Recent Changes for "Reading a log table" - Obsolete Skills Wiki

Reading a log table

2008-02-27 16:14:58

More complicated example whre logarithms actually saves a lot of work

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- 1. Express the two numbers in scientific notation, i.e., of the form A × 10^B^ 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.	+ 1. Express the two numbers in scientific notation, i.e., of the form A × 10^B^ 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.
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- 1. Rewrite 25.2 = 2.52 × 10^1^ and 120 = 1.20 × 10^2^. - 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.	+ 1. Rewrite 25.2 = 2.52 × 10^1^ and 120 = 1.20 × 10^2^. + 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.
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	+ Practical example: The volume of a sphere is given by the formula V = 4πr^3^/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 × 30^3^ / 3 = 4 × 30^3 = 4 × 27,000 ≈ 100,000. + + * Using the rule of logarithms, log(V) = log(4) + log(π) + 3 log(r) - log(3). + * Use the rule of logarithms again to convert log(4) - log(3) = log(4/3) = log(1.33). + * From the table, log(3.08) = .489, log(1.33) = .124. + * Most log tables also provide the logarithms of well-known constants, so you can look up log(π) = .497. + * Calculate 3 log(r) = 3 × .489 = 1.467. This calculation is simple enough that it can be done in your head. + * Add the pieces together. log(1.33) + log(π) + 3 log(r) = .124 + .497 + 1.467 = 2.088. + * Look up .088 in the logarithm table: It is closest to log(1.22) = .08636. + * Therefore, the volume of the sphere is approximately 122,000 cubic meters. (The correct answer is 122,388.541...) + + 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. +
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- * Step 1 can be shortened to "move the decimal point to immediately after the first nonzero digit." - * Replace step 5 with "Move the decimal place to the location that makes sense based on the values being multiplied."	+ * Step 1 can be shortened to "move the decimal point to immediately after the first nonzero digit." + * Replace step 5 with "Move the decimal place to the location that makes sense based on the values being multiplied."
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- 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.)	+ 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.)

Reading a log table

JabberWokky — 2008-02-27 11:37:19

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Reading a log table

2008-02-27 05:13:37

Here's your interpolation

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	+ ===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. + + +

Reading a log table

2008-02-27 05:01:55

Shortcuts

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- 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 is shorthand for ".40993".	+ 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.
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- ===How to use a log table to multiply two numbers===	+ ===How to use a log table to multiply two positive numbers===
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- 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 to the right is of a table of natural logarithms.)	+ 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.)
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- 2. For each number, look it its "A" value in the log table to obtain its logarithm. - 3. Add the two resulting looked-up values. - 4. If the result is greater than 1, then ignore the 1 for now. - 5. Look in the body of the table for the number closest to the sum you computed and use the row and column headers to determine what number it corresponds to. This is the value of "A" for the result. - 6. Add the "B" value together (plus 1 more if you had an extra 1 in step 4). This is the value of "B" for the result.	+ 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.
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- 2. According to the log table, log(2.52) = .40140 and log(1.20) = .07918. - 3. Compute .40140 + .07918 = .48058. - 4. The value is still less than 1, so no extra work is done in this step. - 5. The value .48058 lies between .48001 and .48144, but it's closer to .48001, which corresponds to the value 3.02. - 6. Adding the exponents 1 + 2 = 3. We did not have an extra 1 from step 4, so the total exponent is 3.	+ 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.
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	+ ===Shortcuts=== + + * Step 1 can be shortened to "move the decimal point to immediately after the first nonzero digit." + * Replace step 5 with "Move the decimal place to the location that makes sense based on the values being multiplied." + + 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.)

Reading a log table

2008-02-27 04:46:17

Show how to use a log table to multiply two numbers

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- The idea behind a log table is the following property of ''logarithms'':	+ The principle behind a log table is the following property of ''logarithms'':
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- It's much faster to, on paper, add two numbers than to multiply them.	+ 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 is shorthand for ".40993". + + ===How to use a log table to multiply two 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 to the right is of a table of natural logarithms.) + + 1. Express the two numbers in scientific notation, i.e., of the form A × 10^B^ where A is greater than or equal to 1 and less than 10. + 2. For each number, look it its "A" value in the log table to obtain its logarithm. + 3. Add the two resulting looked-up values. + 4. If the result is greater than 1, then ignore the 1 for now. + 5. Look in the body of the table for the number closest to the sum you computed and use the row and column headers to determine what number it corresponds to. This is the value of "A" for the result. + 6. Add the "B" value together (plus 1 more if you had an extra 1 in step 4). This is the value of "B" for the result. + + Example: Compute 25.2 × 120. + + \|\|<tableborder="1" rowbgcolor='#C0C0C0'>'''N'''\|\| 0 \|\| 1 \|\| 2 \|\| 3 \|\| 4 \|\| 5 \|\| 6 \|\| 7 \|\| 8 \|\| 9 \|\| + \|\| 1.2 \|\| 07918 \|\| 08279 \|\| 08636 \|\| 08991 \|\| 09342 \|\| 09691 \|\| 10037 \|\| 10380 \|\| 10721 \|\| 11059 \|\| + \|\|<-11> ... \|\| + \|\| 2.5 \|\| 39794 \|\| 39967 \|\| 40140 \|\| 40312 \|\| 40483 \|\| 40654 \|\| 40824 \|\| 40993 \|\| 41162 \|\| 41330 \|\| + \|\|<-11> ... \|\| + \|\| 3.0 \|\| 47712 \|\| 47857 \|\| 48001 \|\| 48144 \|\| 48287 \|\| 48430 \|\| 48572 \|\| 48714 \|\| 48855 \|\| 48996 \|\| + + 1. Rewrite 25.2 = 2.52 × 10^1^ and 120 = 1.20 × 10^2^. + 2. According to the log table, log(2.52) = .40140 and log(1.20) = .07918. + 3. Compute .40140 + .07918 = .48058. + 4. The value is still less than 1, so no extra work is done in this step. + 5. The value .48058 lies between .48001 and .48144, but it's closer to .48001, which corresponds to the value 3.02. + 6. Adding the exponents 1 + 2 = 3. We did not have an extra 1 from step 4, so the total exponent is 3. + + The product is therefore approximately 3.02 × 10^3^ = 3020. (The exact answer is 3024.) +

Reading a log table

2008-02-22 02:58:20

Comment added.

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	+ ------ + ''2008-02-22 02:58:20'' [[nbsp]] 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

Reading a log table

PhilipNeustrom — 2008-02-19 03:15:33

all i can add for now

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- \|\| Mathematics \|\|	+ \|\| Applied Mathematics \|\|
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- \|\| When did this skill become obsolete? \|\|	+ \|\| 1960s \|\|
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- \|\| What made this skill obsolete? \|\|	+ \|\| The advent of inexpensive calculators \|\|
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- \|\| What does the reader need to know to use this guide? \|\|	+ \|\| Basic high school math skills \|\|
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- \|\| When is this skill still useful? \|\|	+ \|\| When doing arithmetic without the aid of a calulator \|\|
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- Then, describe the skill! Tell us how to perform the task. Photos really help!	+ The idea behind a log table is the following property of ''logarithms'': + + {{{ + log (xy) = log x + log y + }}} + + It's much faster to, on paper, add two numbers than to multiply them.

Reading a log table

JasonAller — 2008-02-19 01:25:11

Upload of image Log_Table.JPG.

Reading a log table

JasonAller — 2008-02-19 01:24:46

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	+ [[Image(Log_Table.JPG, right, thumbnail, 400)]] + + \|\|<bgcolor='#E0E0FF'>'''Field'''\|\| + \|\| Mathematics \|\| + \|\|<bgcolor='#E0E0FF'>'''Went Obsolete'''\|\| + \|\| When did this skill become obsolete? \|\| + \|\|<bgcolor='#E0E0FF'>'''Made Obsolete By'''\|\| + \|\| What made this skill obsolete? \|\| + \|\|<bgcolor='#E0E0FF'>'''Knowledge Assumed'''\|\| + \|\| What does the reader need to know to use this guide? \|\| + \|\|<bgcolor='#E0E0FF'>'''When useful'''\|\| + \|\| When is this skill still useful? \|\| + + Then, describe the skill! Tell us how to perform the task. Photos really help! + + [[Include(Seed)]] + + == Other references == + + * list other references about this skill here + + [[Comments]]