Question

find the value of the following (using log tables) = 3.98×8.76×0.1718÷0.03×0.526×8.43

Answer 1

Answer

885.324334828




3.98 X 8.76 X 0.1718 
__________________ 

0.03 X 0.526 X 8.43 

On your calculator and using 'logs to base 10'. 

log 3.98 = 0.599883072 
log 8.76 = 0.942504106 
log 0.1718 = -0.76497684 

'ADD' these log values together. 
0.777410377 That is the 'log' value of the top (numerators) together. 

Now do the same for the bottom line 

log 0.03 = -1.522878745 
log 0.526 = -0.279014255 
log 8.43 = 0.925827574 

Again 'ADD' these log values together. 
-0.876065426 That is the 'log' value of the bottom (denominators) together. 

To make the division between the numerators and denominators we 'SUBTRACT' to two log values. 

0.777410377 - -0.876065426 = 1.653475803 

Antilog . Because we are using 'logs to base 10' then the antilog is 10^(1.653475803 ) 
10^(1.653475803 ) = 45.2728933 

However, if you are using Castles Four Figure Tables. 
Then to log 3.98 . 
First there is only a 'units' digit (3) in the number so the log starts as '0.' 

Then ignoring the decimal point in 3'98' so that it is read as '398' look down the extreme left hand column for '39' move along this line to the right until you column to column '8'. The answer at this given juncture is '5999'. 
So the log(10) = 0.5999. 

To log a value like 0.1718, we note that there is no units (0.) so the first digit in the log is '-1'. 
Then from the tables as before read for '1718' , The log value is'2350. 
So the log(10) 0.1718 is -1 + 0.2350 = -0.7650 

Then having logged all the values proceed as above (Add/Subtract) as appropriate for Multiplication/Division respectively. 

Then 'Antilog' 1.653475803 using the tables. 
In the Antilog. table , go down the extreme left column to 0.65 then along to column '3' , and this juncture the value is '4498' . However the next digit in 1.65347 is '47' say '5' , so move to the sub-columns to the right to column '5' . The value at the juncture will be also '5' . 
So this subordinate '5' is added on to '4498' making '4503'. 
The '1' to the left of the decimal point in the 'log value' (1.65347...) indicates that there two digits to the left of the decimal point. Hence '4503' becomes '45.03'. 

45.2728933 Approx. = to 45.03 Allowance has to be made for tabular approximations against calculator approximations!!!! 

NB 
In 'Castle's Four Figure Tables' examples of how to use logs &c., are given in the back of the book. 


Hope that helps!!!!