Finding the log by using the proportional table

[cbarker1]

Logarithms​

[TABLE="class: outer_border, width: 500, align: left"]
[TR]
[TD]N[/TD]
[TD]0[/TD]
[TD]1[/TD]
[TD]2[/TD]
[/TR]
[TR]
[TD]293[/TD]
[TD]46687[/TD]
[TD]46702[/TD]
[TD]46716[/TD]
[/TR]
[TR]
[TD]294[/TD]
[TD]46850[/TD]
[TD]46835[/TD]
[TD]46864[/TD]
[/TR]
[TR]
[TD]295[/TD]
[TD]46982[/TD]
[TD]46997[/TD]
[TD]47012[/TD]
[/TR]
[TR]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[/TR]
[TR]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[/TR]
[/TABLE]

Prop. Pts​

[TABLE="class: grid, width: 500, align: right"]
[TR]
[TD]1[/TD]
[TD]1.6[/TD]
[/TR]
[TR]
[TD]2[/TD]
[TD]3.2[/TD]
[/TR]
[TR]
[TD]3[/TD]
[TD]4.8[/TD]
[/TR]
[TR]
[TD]4[/TD]
[TD]6.4[/TD]
[/TR]
[TR]
[TD]5[/TD]
[TD]8.0[/TD]
[/TR]
[TR]
[TD]6[/TD]
[TD]9.6[/TD]
[/TR]
[TR]
[TD]7[/TD]
[TD]11.2[/TD]
[/TR]
[TR]
[TD]8[/TD]
[TD]12.8[/TD]
[/TR]
[TR]
[TD]9[/TD]
[TD]14.4[/TD]
[/TR]
[/TABLE]

Find the value of $\log\left({29517}\right)$
Work:
4+$\log\left({2.9517}\right)$

Thanks for your help

 

[I like Serena]

Logarithms​

[TABLE="class: outer_border, width: 500, align: left"]
[TR]
[TD]N[/TD]
[TD]0[/TD]
[TD]1[/TD]
[TD]2[/TD]
[/TR]
[TR]
[TD]293[/TD]
[TD]46687[/TD]
[TD]46702[/TD]
[TD]46716[/TD]
[/TR]
[TR]
[TD]294[/TD]
[TD]46850[/TD]
[TD]46835[/TD]
[TD]46864[/TD]
[/TR]
[TR]
[TD]295[/TD]
[TD]46982[/TD]
[TD]46997[/TD]
[TD]47012[/TD]
[/TR]
[TR]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[/TR]
[TR]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[TD][/TD]
[/TR]
[/TABLE]

Prop. Pts​

[TABLE="class: grid, width: 500, align: right"]
[TR]
[TD]1[/TD]
[TD]1.6[/TD]
[/TR]
[TR]
[TD]2[/TD]
[TD]3.2[/TD]
[/TR]
[TR]
[TD]3[/TD]
[TD]4.8[/TD]
[/TR]
[TR]
[TD]4[/TD]
[TD]6.4[/TD]
[/TR]
[TR]
[TD]5[/TD]
[TD]8.0[/TD]
[/TR]
[TR]
[TD]6[/TD]
[TD]9.6[/TD]
[/TR]
[TR]
[TD]7[/TD]
[TD]11.2[/TD]
[/TR]
[TR]
[TD]8[/TD]
[TD]12.8[/TD]
[/TR]
[TR]
[TD]9[/TD]
[TD]14.4[/TD]
[/TR]
[/TABLE]

Find the value of $\log\left({29517}\right)$
Work:
4+$\log\left({2.9517}\right)$

Thanks for your help

Hi Cbarker1,

To find $\log(2.9517)$, we look up row $295$ in the table.
Then we pick the column with 1, where we find $46997$.
For the last digit we consult the 2nd table, where entry $7$ has $11.2$, which we add for a total of $47008.2$.

In the table the decimal points have been left out, which means we need to read this as $0.470082$.
Add the $4$ you found for a total of $4.470082$.

My calculator says $4.470072$.
Presumably the small discrepancy is an approximation error due to the use of the second table.