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5.5 Properties and Laws of Logarithms Objective: 1. Use properties and laws of logarithms to simplify and evaluate expressions. Basic Properties of Logarithms Example #1 Solving with the Basic Properties Use the basic properties of logarithms to solve the equations below. A. log( x 3) 1 log x 3 1 10 10 x 3 10 x 13 A power with base 10 is the inverse of the common log. Since they are inverses they “undo” each other and we are left with the resulting expression. Don’t forget when something is done to one side of an equation, it must also be done to the other. Example #1 Solving with the Basic Properties Use the basic properties of logarithms to solve the equations below. B. 3 ln( 2 x) 2 ln 2 x 1 ln 2 x e e 2x e e x 2 1 Just like common logs, natural logs have an inverse as well, but with powers of base e. Product Law of Logarithms Example #2 Use the Product Law of Logarithms to evaluate each logarithm. A. Given that log 3 0.4771 and log 4 0.6021, find log 12. log 12 log 3 4 log 3 log 4 0.4771 0.6021 1.0792 Example #2 Use the Product Law of Logarithms to evaluate each logarithm. B. Given that log 3 0.4771 and log 4 0.6021, find log 27. log 27 log 3 3 3 log 3 log 3 log 3 3 log 3 3(0.4771) 1.4313 Example #2 Use the Product Law of Logarithms to evaluate each logarithm. C. Given that ln 3 1.0986 and ln 6 1.7918, find ln 54. ln 54 ln 3 3 6 ln 3 ln 3 ln 6 2 ln 3 ln 6 2(1.0986 ) 1.7918 3.989 Quotient Law of Logarithms Example #3 Use the Quotient Law of Logarithms to evaluate each logarithm. A. Given that log 40 1.6021 and log 8 0.9031, find log 5. log 5 40 log 8 log 40 log 8 1.6021 0.9031 0.699 Example #3 Use the Quotient Law of Logarithms to evaluate each logarithm. B. Given that ln 32 3.4657 and ln 8 2.0794, find ln 4. ln 4 32 ln 8 ln 32 ln 8 3.4657 2.0794 1.3863 Power Law of Logarithms Example #4 Use the Power Law of Logarithms to evaluate each logarithm. A. Given that log 25 1.3979, find log 4 25. log 4 25 log 25 1 4 14 log 25 14 1.3979 0.349475 Example #4 Use the Power Law of Logarithms to evaluate each logarithm. B. Given that ln 22 3.0910, find ln 22 . ln 22 ln 22 1 2 12 ln 22 12 3.0910 1.5455 Example #5 Simplifying Expressions Write each expression as a single logarithm. A. log 8x 3 log x log 2 x When condensing logarithms into a single logarithm, you must use the properties of logarithms backwards. Always bring coefficients up as exponents first, then combine addition and subtraction by multiplication and division. 2 log 8 x log x 3 log 2 x 2 8x x3 log 2 2 x 8x 4 log 2 2x log 4 x 2 Example #5 Simplifying Expressions Write each expression as a single logarithm. B. 4 ln( ex ) 4 You cannot change the subtraction into division because they both must be logarithms. 2 4 ln e ln e x ln e ln ex 2 4 4 8 This implies that e 4 x8 ln 4 e 4 ln e 4 ln x 8 4 1 4 & ln e 1 4 Example #5 Simplifying Expressions Write each expression as a single logarithm. B. 4 ln( ex ) 4 2 Alternatively, this problem can also be solve by expanding the expression first before condensing. 4 ln e ln x 2 4 4 ln e 4 ln x 2 4 4 4 ln x 2 4 4 ln x 2 ln x 2 4 ln x 8 Example #5 Simplifying Expressions Write each expression as a single logarithm. C. 2 log 2000 x log 10 y Like the previous problem, this one can also be solved by introducing a logarithm first, but this time use the fact that: log 10 1 2 log 10 log 2000 x log10 y log 10 2 log 2000 x log 10 y 2000 x log 2 10 10 y 2000 x log 1000 y 2x log y Example #5 Simplifying Expressions Write each expression as a single logarithm. C. 2 log 2000 x log 10 y This problem could have been solved by expanding first as well: 2 log 20 x 100 log 10 y 2 log 20 x log 100 log 10 y 2 log 20 x 2 log 10 y log 20 x log 10 y 20 x log 10 y 2x log y Example #6 Simplifying by Substitution Write the expression in terms of u and v given u ln x and v ln y A. 4 3 ln( x y ) Expand the expression first, the substitute. ln x ln y 4 ln x 3 ln y 4u 3v 4 3 Example #6 Simplifying by Substitution Write the expression in terms of u and v given u ln x and v ln y B. 3 x ln 2 ln 3 x ln y 2 y 1 2 3 ln x ln y 1 ln x 2 ln 3 1 u 2 v 3 y