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5. Exponential and Logarithmic Functions. EXPONENTIAL AND LOGARITHMIC FUNCTIONS. 5.5 Exponential and Logarithmic Equations: Problem Solving. Objectives Solve logarithmic equations. Use logarithms to solve application problems involving exponential growth and decay.

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## Exponential and Logarithmic Functions

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**5**Exponential and Logarithmic Functions**EXPONENTIAL AND LOGARITHMIC FUNCTIONS**5.5 Exponential and Logarithmic Equations: Problem Solving Objectives • Solve logarithmic equations. • Use logarithms to solve application problems involving exponential growth and decay. • Solve application problems involving Richter numbers. • Use a change-of-base formula to evaluate a logarithm.**Exponential and Logarithmic Equations:Problem Solving**Property 5.8 If x > 0, y > 0, b > 0, and b 1, then x = y if and only if logb x = logb y**Exponential and Logarithmic Equations:Problem Solving**Example 1 Solve 3x= 5 to the nearest hundredth.**Exponential and Logarithmic Equations:Problem Solving**Example 1 Solution: By using common logarithms, we can proceed as follows: 3x = 5 log 3x = log 5 x log 3 = log 5 x = 1.46 nearest hundredth Property 5.8 log r p = p log r**Exponential and Logarithmic Equations:Problem Solving**Example 1 Check: Because 31.46 4.972754647, we say that, to the nearest hundredth, the solution set for 3x= 5 is {1.46}.**Logarithmic Equations**Example 4 Solve log x + log(x - 15) = 2.**Logarithmic Equations**Example 4 Solution: Because log 100 = 2, the given equation becomes log x + log(x - 15) = log 100 Now we can simplify the left side, apply Property 5.8, and proceed as follows: log [(x)(x – 15)] = log 100 x(x – 15) = 100 x2 – 15x – 100 = 0 (x – 20)(x + 5) = 0 x – 20 = 0 or x + 5 = 0 x = 20 or x = -5 The domain of a logarithmic function must contain only positive numbers, so x and x - 15 must be positive in this problem. Therefore we discard the solution of -5, and the solution set is {20}.**Problem Solving**Problem 1 How long will it take $500 to double if it is invested at 8% compounded quarterly?**Problem Solving**Problem 1 Solution: To double $500 means that the $500 will grow to $1000. We want to find out how long it will take; that is, what is t? Thus = 500(1 + 0.02)4t = 500(1.02)4t We multiply both sides of 1000 = 500(1.02)4tby to get 2 = (1.02)4t Therefore**Problem Solving**Problem 1 log 2 = (1.02)4t = 4t log 1.02 Now let’s solve for t: 4t log 1.02 = log 2 t = 8.8 nearest tenth Therefore we are claiming that $500 invested at 8% interest compounded quarterly will double in approximately 8.8 years. Property 5.8 log r p = p log r**Problem Solving**Problem 1 Check: $500 invested at 8% compounded quarterly for 8.8 years will produce = $500(1.02)35.2 = $1003.91**Richter Numbers**Problem 3 An earthquake that occurred in San Francisco in 1989 was reported to have a Richter number of 6.9. How did its intensity compare to the reference intensity?**Richter Numbers**Problem 3 Solution: I = (106.9)(I0) I 7,943,282I0 Thus its intensity was a little less than 8 million times the reference intensity.**Logarithms with Base Other Than 10 or e**Example 7 Evaluate log3 41.**Logarithms with Base Other Than 10 or e**Example 7 Solution: Let x = log3 41. Changing to exponential form, we obtain 3x= 41 Now we can apply Property 5.8. log 3x = log 41 x log 3 = log 41 x = 3.3802 rounded to four decimal places Therefore we are claiming that 3 raised to the 3.3802 power will produce approximately 41. Check it!**Logarithms with Base Other Than 10 or e**• Using the method of Example 7 to evaluate loga r produces the following formula, which is often referred to as the change-of-base formula for logarithms.**Exponential and Logarithmic Equations:Problem Solving**Property 5.9 If a, b, and r are positive numbers, with a 1 and b 1, then

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