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Logarithmic Functions By Dr. Julia Arnold and Ms. Karen Overman

Review of Logs. Logarithmic Functions By Dr. Julia Arnold and Ms. Karen Overman using Tan’s 5th edition Applied Calculus for the managerial , life, and social sciences text. Concept 1 The logarithmic function is the inverse of the exponential function.

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Logarithmic Functions By Dr. Julia Arnold and Ms. Karen Overman

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  1. Review of Logs Logarithmic Functions By Dr. Julia Arnold and Ms. Karen Overman using Tan’s 5th edition Applied Calculus for the managerial , life, and social sciences text

  2. Concept 1 The logarithmic function is the inverse of the exponential function.

  3. Recall that an Inverse Function reverses the ordered pairs which belong to the original function. i.e. (x,y) becomes (y,x) The log function is the inverse of the exponential function. If the exponential is 2x, then its inverse is log2 (x) (read log x base 2 ) If the exponential is 3x... its inverse is log3(x) (read log x base 3) If the exponential is 10x... its inverse is logx (read log x) Note:Base 10 is considered the common base and thus log x is the common log and as such the base is omitted and understood to be 10.

  4. If the exponential is y=2x, then its inverse is y=log2 (x). (read y = log x base 2 ) Are you getting the hang of it?? If the exponential is y=3x ... its inverse is y=log3(x) If the exponential is y=10x... its inverse is y=logx Note: Base 10 is considered the common base and thus log x is the common log and as such the base is omitted. If the exponential is ex... its inverse is lnx (read l n x or natural log x) Note: Base e is considered the natural base and thus ln x is the natural log and is written ln to distinguish it from log.

  5. Concept 2 How to graph the logarithmic function.

  6. x 2x 0 1 1 2 2 4 -1 1/2 -2 1/4 f(x)=2x Remember this graph from the last section on exponential functions? We used a table of values to graph the exponential y = 2x. Since we know that the y = log2(x) is the inverse of the function above, we can just switch the ordered pairs in the table above and create the log graph for base 2.

  7. x 2x 0 1 1 2 2 4 -1 1/2 -2 1/4 y=2x x log2(x) y = log2(x) is the INVERSE of the function y = 2x. To create the graph we can just switch the ordered pairs in the left table and create the log graph for base 2. 1 0 This is the easy way to do a log graph. 2 1 Switch 4 2 1/2 -1 1/4 -2 y = log2(x)

  8. f(x)=2x Let’s look at the two graphs: The exponential function, f(x)=2x goes through (0, 1), since 1 = 20. The domain is all real numbers. The range is all positive real numbers. The logarithmic function, f(x)=log2(x) goes through (1, 0) which means log2(1) = 0. The domain is all positive real numbers. The range is all real numbers. f(x)=log2(x)

  9. Concept 3 Converting between exponential form and logarithmic form.

  10. Exponential Form Logarithmic Form x = ay becomes y = loga(x) In both forms, a is called the base. In logarithmic form, y equals the log or the log equals the exponent. In exponential form, y is the exponent. log2(32)=5 25 = 32 becomes Read: log 32 base 2 =‘s 5

  11. Let’s practice converting from exponential form to logarithmic form. log2(32) = 5 We saw that, 25 = 32 becomes So then, 34 = 81 becomes log3(81) = 4 log2(1/8) = -3 2-3 = 1/8 becomes 5-2 = 1/25 becomes log5(1/25) = -2 20 = 1 becomes log2(1) = 0 103 = 1000 becomes log(1000) = 3 e1 = e becomes ln (e) = 1

  12. Logarithmic Form Exponential Form y = loga(x) becomes x = ay Recall, in both forms, a is called the base. In logarithmic form, the log equals the exponent, thus y is the exponent in exponential form. For example, log2(32)=5 25 = 32 becomes

  13. Let’s practice converting from logarithmic form to exponential form. We saw that, log2(32) = 5 becomes 25 = 32 So then,log3(81) = 4 becomes 34 = 81 log2(1/8) = -3becomes 2-3 = 1/8 log5(1/25) = -2becomes 5-2 = 1/25 20 = 1 log2(1) = 0becomes log(1000) = 3becomes 103 = 1000 ln (e) = 1becomes e1 = e

  14. Concept 4 Properties for Logs.

  15. There are a few truths about logs which we will call properties: 1. loga(1) = 0 for any a > 0 and not equal to 1 because a0=1 (exponential form of log form) 2. loga(a) = 1 for any a > 0 and not equal to 1 because a1=a (exponential form of log form) 3. loga(ax) = x for any a > 0 and not equal to 1 because ax=ax (exponential form of log form) 4. If loga x = loga y , then x = y.

  16. Practice Problems 1. Solve for x: log3x = log3 4 X = 34 X = 4 Click on the green arrow of the correct answer above.

  17. No, x = 34 is not correct. Use the 4th property: 4. If loga x = loga y , then x = y. log3x = log3 4, then x = 4 Go back.

  18. Way to go! • Using the 4th property: • If loga x = loga y , then x = y. • You concluded correctly that: • If log3x = log3 4, then x = 4.

  19. Practice Problems 2. Solve for x: log21/8 = x X = -3 X = 3 Click on the green arrow of the correct answer above.

  20. No, x = 3 is not correct. Use the 3rd property and the rules for exponents: 3. loga(ax) = x log21/8 = log2 8-1 = log2 (23)-1 = log2 2-3 then x = -3 since the 2’s make a match. Go back.

  21. Way to go! Using the 3rd property: 3. loga(ax) = x log21/8 = log2 8-1 = log2 (23)-1 = log2 2-3 then x = -3 since the 2’s make a match.

  22. Practice Problems 3. Evaluate: ln 1 + log 10 - log2(24) -2 -3 Click on the green arrow of the correct answer above.

  23. No, -2 is not correct. Using properties 1,2 and 3: ln 1 = 0 log10 = 1 -log2 24 = -4 which totals to -3 Go back.

  24. Way to go! Using properties 1,2 and 3: ln 1 = 0 log10 = 1 -log2 24 = -4 which totals to -3

  25. Concept 5 Expansion properties of logs.

  26. Expansion properties of logs 1. loga(uv) = logau + logav 2. 3.

  27. The first expansion property is proven below. 1. loga(uv) = logau + logav Proof: Set logau =x and logav =y then change to exponential form. ax = u and ay = v. ax+y =ax ay = uv so, write ax+y = uv in log form loga(uv) = x + y but that’s logau =x and logav =y , so write loga(uv) = logau + logav which is the result we were looking for. Do you see how this property relates to the exponential property?

  28. Example 1: Expand to single log expressions. Applying property 1 Log 10 = 1 from the 2nd property which we had earlier.

  29. Example 2: Expand the logarithm completely. Applying property 2 Applying property 1 and from before ln e = 1

  30. Example 3 : Expand the logarithm completely. First change the radical to an exponent. Wait, we’re not done yet! Next, apply property 3.

  31. Example 3 continued: After applying property 3. Next, apply property 2 for quotients

  32. Example 3 continued: After applying property 2 for quotients. Next, apply property 3to the exponentials. Now they are single logs.

  33. Your turn: Expand the logarithm completely: The first step is to use property 1 The first step is to use property 3

  34. No, incorrect, return to the previous slide.

  35. The first step is to use property 1 which will expand to: Now we use property 3 This is the final answer.

  36. No, this is not the final answer. Return to the previous slide and click on the correct answer.

  37. Yes, we now use property 3 to expand further to: This is the final answer. This is still not the final answer.

  38. Nope, we are not done yet. Return

  39. Whenever the base of the log matches the number you are taking the log of, the answer is the exponent on the number which is 1 in this case.

  40. is the property. Or from the beginning of the problem you could have said:

  41. We can use the same three expansion properties of logs to take an expanded log and condense it back to a single log expression. 1. logau + logav =loga(uv) 2. 3.

  42. Here’s an example: Condense this expression into a single log. Always begin by reversing property 3: Next use property 2: (Now we have a single log.)

  43. Another example: Condense the expression into a single log. First, reverse property 3. Next, use property 2. (Now we have a single log.)

  44. Concept 6 Finding logs on your calculator for any base number.

  45. On your graphing calculator or scientific calculator, you may find the value of the log (of a number) to the base 10 using the log key or the ln(of a number) to the base e by using the ln key. What if you want to find the value of a log to a different base? For example, how can we find the value of log25 ?

  46. How can we find the value of log25 ? 1. Set log25 = x 2. Change to exponential form 2x = 5 3. Take the log (base 10 ) of both sides. log 2x = log 5 • x log 2 = log 5 using the 3rd expansion property • thus x = This shows us how we can create the change of base formula: Changes the base to c.

  47. The change of base formula: We are given base a, and we change to base c.

  48. Example 1: Find the following value using a calculator: Since the calculator is built to find base 10 or base e, choose either one and use the change of base formula.

  49. Some problems can be done without a calculator, but not all. 1. Find 2. Find

  50. Here’s a problem that can be done without a calculator. 1. Find Click on the correct slide to advance to the next slide. Click on the wrong side and you will remain here.

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