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'base 10 log'
1996\05\29@173907
by
Stuart Allman
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
1996\05\29@205135
by
Steve Hardy

> 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...
1996\05\29@205755
by
Scott Dattalo
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
1996\05\30@100117
by
Stuart Allman
>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
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