>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
studio@halcyon.com