x)=.5ln(x-2)

Last edited by Guest on 28 Oct 2009 03:52:55 pm; edited 1 time in total
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Authentic INTJ

Super Elite (Last Title)

Joined: 25 Nov 2004
Posts: 3976

Posted: 28 Oct 2009 03:58:06 pm    Post subject:

GloryMXE7 wrote:
you can simplify your formula to f(x)=.5ln(x-2)

What can I say? Thanks for testing that?

Last edited by Guest on 28 Oct 2009 03:58:23 pm; edited 1 time in total
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Advanced Member

Joined: 22 Aug 2008
Posts: 472

Posted: 28 Oct 2009 07:16:46 pm    Post subject:

GloryMXE7: check your math.

Flofloflo: The function you're looking for is f(x) = log4 x (which is, of course, equivalent to ln x / ln 4, and can be written in many other ways besides.) In general, if you have some expression like f(2^x) = 0.5x, what you generally want to do is introduce another variable (say, u), define u = 2^x, and then try to replace the expression 0.5x with some expression in terms of u.
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