LESSON 11 EXPONENTIAL AND LOGARITHMIC EQUATIONS Since any exponential function is one-to-one, then b u b v if and only if u v . Examples Solve the following exponential equations. Give exact answers. No decimal approximations. 1. 3 x 81 Using the one-to-one property: 3 x 81 3 x 3 4 x 4 x Using logarithms base 3: 3 x 81 log 3 3 log 3 81 x log 3 3 log 3 81 x log 3 81 4 NOTE: log 3 3 1 Using natural logarithms: 3 x 81 ln 3 x ln 81 x ln 3 ln 81 x ln 81 x 4 (using a calculator) ln 3 Answer: 4 2. 5 t 25 3. 2 3x 11 32 Using the one-to-one property: 2 3 x 11 32 2 3 x Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1320 11 25 3 x 11 5 3 x 16 x Using logarithms base 2: 2 3 x 11 16 3 32 log 2 2 3 x 11 log 2 32 ( 3 x 11 ) log 2 2 log 2 32 3 x 11 5 3 x 16 x 16 3 NOTE: log 2 2 1 Using natural logarithms: 2 3 x 11 32 ln 2 3 x 11 ln 32 ( 3 x 11 ) ln 2 ln 32 3 x ln 2 11 ln 2 ln 32 ln 32 11 ln 2 ln 32 ( 2 11 ) 16 3 x ln 2 ln 32 11 ln 2 x 3 ln 2 ln 8 3 (using a calculator) 16 3 Answer: 3 1 16 4. 4x 5. 16 t 64 6. 9x 7. 6 x 12 4 1 27 Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1320 Using natural logarithms: 6 x 12 ln 6 x ln 12 x ln 6 ln 12 x NOTE: x ln 12 ln 6 ln 12 1.38685 and 6 1.38685 11.99994 ln 6 x Using logarithms base 6: 6 x 12 log 6 6 log 6 12 x log 6 6 log 6 12 x log 6 12 NOTE: log 6 6 1 Since your calculator does not have logarithm base 6 key, you would have to do a change of bases to obtain an approximation for log 6 12 . Since your calculator has a natural logarithm key LN , then we obtain that log 6 12 = log a u ln 12 log u b using the change of base formula that log a b , where ln 6 u 12 , b 6 , and a e. Or, Since your calculator has a common log 12 logarithm key LOG , then we obtain that log 6 12 = using the lo g 6 log a u log u u 12 , b 6 , and b change of base formula that log a b , where a 10 . Answer: x ln 12 ln 6 2 3 8. 5t 9. 7x 3 4 Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1320 10. 8 5 2 x 65 11. 37x 4 49 Examples Solve the following logarithmic equations. Give exact answers. No decimal approximations. 1. log x 16 2 y Using the definition of logarithm ( y log b x if and only if b x ), we will write the logarithmic equation as an exponential equation: log x 16 2 x 2 16 Using square roots to solve the equation x 2 16 , we have that x 2 16 x2 16 x 4 x 4 Since the base of a logarithm can not be negative, then the solution of x 4 can not be used. Thus, the only solution of the equation log x 16 2 is x 4 . Answer: x 4 2. log x 5 3 y Using the definition of logarithm ( y log b x if and only if b x ), we will write the logarithmic equation as an exponential equation: log x 5 3 x 3 5 Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1320 Using cube roots to solve the equation x 3 5 , we have that x3 5 Answer: x 3. 3 3 x3 3 5 x 3 5 5 log 3 x 2 y Using the definition of logarithm ( y log b x if and only if b x ), we will write the logarithmic equation as an exponential equation: log 3 x 2 x 3 2 x 1 9 1 makes the argument of the logarithm 9 positive since the logarithm of a negative number or zero is undefined. The 1 argument of log 3 x is x. Thus, x will be positive when x . Thus, 9 1 is a solution of the equation log 3 x 2 . Of course, it is the only 9 solution. We need to check that the number Answer: x 4. 1 9 log ( 4 t 9 ) 2 Recall that log is the notation for the common logarithm, and the base of the common logarithm is 10. Using the definition of logarithm ( y log b x if y and only if b x ), we will write the logarithmic equation as an exponential equation: Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1320 log ( 4 t 9 ) 2 4 t 9 10 2 4 t 9 100 Solving the equation 4 t 9 100 , we have that t 109 . 4 109 We need to check that the number makes the argument of the 4 logarithm positive since the logarithm of a negative number or zero is undefined. The argument of log ( 4 t 9 ) is 4 t 9 . When t 109 109 9 = 109 9 0 . , we have that 4 t 9 = 4 4 4 109 is a solution of the equation log ( 4 t 9 ) 2 . Of course, it is 4 the only solution. Thus, Answer: t 5. 109 4 log 2 x log 2 ( x 12 ) 6 First, we’ll use the property of logarithms that log b u v = log b u + log b v in order to write log 2 x log 2 ( x 12 ) as log 2 x ( x 12 ) . Thus, log 2 x log 2 ( x 12 ) 6 log 2 x ( x 12 ) 6 y Now, using the definition of logarithm ( y log b x if and only if b x ), we will write the logarithmic equation log 2 x ( x 12 ) 6 as an exponential equation: log 2 x ( x 12 ) 6 x ( x 12 ) 2 6 x ( x 12 ) 64 Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1320 Solving the equation x ( x 12 ) 64 , we have that x ( x 12 ) 64 x 2 12 x 64 x 2 12 x 64 0 ( x 4 ) ( x 16 ) 0 x 4 , x 16 We need to check that the numbers 4 and 16 make the argument of the logarithms positive since the logarithm of a negative number or zero is undefined. The argument of log 2 x is x and the argument of log 2 ( x 12 ) is x 12 . When x 4 , we have that x 4 0 . Thus, when x 4 , we have that log 2 x log 2 ( 4 ) . However, log 2 ( 4 ) is undefined. Thus, 4 is a solution of the equation x ( x 12 ) 64 , but it is not a solution of the equation log 2 x log 2 ( x 12 ) 6 . When x 16 , we have that x 16 0 and x 12 16 12 0 . Thus, 16 is a solution of the equation log 2 x log 2 ( x 12 ) 6 . Answer: x 16 6. ln x ln ( 5 x ) In order to solve this equation, we will use the fact that any logarithm function is one-to-one. Thus, log b u log b v if and only if u v . Thus, by the one-to-one property, we have that ln x ln ( 5 x ) x 5 x Solving the equation x 5 x , we have that x Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1320 5 . 2 5 makes the argument of the logarithms 2 positive since the logarithm of a negative number or zero is undefined. The argument of ln x is x and the argument of ln ( 5 x ) is 5 x . We need to check that the number When x 5 , we have that 2 x 5 0 2 and 5 x 5 5 0. 2 5 Thus, is a solution of the equation ln x ln ( 5 x ) . 2 Answer: x 7. 5 2 log 8 t log 8 ( 7 t 11) In order to solve this equation, we will use the fact that any logarithm function is one-to-one. Thus, log b u log b v if and only if u v . Thus, by the one-to-one property, we have that log 8 t log 8 ( 7 t 11 ) t 7 t 11 Solving the equation t 7 t 11 , we have that t 11 . 6 11 makes the argument of the 6 logarithms positive since the logarithm of a negative number or zero is undefined. The argument of log 8 t is t and the argument of log 8 ( 7 t 11 ) is 7 t 11 . We need to check that the number 11 11 11 0 . Thus, when t , we , we have that t 6 6 6 11 11 . However, log 8 is undefined. have that log 8 t log 8 6 6 When t Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1320 11 is a solution of the equation t 7 t 11 , but it is not a 6 solution of the equation log 8 t log 8 ( 7 t 11) . Thus, Answer: No solution Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1320

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# Lesson 11 Exponential and Logarithmic Equations