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1 | 305-308 | e , m lies between 0 and 1 When log
n has been expressed as p + log m, where p is an integer and 0 log m < 1, we say that p is the
“characteristic” of log n and that log m is the “mantissa of log n Note that characteristic is always an
integer – positive, negative or zero, and mantissa is never negative and is always less than 1 |
1 | 306-309 | , m lies between 0 and 1 When log
n has been expressed as p + log m, where p is an integer and 0 log m < 1, we say that p is the
“characteristic” of log n and that log m is the “mantissa of log n Note that characteristic is always an
integer – positive, negative or zero, and mantissa is never negative and is always less than 1 If we can
find the characteristics and the mantissa of log n, we have to just add them to get log n |
1 | 307-310 | When log
n has been expressed as p + log m, where p is an integer and 0 log m < 1, we say that p is the
“characteristic” of log n and that log m is the “mantissa of log n Note that characteristic is always an
integer – positive, negative or zero, and mantissa is never negative and is always less than 1 If we can
find the characteristics and the mantissa of log n, we have to just add them to get log n Thus to find log n, all we have to do is as follows:
1 |
1 | 308-311 | Note that characteristic is always an
integer – positive, negative or zero, and mantissa is never negative and is always less than 1 If we can
find the characteristics and the mantissa of log n, we have to just add them to get log n Thus to find log n, all we have to do is as follows:
1 Put n in the standard form, say
n = m × 10
p, 1 < m <10
2 |
1 | 309-312 | If we can
find the characteristics and the mantissa of log n, we have to just add them to get log n Thus to find log n, all we have to do is as follows:
1 Put n in the standard form, say
n = m × 10
p, 1 < m <10
2 Read off the characteristic p of log n from this expression (exponent of 10) |
1 | 310-313 | Thus to find log n, all we have to do is as follows:
1 Put n in the standard form, say
n = m × 10
p, 1 < m <10
2 Read off the characteristic p of log n from this expression (exponent of 10) 3 |
1 | 311-314 | Put n in the standard form, say
n = m × 10
p, 1 < m <10
2 Read off the characteristic p of log n from this expression (exponent of 10) 3 Look up log m from tables, which is being explained below |
1 | 312-315 | Read off the characteristic p of log n from this expression (exponent of 10) 3 Look up log m from tables, which is being explained below 4 |
1 | 313-316 | 3 Look up log m from tables, which is being explained below 4 Write log n = p + log m
If the characteristic p of a number n is say, 2 and the mantissa is |
1 | 314-317 | Look up log m from tables, which is being explained below 4 Write log n = p + log m
If the characteristic p of a number n is say, 2 and the mantissa is 4133, then we have log n = 2
+ |
1 | 315-318 | 4 Write log n = p + log m
If the characteristic p of a number n is say, 2 and the mantissa is 4133, then we have log n = 2
+ 4133 which we can write as 2 |
1 | 316-319 | Write log n = p + log m
If the characteristic p of a number n is say, 2 and the mantissa is 4133, then we have log n = 2
+ 4133 which we can write as 2 4133 |
1 | 317-320 | 4133, then we have log n = 2
+ 4133 which we can write as 2 4133 If, however, the characteristic p of a number m is say –2 and the
mantissa is |
1 | 318-321 | 4133 which we can write as 2 4133 If, however, the characteristic p of a number m is say –2 and the
mantissa is 4123, then we have log m = –2 + |
1 | 319-322 | 4133 If, however, the characteristic p of a number m is say –2 and the
mantissa is 4123, then we have log m = –2 + 4123 |
1 | 320-323 | If, however, the characteristic p of a number m is say –2 and the
mantissa is 4123, then we have log m = –2 + 4123 We cannot write this as –2 |
1 | 321-324 | 4123, then we have log m = –2 + 4123 We cannot write this as –2 4123 |
1 | 322-325 | 4123 We cannot write this as –2 4123 (Why |
1 | 323-326 | We cannot write this as –2 4123 (Why ) In order
to avoid this confusion we write 2 for –2 and thus we write log m = 2 |
1 | 324-327 | 4123 (Why ) In order
to avoid this confusion we write 2 for –2 and thus we write log m = 2 4123 |
1 | 325-328 | (Why ) In order
to avoid this confusion we write 2 for –2 and thus we write log m = 2 4123 Now let us explain how to use the table of logarithms to find mantissas |
1 | 326-329 | ) In order
to avoid this confusion we write 2 for –2 and thus we write log m = 2 4123 Now let us explain how to use the table of logarithms to find mantissas A table is appended at the
end of this Appendix |
1 | 327-330 | 4123 Now let us explain how to use the table of logarithms to find mantissas A table is appended at the
end of this Appendix Observe that in the table, every row starts with a two digit number, 10, 11, 12, |
1 | 328-331 | Now let us explain how to use the table of logarithms to find mantissas A table is appended at the
end of this Appendix Observe that in the table, every row starts with a two digit number, 10, 11, 12, 97, 98, 99 |
1 | 329-332 | A table is appended at the
end of this Appendix Observe that in the table, every row starts with a two digit number, 10, 11, 12, 97, 98, 99 Every
column is headed by a one-digit number, 0, 1, 2, |
1 | 330-333 | Observe that in the table, every row starts with a two digit number, 10, 11, 12, 97, 98, 99 Every
column is headed by a one-digit number, 0, 1, 2, 9 |
1 | 331-334 | 97, 98, 99 Every
column is headed by a one-digit number, 0, 1, 2, 9 On the right, we have the section called “Mean
differences” which has 9 columns headed by 1, 2 |
1 | 332-335 | Every
column is headed by a one-digit number, 0, 1, 2, 9 On the right, we have the section called “Mean
differences” which has 9 columns headed by 1, 2 9 |
1 | 333-336 | 9 On the right, we have the section called “Mean
differences” which has 9 columns headed by 1, 2 9 1
2
3
4
5
6
7
8
9 |
1 | 334-337 | On the right, we have the section called “Mean
differences” which has 9 columns headed by 1, 2 9 1
2
3
4
5
6
7
8
9 1
1
2
3
4
4
5
6
6
1
1
2
3
3
4
5
6
6
1
1
2
3
3
4
5
6
6 |
1 | 335-338 | 9 1
2
3
4
5
6
7
8
9 1
1
2
3
4
4
5
6
6
1
1
2
3
3
4
5
6
6
1
1
2
3
3
4
5
6
6 0
1
2
3
4
5
6
7
8
9 |
1 | 336-339 | 1
2
3
4
5
6
7
8
9 1
1
2
3
4
4
5
6
6
1
1
2
3
3
4
5
6
6
1
1
2
3
3
4
5
6
6 0
1
2
3
4
5
6
7
8
9 61
7853
7860 7868
7875
7882 7889
7896
7803 7810
7817
62
7924
7931 7935
7945
7954 7959
7966
7973 7980
7987
63
7993
8000 8007
8014
8021 8028
8035
8041 8048
8055 |
1 | 337-340 | 1
1
2
3
4
4
5
6
6
1
1
2
3
3
4
5
6
6
1
1
2
3
3
4
5
6
6 0
1
2
3
4
5
6
7
8
9 61
7853
7860 7868
7875
7882 7889
7896
7803 7810
7817
62
7924
7931 7935
7945
7954 7959
7966
7973 7980
7987
63
7993
8000 8007
8014
8021 8028
8035
8041 8048
8055 Rationalised 2023-24
148
Chemistry
Now suppose we wish to find log (6 |
1 | 338-341 | 0
1
2
3
4
5
6
7
8
9 61
7853
7860 7868
7875
7882 7889
7896
7803 7810
7817
62
7924
7931 7935
7945
7954 7959
7966
7973 7980
7987
63
7993
8000 8007
8014
8021 8028
8035
8041 8048
8055 Rationalised 2023-24
148
Chemistry
Now suppose we wish to find log (6 234) |
1 | 339-342 | 61
7853
7860 7868
7875
7882 7889
7896
7803 7810
7817
62
7924
7931 7935
7945
7954 7959
7966
7973 7980
7987
63
7993
8000 8007
8014
8021 8028
8035
8041 8048
8055 Rationalised 2023-24
148
Chemistry
Now suppose we wish to find log (6 234) Then look into the row starting with 62 |
1 | 340-343 | Rationalised 2023-24
148
Chemistry
Now suppose we wish to find log (6 234) Then look into the row starting with 62 In this row, look
at the number in the column headed by 3 |
1 | 341-344 | 234) Then look into the row starting with 62 In this row, look
at the number in the column headed by 3 The number is 7945 |
1 | 342-345 | Then look into the row starting with 62 In this row, look
at the number in the column headed by 3 The number is 7945 This means that
log (6 |
1 | 343-346 | In this row, look
at the number in the column headed by 3 The number is 7945 This means that
log (6 230) = 0 |
1 | 344-347 | The number is 7945 This means that
log (6 230) = 0 7945*
But we want log (6 |
1 | 345-348 | This means that
log (6 230) = 0 7945*
But we want log (6 234) |
1 | 346-349 | 230) = 0 7945*
But we want log (6 234) So our answer will be a little more than 0 |
1 | 347-350 | 7945*
But we want log (6 234) So our answer will be a little more than 0 7945 |
1 | 348-351 | 234) So our answer will be a little more than 0 7945 How much more |
1 | 349-352 | So our answer will be a little more than 0 7945 How much more We look
this up in the section on Mean differences |
1 | 350-353 | 7945 How much more We look
this up in the section on Mean differences Since our fourth digit is 4, look under the column headed
by 4 in the Mean difference section (in the row 62) |
1 | 351-354 | How much more We look
this up in the section on Mean differences Since our fourth digit is 4, look under the column headed
by 4 in the Mean difference section (in the row 62) We see the number 3 there |
1 | 352-355 | We look
this up in the section on Mean differences Since our fourth digit is 4, look under the column headed
by 4 in the Mean difference section (in the row 62) We see the number 3 there So add 3 to 7945 |
1 | 353-356 | Since our fourth digit is 4, look under the column headed
by 4 in the Mean difference section (in the row 62) We see the number 3 there So add 3 to 7945 We
get 7948 |
1 | 354-357 | We see the number 3 there So add 3 to 7945 We
get 7948 So we finally have
log (6 |
1 | 355-358 | So add 3 to 7945 We
get 7948 So we finally have
log (6 234) = 0 |
1 | 356-359 | We
get 7948 So we finally have
log (6 234) = 0 7948 |
1 | 357-360 | So we finally have
log (6 234) = 0 7948 Take another example |
1 | 358-361 | 234) = 0 7948 Take another example To find log (8 |
1 | 359-362 | 7948 Take another example To find log (8 127), we look in the row 81 under column 2, and we find 9096 |
1 | 360-363 | Take another example To find log (8 127), we look in the row 81 under column 2, and we find 9096 We continue in the same row and see that the mean difference under 7 is 4 |
1 | 361-364 | To find log (8 127), we look in the row 81 under column 2, and we find 9096 We continue in the same row and see that the mean difference under 7 is 4 Adding this to 9096, and
we get 9100 |
1 | 362-365 | 127), we look in the row 81 under column 2, and we find 9096 We continue in the same row and see that the mean difference under 7 is 4 Adding this to 9096, and
we get 9100 So, log (8 |
1 | 363-366 | We continue in the same row and see that the mean difference under 7 is 4 Adding this to 9096, and
we get 9100 So, log (8 127) = 0 |
1 | 364-367 | Adding this to 9096, and
we get 9100 So, log (8 127) = 0 9100 |
1 | 365-368 | So, log (8 127) = 0 9100 Finding N when log N is given
We have so far discussed the procedure for finding log n when a positive number n given |
1 | 366-369 | 127) = 0 9100 Finding N when log N is given
We have so far discussed the procedure for finding log n when a positive number n given We now turn
to its converse i |
1 | 367-370 | 9100 Finding N when log N is given
We have so far discussed the procedure for finding log n when a positive number n given We now turn
to its converse i e |
1 | 368-371 | Finding N when log N is given
We have so far discussed the procedure for finding log n when a positive number n given We now turn
to its converse i e , to find n when log n is given and give a method for this purpose |
1 | 369-372 | We now turn
to its converse i e , to find n when log n is given and give a method for this purpose If log n = t, we
sometimes say n = antilog t |
1 | 370-373 | e , to find n when log n is given and give a method for this purpose If log n = t, we
sometimes say n = antilog t Therefore our task is given t, find its antilog |
1 | 371-374 | , to find n when log n is given and give a method for this purpose If log n = t, we
sometimes say n = antilog t Therefore our task is given t, find its antilog For this, we use the ready-
made antilog tables |
1 | 372-375 | If log n = t, we
sometimes say n = antilog t Therefore our task is given t, find its antilog For this, we use the ready-
made antilog tables Suppose log n = 2 |
1 | 373-376 | Therefore our task is given t, find its antilog For this, we use the ready-
made antilog tables Suppose log n = 2 5372 |
1 | 374-377 | For this, we use the ready-
made antilog tables Suppose log n = 2 5372 To find n, first take just the mantissa of log n |
1 | 375-378 | Suppose log n = 2 5372 To find n, first take just the mantissa of log n In this case it is |
1 | 376-379 | 5372 To find n, first take just the mantissa of log n In this case it is 5372 |
1 | 377-380 | To find n, first take just the mantissa of log n In this case it is 5372 (Make sure it is positive |
1 | 378-381 | In this case it is 5372 (Make sure it is positive )
Now take up antilog of this number in the antilog table which is to be used exactly like the log table |
1 | 379-382 | 5372 (Make sure it is positive )
Now take up antilog of this number in the antilog table which is to be used exactly like the log table In the antilog table, the entry under column 7 in the row |
1 | 380-383 | (Make sure it is positive )
Now take up antilog of this number in the antilog table which is to be used exactly like the log table In the antilog table, the entry under column 7 in the row 53 is 3443 and the mean difference for the
last digit 2 in that row is 2, so the table gives 3445 |
1 | 381-384 | )
Now take up antilog of this number in the antilog table which is to be used exactly like the log table In the antilog table, the entry under column 7 in the row 53 is 3443 and the mean difference for the
last digit 2 in that row is 2, so the table gives 3445 Hence,
antilog ( |
1 | 382-385 | In the antilog table, the entry under column 7 in the row 53 is 3443 and the mean difference for the
last digit 2 in that row is 2, so the table gives 3445 Hence,
antilog ( 5372) = 3 |
1 | 383-386 | 53 is 3443 and the mean difference for the
last digit 2 in that row is 2, so the table gives 3445 Hence,
antilog ( 5372) = 3 445
Now since log n = 2 |
1 | 384-387 | Hence,
antilog ( 5372) = 3 445
Now since log n = 2 5372, the characteristic of log n is 2 |
1 | 385-388 | 5372) = 3 445
Now since log n = 2 5372, the characteristic of log n is 2 So the standard form of n is given by
n = 3 |
1 | 386-389 | 445
Now since log n = 2 5372, the characteristic of log n is 2 So the standard form of n is given by
n = 3 445 × 10
2
or n = 344 |
1 | 387-390 | 5372, the characteristic of log n is 2 So the standard form of n is given by
n = 3 445 × 10
2
or n = 344 5
Illustration 1:
If log x = 1 |
1 | 388-391 | So the standard form of n is given by
n = 3 445 × 10
2
or n = 344 5
Illustration 1:
If log x = 1 0712, find x |
1 | 389-392 | 445 × 10
2
or n = 344 5
Illustration 1:
If log x = 1 0712, find x Solution: We find that the number corresponding to 0712 is 1179 |
1 | 390-393 | 5
Illustration 1:
If log x = 1 0712, find x Solution: We find that the number corresponding to 0712 is 1179 Since characteristic of log x is 1, we
have
x = 1 |
1 | 391-394 | 0712, find x Solution: We find that the number corresponding to 0712 is 1179 Since characteristic of log x is 1, we
have
x = 1 179 × 10
1
= 11 |
1 | 392-395 | Solution: We find that the number corresponding to 0712 is 1179 Since characteristic of log x is 1, we
have
x = 1 179 × 10
1
= 11 79
Illustration 2:
If log10 x = 2 |
1 | 393-396 | Since characteristic of log x is 1, we
have
x = 1 179 × 10
1
= 11 79
Illustration 2:
If log10 x = 2 1352, find x |
1 | 394-397 | 179 × 10
1
= 11 79
Illustration 2:
If log10 x = 2 1352, find x Solution: From antilog tables, we find that the number corresponding to 1352 is 1366 |
1 | 395-398 | 79
Illustration 2:
If log10 x = 2 1352, find x Solution: From antilog tables, we find that the number corresponding to 1352 is 1366 Since the
characteristic is 2 i |
1 | 396-399 | 1352, find x Solution: From antilog tables, we find that the number corresponding to 1352 is 1366 Since the
characteristic is 2 i e |
1 | 397-400 | Solution: From antilog tables, we find that the number corresponding to 1352 is 1366 Since the
characteristic is 2 i e , –2, so
x = 1 |
1 | 398-401 | Since the
characteristic is 2 i e , –2, so
x = 1 366 × 10
–2 = 0 |
1 | 399-402 | e , –2, so
x = 1 366 × 10
–2 = 0 01366
Use of Logarithms in Numerical Calculations
Illustration 1:
Find 6 |
1 | 400-403 | , –2, so
x = 1 366 × 10
–2 = 0 01366
Use of Logarithms in Numerical Calculations
Illustration 1:
Find 6 3 × 1 |
1 | 401-404 | 366 × 10
–2 = 0 01366
Use of Logarithms in Numerical Calculations
Illustration 1:
Find 6 3 × 1 29
Solution: Let x = 6 |
1 | 402-405 | 01366
Use of Logarithms in Numerical Calculations
Illustration 1:
Find 6 3 × 1 29
Solution: Let x = 6 3 × 1 |
1 | 403-406 | 3 × 1 29
Solution: Let x = 6 3 × 1 29
Then log10 x = log (6 |
1 | 404-407 | 29
Solution: Let x = 6 3 × 1 29
Then log10 x = log (6 3 × 1 |
Subsets and Splits