Post by thread:base 10 log

Thanks for everyone's help. I found a pretty easy way to find the log10. I'll just going to scale the numbers to between 0 and 1/2 (the spec has changed since my first post). Then use 3.33*(sqrt(sqrt(x))-3 That should be just enough accuracy, considering that I am only dealing with four decimal places. This also works for numbers between 0 and 1 in case anyone is interested; I didn't bother to test anything beyond that. I hope this helps other people. Stuart Allman Studio Sound Design spam_OUTstudioTakeThisOuThalcyon.com

> From: Stuart Allman <.....studioKILLspam@spam@HALCYON.COM> > > Thanks for everyone's help. I found a pretty easy way to find the log10. > I'll just going to scale the numbers to between 0 and 1/2 (the spec has > changed since my first post). Then use 3.33*(sqrt(sqrt(x))-3 That should > be just enough accuracy, considering that I am only dealing with four > decimal places. > > This also works for numbers between 0 and 1 in case anyone is interested; I > didn't bother to test anything beyond that. I hope this helps other people. > > > Stuart Allman > Studio Sound Design > studioKILLspamhalcyon.com > Now I'm confused: isn't log10(0) roughly -infinity? According to your formula (interpreted 2 ways due to missing close parenthesis) 3.33*(sqrt(sqrt(0))) - 3 = 0.33 or 3.33*(sqrt(sqrt(0)) - 3) = -9.99 Taking the second interpretation, since it's closer to the expected value, try substituting 0.5 in the 4th root argument. The real value of log10(0.5) is -0.3010 to 4 places. The above formulae give -0.1998 and -7.1898 respectively. Even if 0.5 actually represents 65536, I can't work it out. Please help! - otherwise I'm going back to my table lookup. As Edwin Park <.....eparkKILLspam.....milton.viasat.com> just mentioned, I would also like to avoid the sqrts even if your DSP eats them for breakfast. Regards, SJH Canberra, Australia PS: I'd work it out for sure if I had more time, but I'm at work...

Stuart Allman wrote: > > Thanks for everyone's help. I found a pretty easy way to find the log10. > I'll just going to scale the numbers to between 0 and 1/2 (the spec has > changed since my first post). Then use 3.33*(sqrt(sqrt(x))-3 That should > be just enough accuracy, considering that I am only dealing with four > decimal places. > > This also works for numbers between 0 and 1 in case anyone is interested; I > didn't bother to test anything beyond that. I hope this helps other people. > So, are you saying log10(x) ~ 3.33*sqrt(sqrt(x)) - 3, for 0 < x <= .5 ? I created a little table over the range of values. This approximation does not provide four decimal places of accuracy. x log10(x) approximation error --------------------------------------------- .1 -1 -1.127 -12.7% .2 -.6990 -0.7731 -10.6% .3 -.5229 -0.5355 -2.4% .4 -.3979 -0.3517 11.6% .5 -.3010 -0.1998 33.6% Did I miss something? Scott

>So, are you saying >log10(x) ~ 3.33*sqrt(sqrt(x)) - 3, for 0 < x <= .5 ? > >I created a little table over the range of values. This approximation does not >provide four decimal places of accuracy. > > x log10(x) approximation error >--------------------------------------------- > .1 -1 -1.127 -12.7% > .2 -.6990 -0.7731 -10.6% > .3 -.5229 -0.5355 -2.4% > .4 -.3979 -0.3517 11.6% > .5 -.3010 -0.1998 33.6% Yes, this is just enough accuracy. The idea is to come pretty close to the log. If you know of a quicker way to get the log more accurately please say so. I think this could help a lot of engineers. Plot the estimated equation on your calculator. The idea is that hearing is pretty much log10 based. A close enough estimation is good enough. As you will see, the estimation has pretty much the same shape as log10. Stuart Allman Studio Sound Design EraseMEstudiospam_OUTTakeThisOuThalcyon.com