Reading a log table

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Log_Table.JPG

Field
Applied Mathematics
Went Obsolete
1960s
Made Obsolete By
The advent of inexpensive calculators
Knowledge Assumed
Basic high school math skills
When useful
When doing arithmetic without the aid of a calulator

The principle behind a log table is the following property of logarithms:

log (xy) = log x + log y

When computing by hand, it's much faster to add two numbers than to multiply them.

Reading the log table

A typical log table has ten columns and, depending on the size of the log table, 90 or 900 rows. The rows provide the initial digits of the value to look up and the column provides the final digit. For example, to look up the logarithm for the number 2.56 in a 90-row log table, go to the row labelled "2.5" and read the entry in the column labelled "6".

Decimal points are frequently omitted in tables of common logarithms to save space. The implied decimal point in the rows is immediately after the first digit. The implied decimal point in the table entries is before the first digit. For example, the entry in the row labelled "25", under column 7 is "40993", which indicates that the logarithm of 2.57 is .40993.

How to use a log table to multiply two positive numbers

First verify that you are using a common logarithm table. In a common logarithm table, the values for 1.0 through 9.99 will range from 0 to 1. (In a natural logarithm table, the values for 1.0 through 9.99 will range from 0 to approximately 2.3. The illustration above is of a table of natural logarithms.)

  1. Express the two numbers in scientific notation, i.e., of the form A × 10B where A is greater than or equal to 1 and less than 10.

  2. For each number, look up the "A" value in the table of logarithms.

  3. Add the two values you looked up. If the result is greater than one, then subtract one.

  4. Look in the body of the table for the number closest to the result and use the row and column headers to determine what number it corresponds to. This is the value of "A" for the product.

  5. Add the two "B" values, plus 1 more if you subtracted 1 in step 3. This is the value of "B" for the product.

Example: Compute 25.2 × 120.

N 0 1 2 3 4 5 6 7 8 9
1.2 07918 08279 08636 08991 09342 09691 10037 10380 10721 11059
...
2.5 39794 39967 40140 40312 40483 40654 40824 40993 41162 41330
...
3.0 47712 47857 48001 48144 48287 48430 48572 48714 48855 48996
  1. Rewrite 25.2 = 2.52 × 101 and 120 = 1.20 × 102.

  2. According to the table, log(2.52) = .40140 and log(1.20) = .07918.

  3. Compute .40140 + .07918 = .48058. The result is less than 1 so no adjustment is necessary.

  4. The value .48058 lies between .48001 and .48144, but it's closer to .48001, which corresponds to the value 3.02.

  5. Adding the exponents 1 + 2 = 3. We did not have an extra 1 from step 3, so the total exponent is 3.

The product is therefore approximately 3.02 × 103 = 3020. (The exact answer is 3024.)

Practical example: The volume of a sphere is given by the formula V = 4πr3/3. What is the approximate volume of a sphere whose radius is 30.8 meters?

Since the measurement is accurate to only three significant digits, we can perform our intermediate computations to only three significant digits. Since π is approximately 3, the value we expect is approximately 4 × 3 × 303 / 3 = 4 × 30^3 = 4 × 27,000 ≈ 100,000.

We have arrived at the correct answer (to three significant digits) using only addition and multiplication by 3. Computing this result by hand without the use of logarithms would have been significantly more difficult.

Shortcuts

Example: Compute 25.2 × 120.

  1. Rewrite 25.2 as 2.52 and 120 as 1.20.

  2. According to the table, log(2.52) = .40140 and log(1.20) = .07918.

  3. Compute .40140 + .07918 = .48058. The result is less than 1 so no adjustment is necessary.

  4. The value .48058 lies between .48001 and .48144, but it's closer to .48001, which corresponds to the value 3.02.

  5. Make a rough estimate of the expected answer. 25.2 × 120 will be higher than 25 × 100 = 2500 but smaller than 30 × 200 = 6000. Starting with the value 3.02, moving the decimal three places to the right results in a value in this range: 3020. (The exact answer is 3024.)

Interpolation

Interpolation is an advanced technique that generates an additional significant digit from the logarithm table.

To read an interpolated value, find the two values closest to the value you wish to read and estimate the final digit based on how close the target value is to the two choices.

Example: Compute the logarithm of 2.523.

According to the table above, log(2.52) = .40140 and log(2.53) = .40312. The value 2.523 is three tenths of the way between 2.52 and 2.53, so the logarithm of 2.523 is approximately three tenths of the way between .40140 and .40312 = .40140 + .3 × (.40312 - .40140) = .40140 + .3 × .00172 = .40140 + .000516 = .401916.

In fact, log(2.523) = 0.40191725..., so the approximate value is good to within 4 parts per million.

Interpolation can also be used in the reverse direction.

Example: Compute the anti-logarithm of .48058.

According to the table above, .48058 lies between log(3.02)=.48001 and log(3.03)=.48144. What fraction of the way between the two values is it? (.48058 - .48001) / (.48144 - .48001) = .398, or .4 to one significant digit. Therefore, the anti-logarithm of .48058 is approximately 3.024. This agrees with the exact answer we computed when we calculated 25.2 × 120.

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2008-02-22 02:58:20   Another skill that has been lost is interpolation — estimating how far between two entries in a table to find the value you seek. —71.168.226.242

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