Solving Logarithm with table.

[math]b.~\frac{ 118.6 }{ 34.95 } - \frac{ 2837 }{ 1964 }[/math][math]= \frac{ 118 \frac{ 6 }{ 10 } }{ 34 \frac{ 95 }{ 100 } } - \frac{ 2837 }{ 1964 }[/math][math]= \frac{ 1186 }{ 10 } \div \left( \frac{ 3495 }{ 100 } \right) - \frac{ 2837 }{ 1964 }[/math][math]= \frac{ 1186 }{ \cancel{10} } \times \frac{ \cancel{100} }{ 3495 } - \frac{ 2837 }{ 1964 }[/math][math]= \frac{ 11860 }{ 3495 } - \frac{ 2837 }{ 1964 }[/math][math]= \frac{ 2675545 }{ 1372836 }[/math]
[math]\begin{array}{|l|r|} \hline \text{No.} & \text{Log} \\ \hline \frac{ 2675545 }{ 1372836 } & 6 .4274 \\ & - \\ & ~6.1376 \\ \hline & 0.2898 \\ & \overline{\text{antilog of}~ 0.2898} \\ & 1.9489 \end{array}[/math]
[math]\therefore[/math][math]\frac{ 118.6 }{ 34.95 } - \frac{ 2837 }{ 1964 }[/math][math]= 1.9489[/math]
 
chijioke said:
[math]b.~\frac{ 118.6 }{ 34.95 } - \frac{ 2837 }{ 1964 }[/math][math]= \frac{ 118 \frac{ 6 }{ 10 } }{ 34 \frac{ 95 }{ 100 } } - \frac{ 2837 }{ 1964 }[/math][math]= \frac{ 1186 }{ 10 } \div \left( \frac{ 3495 }{ 100 } \right) - \frac{ 2837 }{ 1964 }[/math][math]= \frac{ 1186 }{ \cancel{10} } \times \frac{ \cancel{100} }{ 3495 } - \frac{ 2837 }{ 1964 }[/math][math]= \frac{ 11860 }{ 3495 } - \frac{ 2837 }{ 1964 }[/math][math]= \frac{ 2675545 }{ 1372836 }[/math]
[math]\begin{array}{|l|r|} \hline \text{No.} & \text{Log} \\ \hline \frac{ 2675545 }{ 1372836 } & 6 .4274 \\ & - \\ & ~6.1376 \\ \hline & 0.2898 \\ & \overline{\text{antilog of}~ 0.2898} \\ & 1.9489 \end{array}[/math]
[math]\therefore[/math][math]\frac{ 118.6 }{ 34.95 } - \frac{ 2837 }{ 1964 }[/math][math]= 1.9489[/math]
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[math]c.~\frac{ 2.874 ^ { 3 } + 6 }{ 2.874 ^ { 3 } - 6 }[/math]
[math]\begin{array}{|l|r|} \hline \text{No.} & \text{Log} \\ \hline 2.874^3 & 3×(0.4585) \\ & = 1.3755 \\ \hline 6 & 0.7782 \\ \hline \text{antilog of}~1.3755 & \\ 23.7411 & \\ \hline \text{antilog of}~ 0.7782 & \\ = 6.0001 & \\ \hline \end{array}[/math]
[math]\frac{ 2.874 ^ { 3 } + 6 }{ 2.874 ^ { 3 } - 6 }[/math][math]=\frac{ 23.7411 + 6.0001 }{ 23.7411 - 6.0001 }[/math][math]=\frac{ 29.7411 }{ 17.741 }[/math]
[math]\begin{array}{|l|r|} \hline \text{No.} & \text{Log} \\ \hline \\[8pt] \Huge{\frac{ 29.7411 }{ 17.741 }} & 1.4734 \\ & - \\ & 1.2490 \\ \hline & 0.2244 \\ \hline \text{antilog} & \text{log} \\ \hline 1.676 & 0.2244 \\ \hline \end{array}[/math]
[math]\therefore ~ \frac{ 29.7411 }{ 17.741 }[/math][math]= 1.676[/math]
 
chijioke said:
[math]c.~\frac{ 2.874 ^ { 3 } + 6 }{ 2.874 ^ { 3 } - 6 }[/math]
[math]\begin{array}{|l|r|} \hline \text{No.} & \text{Log} \\ \hline 2.874^3 & 3×(0.4585) \\ & = 1.3755 \\ \hline 6 & 0.7782 \\ \hline \text{antilog of}~1.3755 & \\ 23.7411 & \\ \hline \text{antilog of}~ 0.7782 & \\ = 6.0001 & \\ \hline \end{array}[/math]
[math]\frac{ 2.874 ^ { 3 } + 6 }{ 2.874 ^ { 3 } - 6 }[/math][math]=\frac{ 23.7411 + 6.0001 }{ 23.7411 - 6.0001 }[/math][math]=\frac{ 29.7411 }{ 17.741 }[/math]
[math]\begin{array}{|l|r|} \hline \text{No.} & \text{Log} \\ \hline \\[8pt] \Huge{\frac{ 29.7411 }{ 17.741 }} & 1.4734 \\ & - \\ & 1.2490 \\ \hline & 0.2244 \\ \hline \text{antilog} & \text{log} \\ \hline 1.676 & 0.2244 \\ \hline \end{array}[/math]
[math]\therefore ~ \frac{ 29.7411 }{ 17.741 }[/math][math]= 1.676[/math]
Click to expand...
Do you notice that by taking the log of 6, and then its antilog, you have only made the number less accurate??

1665332634168.png​

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.
 
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