If b > 0 and b  1, and m and n are real numbers, then b n = b m if and only if n = m.

1. By using the property below, we are able to solve this equation by matching the bases. Property 1. If b > 0 and b 1, and m and n are real numbers, then bn = bm if and only if n = m. Fortunately, we have another property to take care of this type of equation. Property of Logarithms (taking the log of both sides) . If x > 0, y > 0, b > 0, and b 1, then x = y if and only if logbx= logby. 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) Procedure: To solve equations of the form ax = b • Take the log of both sides. (Just write log in front of both sides of the equation.) • Use the Power Rule of Logarithms to bring the exponent in front of the logarithm. 3. Solve the equation.

2. Check: The check for every problem will not be shown. It is recommended to do the check for the problems. Your Turn Problem #1 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

3. 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: 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… 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. 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 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.

5. 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 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.

6. Get the x terms on one side. Factor out the x. x = 34.34 to the nearest hundredth The solution set is {34.34}. Your Turn Problem #4 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.

7. Recall that the natural logarithm has a base of e. Reason: Therefore x = 1. The next example will be of the form ex = b. Therefore, instead of writing log in front of both sides, write ln in front of both sides. The solution set is {5.04}. x = 5.04 Your Turn Problem #5 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.

8. The solution set is {– 0.61}. Your Turn Problem #6 Before we can take the log of both sides, ax or ex must be by itself. Therefore, subtract the 7, then divide by 3 before taking the ln of both sides.

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 logbx= logby. The previous problems done used the property by taking the log of both sides. Now we will use the property in the opposite direction. If logbx= logby, then x = y. logbx= logby. If x = y then The property is true in either direction because of the words “if and only if.” Next Slide

10. 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. The solution set is {7}. Your Turn Problem #7 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.

11. In this example, x = – 4 is not a solution because it leads to the logarithm of a negative number in the original equation. The only solution is 25, and the solution set is {25}. Note: Fortunately the quadratic equation was factorable. If it were not factorable, we would use the quadratic formula to solve. Your Turn Problem #8 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.

12. In this example, x = 4/3 a solution because it does not lead to the logarithm of a negative number in the original equation. The solution set is {4/3}. Your Turn Problem #9 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.

13. In this example, x = – 5 is not a solution because it leads to the logarithm of a negative number in the original equation. The solution set is {20}. Your Turn Problem #10 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.

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 log216 We evaluated this by setting the log216 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. x = 4 Thus, we have log216 = 4 Next Slide log216 = x 2x=16 2x=24

15. Your Turn Problem #11 Evaluate log512. (Round to 3 decimal places) 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 log315. (Round to 3 decimal places) Solution: Set the logarithm equal to x and then change the equation from logarithmic form to exponential form. log315 = x 3x=15 Now take the log of both sides. log3x = 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 Thus, log315 = 2.465 Then divide by log3 on both sides and calculate.

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. Example 12. Evaluate log521. (Round to 3 decimal places) Solution: Your Turn Problem #12 Evaluate log47.4 (Round to 3 decimal places) 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. Thus, log521 = 1.892 rounded to three decimal places.

18. Solving Application Problems Previously, we covered Applications of Exponential Functions. We will now revisit application problems with expanded capabilities using the properties and rules of this section. 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. A = accumulated amount P = principle invested r = interest rate n = number of times compounded per year t = years Next Slide

19. A = $2800 P = $2000 r = .08 n = 4 t = ? 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) Example 13. How long will it take $2000 to be worth $2800 if it is invested at 8% compounded quarterly? 1. Write down the formula and the given information. Solution: • Now, substitute the values • into the given equation. 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 Answer: 6.9 years

20. A0 = 5000 A = 2500 t = ? Your Turn Problem #14 For a certain strain of bacteria, the number of bacteria present after t hours is given by the equation A = A0e0.45t, where A0 represents the initial number of bacteria. How long would it take 600 bacteria to increase to 3000 bacteria? Example 14. The number of grams of a certain radioactive substance present after t hours is given by the equation A = A0e-0.025t, where A0 represents the initial number of grams. How long would it take 5000 grams to be reduced to 2500 grams? (i.e., What is the half-life?) 1. Write down the formula and the given information. Solution: • Now, substitute the values • into the given equation. 3. Divide by 5000 on both sides. 4. Take the ln of both sides to solve. • Solve for t by dividing by -0.025 on both sides. t = 27.7 hours to the nearest tenth The End B.R. 1-14-07 Answer: 3.6 years