| 28 Oct 2009 03:23:09 pm by Flofloflo |
|
|
Hello,
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... -.- |
| 28 Oct 2009 03:50:17 pm by GloryMXE7 |
|
|
you can simplify your formula to f(x)=.5ln(x-2)
proof
ln(x)/ln(2)=ln(x-2) |
| 28 Oct 2009 03:58:06 pm by Weregoose |
|
|
| GloryMXE7 wrote: |
you can simplify your formula to f(x)=.5ln(x-2)
proof
ln(x)/ln(2)=ln(x-2) |
*facepalm*
What can I say? Thanks for testing that? |
| 28 Oct 2009 07:16:46 pm by FloppusMaximus |
|
|
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. |
