c. 2.8743−62.8743+6
No.2.87436antilog of 1.375523.7411antilog of 0.7782=6.0001Log3×(0.4585)=1.37550.7782
2.8743−62.8743+6=23.7411−6.000123.7411+6.0001=17.74129.7411
No.17.74129.7411antilog1.676Log1.4734−1.24900.2244log0.2244
∴ 17.74129.7411=1.676
Do you notice that by taking the
log of 6, and then its
antilog, you have only made the number
less accurate??

The point of using logarithms isn't to be
less accurate; it's to make approximate answers
easier when there is no calculator available. (By the way, if you are using a computer, there
is a calculator available! And even if I were doing this on paper, I would probably not make such an elaborate table, with all the rewriting of numbers it involves.)
The tiny error, of course, is beyond the four digits of accuracy you have in your work, so it really doesn't make a difference; but surely it's a waste of time to use the table twice to do
nothing. Labor-saving devices have to be used with intelligence, not blindly, or they become labor-increasing devices.
But your work is correct.