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List of forum topicsenRe: Floor of the log equation: s = (floor(log10(x))+1)*x - round((10^(floor(log10(x))+1)-10)/9)
http://mathforum.org/kb/thread.jspa?messageID=9126019&tstart=0#9126019
Since I don't have log(x) I calculate x using log(s).]]>May 22, 2013 11:09:34 PMMay 22, 2013 11:09:34 PMcarlosnepomuceno@outlook.com0Re: Floor of the log equation: s = (floor(log10(x))+1)*x - round((10^(floor(log10(x))+1)-10)/9)
http://mathforum.org/kb/thread.jspa?messageID=9125993&tstart=0#9125993
less than a minute (9:31 PM) Thank you Jim! I ended up developing a search algorithm fro that like you've suggested.

Since I don't have]]>May 22, 2013 8:32:36 PMMay 22, 2013 8:32:36 PMcarlosnepomuceno@outlook.com0Re: Floor of the log equation: s = (floor(log10(x))+1)*x - round((10^(floor(log10(x))+1)-10)/9)
http://mathforum.org/kb/thread.jspa?messageID=9125991&tstart=0#9125991
Since I don't have log(x) I calculate x using log(s).]]>May 22, 2013 8:31:27 PMMay 22, 2013 8:31:27 PMcarlosnepomuceno@outlook.com0Re: Floor of the log equation: s = (floor(log10(x))+1)*x -<br> round((10^(floor(log10(x))+1)-10)/9)
http://mathforum.org/kb/thread.jspa?messageID=9125266&tstart=0#9125266
> Hi guys! I need your help to solve this]]>May 20, 2013 12:23:30 PMMay 20, 2013 12:23:30 PMnot@valid.invalid2Floor of the log equation: s = (floor(log10(x))+1)*x - round((10^(floor(log10(x))+1)-10)/9)
http://mathforum.org/kb/thread.jspa?messageID=9124969&tstart=0#9124969
I need to find 'x' for a given 's'. Both of them are natural numbers (>0).

I don't]]>May 19, 2013 2:20:13 PMMay 19, 2013 2:20:13 PMnepomucenocarlos68@gmail.com4