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Properties of Logarithms

Properties of Logarithms. Product property of logarithms. For all positive numbers m, n, and b, where b  1, log b mn = log b m + log b n. Example1. Given log 3 5  1.4650, find each logarithm. a. log 3 45. log 3 (9•5). log 3 9 + log 3 5. 2 + 1.4650 = 3.4650.

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Properties of Logarithms

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  1. Properties of Logarithms

  2. Product property of logarithms For all positive numbers m, n, and b, where b  1, logbmn = logbm + logbn.

  3. Example1. Given log35  1.4650, find each logarithm. a. log345 log3(9•5) log39 + log35 2 + 1.4650 = 3.4650

  4. Example1. Given log35  1.4650, find each logarithm. b. log325 log3(5•5) log35 + log35 1.4650 + 1.4650 = 2.9300

  5. Quotient property of logarithms For all positive numbers m, n, and b, where b  1, logbm/n = logbm - logbn.

  6. Example2. Given log45  1.1610 and log415  1.9534, find each logarithm. a. log45/16 log45 - log416 1.1610 - 2 = -0.8390

  7. Example2. Given log45  1.1610 and log415  1.9534, find each logarithm. b. log43 How can I rewrite 3 using 5 and 15? log415/5 log415 - log45 1.9534 - 1.1610 = 0.7924

  8. Example3. The pH of a substance is the concentration of hydrogen ions, [H+], measured in moles of hydrogen per liter of substance. It is given by the formula, pH = log10(1/[H+]) Find the amount of hydrogen in a liter of acid rain that has a pH of 4.2.

  9. Example3. The pH of acid rain. It is given by the formula, pH = log10(1/[H+]) Find the amount of hydrogen in a liter of acid rain that has a pH of 4.2. 4.2 = log10(1/[H+]) 4.2 = log101 - log10[H+]

  10. Example3. The pH of acid rain. Find the amount of hydrogen in a liter of acid rain that has a pH of 4.2. 4.2 = log10(1/[H+]) 4.2 = log101 - log10[H+] 4.2 = 0 - log10[H+] 4.2 = -log10[H+]

  11. Example3. The pH of acid rain. 4.2 = log101 - log10[H+] 4.2 = 0 - log10[H+] 4.2 = -log10[H+] -4.2 = log10[H+] 10-4.2 = [H+] [H+] = 10-4.2 0.000063

  12. Power property of logarithms For any real number p and positive numbers m, and b, where b  1, logbmp = p•logbm .

  13. Example4. Solve a. 2log36 - (1/4)log316 = log3x b. log10z + log10(z+3) = 1

  14. Example4. Solve a. 2log36 - (1/4)log316 = log3x log362 - log3161/4 = log3x log336 - log32 = log3x log336/2 = log3x log318 = log3x x = 18

  15. Example4. Solve b. log10z + log10(z+3) = 1 log10z(z+3) = 1 z(z+3) = 101 z2 + 3z - 10 = 0 z = 2 is the solution. (z+5)(z-2) = 0 z = -5 or z = 2

  16. Example5. Rewrite as one logarithm log102 + log10(x+9) + log10(y+6) The properties allow us to rewrite these two additions as a single multiplication problem. log10[2(x+9)(y+6)] log10(2xy+12x+18y+108)

  17. 47 69 47 69 47 69 loga √ 1 2 1 2 1 2 1 2 1 2 loga loga loga47-loga69 loga47 - loga69 Example6. Rewrite as an equivalent logarithmic expression

  18. x4y9 z5 x4y9 z5 x4y9 z5 6 loga loga √ logax4y9-logaz5 loga 1 6 1 6 1 6 1 6 logax4 + logay9-logaz5 Example7. Rewrite as an equivalent logarithmic expression

  19. 4logax+ 9logay-5logaz 1 6 1 6 2 3 3 2 5 6 logax+ logay- logaz logax4 + logay9-logaz5 Example7. Rewrite as an equivalent logarithmic expression

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