Question 1210626
A better computation set up should be this.

<pre>
input         log
5.600         0.748188
5.601           ?
5.610         0.748963
</pre>
letting d be how far from 0.748188 to the unknown part of the table, then

{{{d/0.000775=0.001/0.01}}}

{{{d=0.0000775}}}



{{{log(5.601x10^3)}}}

{{{3+log(5.601)}}}

{{{3+0.748188+0.000075}}}

{{{3.7482655}}}

or better {{{3.74827}}}