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'base 10 log'
1996\05\29@173907 by

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
studiohalcyon.com
> From: Stuart Allman <studioHALCYON.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
> studiohalcyon.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.

As Edwin Park <eparkmilton.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
studiohalcyon.com

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