Question

use log tables to find the value of
0.678*9.01
---------
0.0234

Answer 1

Hi, let's take a crack at it. I'm using the appendix from my old college algebra book by Sobel and Lerner.
First I just write my law of logs Log (0.678*9.01) = Log (0.678) + Log(9.01)

I think the trick here is that you have a log of a number between 0 and 1, so your characteristic value is -1. So go to 6.7 in the leftmost column and read across the row till you get to 8. That value is .8312. Add this value to your characteristic value of -1

-1 + .8312 = -.1688 So , Log (.678) = -.1688

Read the lefthand column at 9.0 and read across the row till you get to the column under 1
That value is .9547. Your characteristic value is 0, because 9.01 is between 1 and 10.
So Log(9.01)=.9547

Log(.678) + Log(9.01) = -.1688 + .9547= .7859 which equals Log (.678*9.01)

Start reading down the columns to find a number close to 0.7859.
My tables have the value 0.7860, in the 1 column along the row from 6.1.
So the answer is 6.11

(Since the characteristic value of .7859 is 0, this means that this is the log of a number between 1 and 10.)

Have a good one!