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Topic: Floor of the log equation: s = (floor(log10(x))+1)*x - round((10^(floor(log10(x))+1)-10)/9)
Replies: 4   Last Post: May 22, 2013 11:09 PM

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James Waldby

Posts: 355
Registered: 1/27/11
Re: Floor of the log equation: s = (floor(log10(x))+1)*x -
round((10^(floor(log10(x))+1)-10)/9)

Posted: May 20, 2013 12:23 PM
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On Sun, 19 May 2013 11:20:13 -0700, nepomucenocarlos68 wrote:

> Hi guys! I need your help to solve this equation:
>
> I need to find 'x' for a given 's'. Both of them are natural numbers (>0).
>
> I don't know how to handle the floor term.
>
> [octave/matlab format]
> s = (floor(log10(x))+1)*x - round((10^(floor(log10(x))+1)-10)/9);
>
> [or TeX}
> s = \left( \lfloor\log(x)\rfloor+1 \right)x - \frac{10^{\lfloor\log(x)\rfloor+1}-10}{9}
>
> [or image]
> http://postimg.org/image/r5fd2enll/
>
>
> Exact or approximate values are good.
>
> Is there a solution? How do I solve it?


Your image and tex forms show log() in some cases where the octave/matlab
form shows log10(); in following I ignore that inconsistency (which
doesn't affect the method outlined) and write l() to stand for log to
some base.

A closed-form solution giving x in terms of s may be difficult to find,
but if you can get some reasonably-close estimates x1, x2 of x such that
s is bounded by s(x1) and s(x2), then you can do a binary or other search
to find an x0 with s = s(x0), or perhaps the range of x values that yield
s. Anyhow, ignoring the floor and round functions and supposing logs are
base 10, the expression may be like

s = x*l(x) + x - ( 10*10^l(x) - 10 )/9 [5]

so s ~ x*l(x) + x - 1.1*x - 1.1 [6]

so s ~ x*l(x). [7]

According to <http://en.wikipedia.org/wiki/Lambert_W_function#Example_4>,
approximation [7] is solved by

x = e^W(s*ln(10)) [8]

You could compute values of x(s) using [8] for a couple of values of
s, and then search. Or via some of the other Lambert-function examples,
solve [6] to get more accurate starting values of x before the search.

--
jiw



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