1. 3.5 Exponential Equations, Logarithmic Equations, and Problem Solving 1 If b > 0 and b  1, and m and n are real numbers, then b n = b m if and only if n = m. Property 1. By using the property below, we are able to solve this equation by matching the bases. Fortunately, we have another property to take care of this type of equation. Property of Logarithms (taking the log of both sides). This is often referred to as “taking the log of both sides”. Also, we will use the common logarithm for the base. We can than perform the calculations using our calculator (i.e., b = 10) If x > 0, y > 0, b > 0, and b  1, then x = y if and only if log b x= log b y. Procedure: To solve equations of the form a x = b 1.Take the log of both sides. (Just write log in front of both sides of the equation.) 2.Use the Power Rule of Logarithms to bring the exponent in front of the logarithm. 3.Solve the equation.

2. 3.5 Exponential Equations, Logarithmic Equations, and Problem Solving 2 1.Take the log of both sides. 2.Bring the exponent in front of the logarithm. 3.Solve the equation by dividing by log 3 on both sides. x = 2.26 to the nearest hundredth Check: The check for every problem will not be shown. It is recommended to do the check for the problems. Your Turn Problem #1

3. 3.5 Exponential Equations, Logarithmic Equations, and Problem Solving 3 a. Use parentheses around the numerator. Then divide by the denominator. The TI-30XIIS will automatically place parentheses after log. The casio fx-300ES will not. (log(15)–log(2))/log(3) using the TI-30XIIS. (log15 – log2)/log3 using Casios or graphing calculators. You should get 1.8340… Calculating Logarithmic Expressions There are different methods for calculating these expressions. Hopefully, you have a calculator where the screen will show the operation you want to perform. A calculator such as the TI- 30xa will not do this. The TI-30XIIS ($15) or the casio fx-300ES will be easier to use. Any graphing calculator would be great, but I don’t like to recommend spending $100. Examples: b. (log(8)–log(5))/(log(3)+log(2)). (log8–log5)/(log3+log2). You should get 0.26231… c. (2log(3)+log(5))/(4log(2) – log(3)). (2log3+log5)/(4log2 – log3). You should get 2.274023…

4. 3.5 Exponential Equations, Logarithmic Equations, and Problem Solving 4 1.Take the log of both sides. 2.Bring the exponent in front of the logarithm. 3.Solve the equation. This equation can be solved by distributing the log5 or dividing by log5 on both sides. x = 5.32 The solution set is {5.32}. Note: This equation could have been solved by dividing by log 5 first. Your Turn Problem #2

5. 3.5 Exponential Equations, Logarithmic Equations, and Problem Solving 5 1.Take the log of both sides. 2.Bring the exponent in front of the logarithm. 3.Solve the equation. This equation can be solved by distributing the log2 or dividing by log2 on both sides. x = 2.98 The solution set is {2.98}. Note: This equation could have been solved by dividing by log2 first. Your Turn Problem #3

6. 3.5 Exponential Equations, Logarithmic Equations, and Problem Solving 6 1.Take the log of both sides. 2. Bring the exponent in front of the logarithms. 3.Solve the equation. This equation can be solved by distributing the log2 and log7. x = 34.34 to the nearest hundredth The solution set is {34.34}. Get the x terms on one side. Factor out the x. Your Turn Problem #4

7. 3.5 Exponential Equations, Logarithmic Equations, and Problem Solving 7 Recall that the natural logarithm has a base of e. Reason: Therefore x = 1. The next example will be of the form e x = b. Therefore, instead of writing log in front of both sides, write ln in front of both sides. 1.Take the ln of both sides. 2. Bring the exponent in front of the logarithm. 3.Solve the equation. Use the fact that lne=1. x = 5.04 The solution set is {5.04}. Your Turn Problem #5

8. 3.5 Exponential Equations, Logarithmic Equations, and Problem Solving 8 Before we can take the log of both sides, a x or e x must be by itself. Therefore, subtract the 7, then divide by 3 before taking the ln of both sides. The solution set is {– 0.61}. Your Turn Problem #6

9. 3.5 Exponential Equations, Logarithmic Equations, and Problem Solving 9 Previously, we used the Product Rule and Quotient Rule of Logarithms to solve equations such as: With the property from this section, our equation solving capabilities have been expanded. If x > 0, y > 0, b > 0, and b  1, then x = y if and only if log b x= log b y. The previous problems done used the property by taking the log of both sides. If x = y then log b x= log b y. Now we will use the property in the opposite direction. If log b x= log b y, then x = y. The property is true in either direction because of the words “if and only if.” Next Slide

10. 3.5 Exponential Equations, Logarithmic Equations, and Problem Solving 10 If log x = log y, then x = y. Then solve the equation. Recall: The domain of a logarithmic function must contain only positive numbers. Therefore, we need to check if x+2 and 3x – 12 are positive when x is replaced by 7. The solution set is {7}. In this example, the equation contains logarithms of positive numbers, therefore x = 7 is the solution. If the equation contained a logarithm of a negative number, x = 7 would not be a solution. Your Turn Problem #7

11. 3.5 Exponential Equations, Logarithmic Equations, and Problem Solving 11 First use the Product Rule of Logarithms to create a single logarithm on each side. Then use the property, if log x = log y, then x = y. The only solution is 25, and the solution set is {25}. In this example, x = – 4 is not a solution because it leads to the logarithm of a negative number in the original equation. Note: Fortunately the quadratic equation was factorable. If it were not factorable, we would use the quadratic formula to solve. Your Turn Problem #8

12. 3.5 Exponential Equations, Logarithmic Equations, and Problem Solving 12 First use the Quotient Rule of Logarithms to create a single logarithm on each side. Then use the property, if log x = log y, then x = y. The solution set is {4/3}. In this example, x = 4/3 a solution because it does not lead to the logarithm of a negative number in the original equation. Your Turn Problem #9

13. 3.5 Exponential Equations, Logarithmic Equations, and Problem Solving 13 The solution set is {20}. In this example, x = – 5 is not a solution because it leads to the logarithm of a negative number in the original equation. First use the Product Rule of Logarithms to create a single logarithm on each side. Then solve using the definition of a logarithm to change from logarithmic form to exponential form. Remember the base is 10 if it is not written. Your Turn Problem #10

14. 3.5 Exponential Equations, Logarithmic Equations, and Problem Solving 14 Previously, we used the definition of the logarithm to evaluate a logarithm when then base is a positive number other than 10. Example:Evaluate log 2 16 We evaluated this by setting the log 2 16 equal to x and then using the definition of a logarithm to change the equation from logarithmic form to exponential form. Then the equation can be solved by matching the bases. log 2 16 = x 2 x =16 2 x =2 4 x = 4 Thus, we have log 2 16 = 4 Next Slide

15. 3.5 Exponential Equations, Logarithmic Equations, and Problem Solving 15 Set the logarithm equal to x and then change the equation from logarithmic form to exponential form. With the new property given in this section, we now have the capability to solve equations where the bases can not be matched. Example 11. Evaluate log 3 15. (Round to 3 decimal places) Solution: log 3 15 = x 3 x =15 Now take the log of both sides. log3 x = log15 By taking the log of both sides, we can now make use of power rule of logarithms; bring the exponent in front of the logarithm. xlog3 = log15 Then divide by log3 on both sides and calculate. Thus, log 3 15 = 2.465 Your Turn Problem #11 Evaluate log 5 12. (Round to 3 decimal places)

16. 3.5 Exponential Equations, Logarithmic Equations, and Problem Solving 16 Notice the pattern in Example 11 and Your Turn Problem 11. and This leads us the the change-of-base formula for logarithms. (It is just a shortcut to the previous examples.) Change-of-Base Rule If a > 0, b > 0, r > 0, with a  1 and b  1, then Note: any positive number other than 1 can be used for the base ‘b’ in the change-of-base rule, but usually the only practical bases are ‘e’ and 10 since calculators give logarithms only for these two bases. Next Slide

17. 3.5 Exponential Equations, Logarithmic Equations, and Problem Solving 17 Use the change-of-base formula to evaluate Also, use the common logarithm for the base so the calculation can be performed on the calculator. Example 12. Evaluate log 5 21. (Round to 3 decimal places) Solution: Thus, log 5 21 = 1.892 rounded to three decimal places. Your Turn Problem #12 Evaluate log 4 7.4 (Round to 3 decimal places)

18. 3.5 Exponential Equations, Logarithmic Equations, and Problem Solving 18 A = accumulated amount P = principle invested r = interest rate n = number of times compounded per year t = years Next Slide where ‘A’ is the amount of money accumulated at the end of ‘t’ years if ‘P’ dollars is invested at rate of interest ‘r’ compounded ‘n’ times per year. Previously, we covered Applications of Exponential Functions. We will now revisit application problems with expanded capabilities using the properties and rules of this section. Solving Application Problems

19. 3.5 Exponential Equations, Logarithmic Equations, and Problem Solving 19 Example 13. How long will it take $2000 to be worth $2800 if it is invested at 8% compounded quarterly? Solution: 1.Write down the formula and the given information. 2.Now, substitute the values into the given equation. A = $2800 P = $2000 r =.08 n = 4 t = ? 3.Simplify inside the parentheses, then divide by 2000 on both sides. 4.Take the log of both sides to solve. 5.Solve for t by dividing by 4log(1.02) on both sides. t = 4.2 years to the nearest tenth Your Turn Problem #13 How long will it take $2000 to be worth $4500 if it is invested at 12% compounded quarterly? (Round to the nearest tenth) Answer: 6.9 years

20. 3.5 Exponential Equations, Logarithmic Equations, and Problem Solving 20 Example 14. The number of grams of a certain radioactive substance present after t hours is given by the equation A = A 0 e -0.025t, where A 0 represents the initial number of grams. How long would it take 5000 grams tobe reduced to 2500 grams? (i.e., What is the half-life?) Solution: 1.Write down the formula and the given information. 2.Now, substitute the values into the given equation. A 0 = 5000 A = 2500 t = ? 3.Divide by 5000 on both sides. 5.Solve for t by dividing by -0.025 on both sides. t = 27.7 hours to the nearest tenth 4.Take the ln of both sides to solve. Your Turn Problem #14 For a certain strain of bacteria, the number of bacteria present after t hours is given by the equation A = A 0 e 0.45t, where A 0 represents the initial number of bacteria. How long would it take 600 bacteria to increase to 3000 bacteria? Answer: 3.6 years The End B.R. 1-14-07