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- 1. Properties of Logarithm
- 2. Properties of LogarithmRecall the following The correspondingRules of Exponents: Rules of Logs are:
- 3. Properties of LogarithmRecall the following The correspondingRules of Exponents: Rules of Logs are:1. b0 = 1
- 4. Properties of LogarithmRecall the following The correspondingRules of Exponents: Rules of Logs are:1. b0 = 12. br · bt = br+t
- 5. Properties of LogarithmRecall the following The correspondingRules of Exponents: Rules of Logs are:1. b0 = 12. br · bt = br+t br = br-t3. t b
- 6. Properties of LogarithmRecall the following The correspondingRules of Exponents: Rules of Logs are:1. b0 = 12. br · bt = br+t3. tbr = br-t b4. (br)t = brt
- 7. Properties of LogarithmRecall the following The correspondingRules of Exponents: Rules of Logs are:1. b0 = 1 1. logb(1) = 02. br · bt = br+t3. tbr = br-t b4. (br)t = brt
- 8. Properties of LogarithmRecall the following The correspondingRules of Exponents: Rules of Logs are:1. b0 = 1 1. logb(1) = 02. br · bt = br+t 2. logb(x·y) = logb(x)+logb(y)3. tbr = br-t b4. (br)t = brt
- 9. Properties of LogarithmRecall the following The correspondingRules of Exponents: Rules of Logs are:1. b0 = 1 1. logb(1) = 02. br · bt = br+t 2. logb(x·y) = logb(x)+logb(y)3. tbr = br-t x b 3. logb( y ) = logb(x) – logb(y)4. (br)t = brt
- 10. Properties of LogarithmRecall the following The correspondingRules of Exponents: Rules of Logs are:1. b0 = 1 1. logb(1) = 02. br · bt = br+t 2. logb(x·y) = logb(x)+logb(y)3. tbr = br-t x b 3. logb( y ) = logb(x) – logb(y)4. (br)t = brt 4. logb(xt) = t·logb(x)
- 11. Properties of LogarithmRecall the following The correspondingRules of Exponents: Rules of Logs are:1. b0 = 1 1. logb(1) = 02. br · bt = br+t 2. logb(x·y) = logb(x)+logb(y)3. tbr = br-t x b 3. logb( y ) = logb(x) – logb(y)4. (br)t = brt 4. logb(xt) = t·logb(x)We verify part 2: logb(xy) = logb(x) + logb(y), x, y > 0.Proof:
- 12. Properties of LogarithmRecall the following The correspondingRules of Exponents: Rules of Logs are:1. b0 = 1 1. logb(1) = 02. br · bt = br+t 2. logb(x·y) = logb(x)+logb(y)3. tbr = br-t x b 3. logb( y ) = logb(x) – logb(y)4. (br)t = brt 4. logb(xt) = t·logb(x)We verify part 2: logb(xy) = logb(x) + logb(y), x, y > 0.Proof:Let x and y be two positive numbers.
- 13. Properties of LogarithmRecall the following The correspondingRules of Exponents: Rules of Logs are:1. b0 = 1 1. logb(1) = 02. br · bt = br+t 2. logb(x·y) = logb(x)+logb(y)3. tbr = br-t x b 3. logb( y ) = logb(x) – logb(y)4. (br)t = brt 4. logb(xt) = t·logb(x)We verify part 2: logb(xy) = logb(x) + logb(y), x, y > 0.Proof:Let x and y be two positive numbers. Let logb(x) = rand logb(y) = t, which in exp-form are x = br and y = bt.
- 14. Properties of LogarithmRecall the following The correspondingRules of Exponents: Rules of Logs are:1. b0 = 1 1. logb(1) = 02. br · bt = br+t 2. logb(x·y) = logb(x)+logb(y)3. tbr = br-t x b 3. logb( y ) = logb(x) – logb(y)4. (br)t = brt 4. logb(xt) = t·logb(x)We verify part 2: logb(xy) = logb(x) + logb(y), x, y > 0.Proof:Let x and y be two positive numbers. Let logb(x) = rand logb(y) = t, which in exp-form are x = br and y = bt.Therefore x·y = br+t,
- 15. Properties of LogarithmRecall the following The correspondingRules of Exponents: Rules of Logs are:1. b0 = 1 1. logb(1) = 02. br · bt = br+t 2. logb(x·y) = logb(x)+logb(y)3. tbr = br-t x b 3. logb( y ) = logb(x) – logb(y)4. (br)t = brt 4. logb(xt) = t·logb(x)We verify part 2: logb(xy) = logb(x) + logb(y), x, y > 0.Proof:Let x and y be two positive numbers. Let logb(x) = rand logb(y) = t, which in exp-form are x = br and y = bt.Therefore x·y = br+t, which in log-form islogb(x·y) = r + t = logb(x)+logb(y).
- 16. Properties of LogarithmRecall the following The correspondingRules of Exponents: Rules of Logs are:1. b0 = 1 1. logb(1) = 02. br · bt = br+t 2. logb(x·y) = logb(x)+logb(y)3. tbr = br-t x b 3. logb( y ) = logb(x) – logb(y)4. (br)t = brt 4. logb(xt) = t·logb(x)We verify part 2: logb(xy) = logb(x) + logb(y), x, y > 0.Proof:Let x and y be two positive numbers. Let logb(x) = rand logb(y) = t, which in exp-form are x = br and y = bt.Therefore x·y = br+t, which in log-form isThe(x·y) = rules = logbbe verified similarly.logb other r + t may (x)+logb(y).
- 17. Properties of LogarithmExample A:a. Write log( 3x2 ) in terms of log(x) and log(y). √y
- 18. Properties of LogarithmExample A:a. Write log( 3x2 ) in terms of log(x) and log(y). √ylog( 3x2 ) = log( 3x2 ), √y y1/2
- 19. Properties of LogarithmExample A:a. Write log( 3x2 ) in terms of log(x) and log(y). √ylog( 3x2 ) = log( 3x2 ), by the quotient rule √y y1/2 = log (3x2) – log(y1/2)
- 20. Properties of LogarithmExample A:a. Write log( 3x2 ) in terms of log(x) and log(y). √ylog( 3x2 ) = log( 3x2 ), by the quotient rule √y y1/2 = log (3x2) – log(y1/2) product rule = log(3) + log(x2)
- 21. Properties of LogarithmExample A:a. Write log( 3x2 ) in terms of log(x) and log(y). √ylog( 3x2 ) = log( 3x2 ), by the quotient rule √y y1/2 = log (3x2) – log(y1/2) product rule power rule = log(3) + log(x2) – ½ log(y)
- 22. Properties of LogarithmExample A:a. Write log( 3x2 ) in terms of log(x) and log(y). √ylog( 3x2 ) = log( 3x2 ), by the quotient rule √y y1/2 = log (3x2) – log(y1/2) product rule power rule = log(3) + log(x2) – ½ log(y) = log(3) + 2log(x) – ½ log(y)
- 23. Properties of LogarithmExample A:a. Write log( 3x2 ) in terms of log(x) and log(y). √ylog( 3x2 ) = log( 3x2 ), by the quotient rule √y y1/2 = log (3x2) – log(y1/2) product rule power rule = log(3) + log(x2) – ½ log(y) = log(3) + 2log(x) – ½ log(y)b. Combine log(3) + 2log(x) – ½ log(y) into one log.
- 24. Properties of LogarithmExample A:a. Write log( 3x2 ) in terms of log(x) and log(y). √ylog( 3x2 ) = log( 3x2 ), by the quotient rule √y y1/2 = log (3x2) – log(y1/2) product rule power rule = log(3) + log(x2) – ½ log(y) = log(3) + 2log(x) – ½ log(y)b. Combine log(3) + 2log(x) – ½ log(y) into one log.log(3) + 2log(x) – ½ log(y) power rule= log(3) + log(x2) – log(y1/2)
- 25. Properties of LogarithmExample A:a. Write log( 3x2 ) in terms of log(x) and log(y). √ylog( 3x2 ) = log( 3x2 ), by the quotient rule √y y1/2 = log (3x2) – log(y1/2) product rule power rule = log(3) + log(x2) – ½ log(y) = log(3) + 2log(x) – ½ log(y)b. Combine log(3) + 2log(x) – ½ log(y) into one log.log(3) + 2log(x) – ½ log(y) power rule= log(3) + log(x2) – log(y1/2) product rule= log (3x2) – log(y1/2)
- 26. Properties of LogarithmExample A:a. Write log( 3x2 ) in terms of log(x) and log(y). √ylog( 3x2 ) = log( 3x2 ), by the quotient rule √y y1/2 = log (3x2) – log(y1/2) product rule power rule = log(3) + log(x2) – ½ log(y) = log(3) + 2log(x) – ½ log(y)b. Combine log(3) + 2log(x) – ½ log(y) into one log.log(3) + 2log(x) – ½ log(y) power rule= log(3) + log(x2) – log(y1/2) product rule 2 1/2 3x2)= log (3x ) – log(y )= log( 1/2 y
- 27. Properties of LogarithmThe following url is a base 10 log table.http://www.sosmath.com/tables/logtable/logtable.html
- 28. Properties of LogarithmThe following url is a base 10 log table.http://www.sosmath.com/tables/logtable/logtable.htmlThe table was created in early 17th century by HenryBrigg who converted the Napier log into a workableform.
- 29. Properties of LogarithmThe following url is a base 10 log table.http://www.sosmath.com/tables/logtable/logtable.htmlThe table was created in early 17th century by HenryBrigg who converted the Napier log into a workableform. We will demonstrate how numerical calculationsmay be carried out with this table.
- 30. Properties of LogarithmThe following url is a base 10 log table.http://www.sosmath.com/tables/logtable/logtable.htmlThe table was created in early 17th century by HenryBrigg who converted the Napier log into a workableform. We will demonstrate how numerical calculationsmay be carried out with this table. Given a calculationproblem we do two steps.
- 31. Properties of LogarithmThe following url is a base 10 log table.http://www.sosmath.com/tables/logtable/logtable.htmlThe table was created in early 17th century by HenryBrigg who converted the Napier log into a workableform. We will demonstrate how numerical calculationsmay be carried out with this table. Given a calculationproblem we do two steps.1. Use the Rules of Logs to take apart the originalproblems into pieces that correspond to the entries inthe table to find the log of the outcome (which wedon’t know yet).
- 32. Properties of LogarithmThe following url is a base 10 log table.http://www.sosmath.com/tables/logtable/logtable.htmlThe table was created in early 17th century by HenryBrigg who converted the Napier log into a workableform. We will demonstrate how numerical calculationsmay be carried out with this table. Given a calculationproblem we do two steps.1. Use the Rules of Logs to take apart the originalproblems into pieces that correspond to the entries inthe table to find the log of the outcome (which wedon’t know yet).2. Use the answer from step 1 to construct theanswer, again with the help of the table.
- 33. Properties of LogarithmLet’s find 2.58100.
- 34. Properties of LogarithmLet’s find 2.58100.Step 1. Calculate log of the answer.log(2.5100) =
- 35. Properties of LogarithmLet’s find 2.58100.Step 1. Calculate log of the answer.log(2.5100) = 100 log(2.5) rules of logs
- 36. Properties of LogarithmLet’s find 2.58100.Step 1. Calculate log of the answer.log(2.5100) = 100 log(2.5) rules of logs = 100 (0.39794) use the table
- 37. Properties of LogarithmLet’s find 2.58100.Step 1. Calculate log of the answer.log(2.5100) = 100 log(2.5) rules of logs = 100 (0.39794) use the table = 39.794
- 38. Properties of LogarithmLet’s find 2.58100.Step 1. Calculate log of the answer.log(2.5100) = 100 log(2.5) rules of logs = 100 (0.39794) use the table = 39.794Step 2. Construct the answer. ≈
- 39. Properties of LogarithmLet’s find 2.58100.Step 1. Calculate log of the answer.log(2.5100) = 100 log(2.5) rules of logs = 100 (0.39794) use the table = 39.794Step 2. Construct the answer. ≈From step 1, log(2.5100) = 39.794,
- 40. Properties of LogarithmLet’s find 2.58100.Step 1. Calculate log of the answer.log(2.5100) = 100 log(2.5) rules of logs = 100 (0.39794) use the table = 39.794Step 2. Construct the answer. ≈From step 1, log(2.5100) = 39.794,therefore 2.5100 ≈ 1039.794
- 41. Properties of LogarithmLet’s find 2.58100.Step 1. Calculate log of the answer.log(2.5100) = 100 log(2.5) rules of logs = 100 (0.39794) use the table = 39.794Step 2. Construct the answer. ≈From step 1, log(2.5100) = 39.794,therefore 2.5100 ≈ 1039.794 write it in two parts ≈ 1039 100.794
- 42. Properties of LogarithmLet’s find 2.58100.Step 1. Calculate log of the answer.log(2.5100) = 100 log(2.5) rules of logs = 100 (0.39794) use the table = 39.794Step 2. Construct the answer. ≈From step 1, log(2.5100) = 39.794,therefore 2.5100 ≈ 1039.794 write it in two parts ≈ 1039 100.794 use the table ≈ 6.22x1039
- 43. Properties of LogarithmThe introduction of the Brigg’s table in 17th centuryenabled humans to perform complicated calculations.
- 44. Properties of LogarithmThe introduction of the Brigg’s table in 17th centuryenabled humans to perform complicated calculations. It is analogous to the modern introduction ofcomputers to handle large amounts of data which wasimpossible for humans to handle before.
- 45. Properties of LogarithmExercise A. Use the Rules of logs to take apart thefollowing expressions as much as possible. y x2y )1. log (3x y) 2 2. log( 2) 3. log( 3x 34. log (y3√x) 5. log (√y3x) 6. log( 3 ) √y3x7. log (10y) 8. log (10y+√x) 9. log (k*100.89t)10. ln (5,000e0.05t) 11. ln (Pe40r) 12. ln(k*x5.46ty0.89t)13. log (√y +x) 3 14. ln( 3 + x ) 15. ln( a + bx ) 3√y3 c – dx
- 46. Properties of LogarithmExercise A. Use the Rules of Logs to assemble thefollowing expressions into a single log–expression.16. log(3)+ 2log(x)+log(y) 17. –log(3)–2log(x)+log(y)18. –log(3)+2log(x)+log(y) 19. ½log(x)+3log(y)20. ½log(x)+3log(y)/2 21. log(3)–½log(x)–3log(y)/222. log (3)+log(c+d) 23. ln (a+b)–ln(3)24. log (a+b)–log(c+d) 25. log(a+b)+ln(c+d)26. ln (a2–b2)–log(a–b) 27. ln (x2–3x–4)–ln(x–4)28. 1+log(x–2) 29. 1–log(x–2)30. 1–½ log(x)

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