Question 1210603
A table with entries like this:
{{{matrix(3,2,number,mantisa,6250,7959,6360,7966)}}}
can be used to find the mantissa (sort-of the decimal part) of a logarithm in base 10 of a number with 4 significant digits.
The table does not have a mantissa for 6253, so you have to interpolate.
Not worrying for now about decimal places, we are looking for a mantissa between 7959 and 7966.
For an increase between 6250 and 6260 of 10, there is an increase of 7=7966-7959
for the mantissa.
That is a 0.7 increase in mantissa for each unit increase in the number.
The increase in mantissa corresponding the difference between 6250 and 6253 is
3x0.7=2.1, but we round it to 2.
So the mantissa for 6253 is 7959+2=7961
Not worrying for now about decimal places, we are looking for a mantissa between 7959 and 7966.
The mantissa is the same, for {{{log(6250)}}} , {{{log(6.250)=0.7959}}} , and and {{{log(0.006250)}}} . Those logarithms 
No matter where the decimal point is located, those logarithms are all found by adding {{{0.7959}}} plus some integer.
The integer depends on how the number of digits to the left of the decimal point
6250 and 6250.00 have 4 digits to the left, just like 1000=10^3
{{{log(1000)=log(10^3)=3}}}
{{{log(10000)=log(10^4)=4}}}
6253 is more than 1000 and less than 10000, so {{{log(6253)}}} is more than 3 but less than 4.
To find {{{log(6253)}}} we add {{{3}}} to the mantissa {{{0.7961}}}
{{{log(6253)=3+0.7961}}} --> {{{highlight(log(6253=3.761))}}}
 
What about {{{log(0.6253)}}} ?
0.6253 is between 1 and 0.1, with {{{log(1)=log(10^0)=0}}} and {{{log(0.1)=log(1/10)=log(10^-1)=-1}}} ,
so, {{{log(0.6253)}}} must be more than -1 and less than 0.
{{{log(0.6253)=-1+0.7961}}} --> {{{log(0.6253)=-0.2039}}}