5.6 Laws of Logarithms 5.7 Exponential Equations; Changing Base

1. 5.6 Laws of Logarithms5.7 Exponential Equations; Changing Base Objectives: 1. Compare & recall the properties of exponents 2. Deduce the properties of logarithms from/by comparing the properties of exponents • Use the properties of logarithms • Solve the exponential equations

2. Since the logarithmic function y = logb x is the inverse of the exponential function y = bx, the laws of logarithms are very closely related to the laws of exponent. Pre-Knowledge For any b, c, u, v +, and b≠ 1, c ≠ 1, there exists some x, y , such that u = bx, v = by By the previous section knowledge, as long as taking x = logbu, y = logbv

3. 1. Product of Power am an = am+n Pre-Knowledge For any b, c, u, v +, and b≠ 1, c ≠ 1, there exists some x, y , such that u = bx, v = by By the previous section knowledge, as long as taking x = logbu, y = logbv 1. Product Property logbuv = logbu + logbv Proof logbuv = logb(bxby)= logbb x+y = x + y = logbu + logbv

4. 2. Quotient of Power Pre-Knowledge For any b, c, u, v +, and b≠ 1, c ≠ 1, there exists some x, y , such that u = bx, v = by By the previous section knowledge, as long as taking x = logbu, y = logbv 2. Quotient Property Proof

5. Equal Power • am = an iff m = n • Equal Property • logbu = logbv iff u = v Proof logbu – logbv = 0

6. 4. Power of Power (am)n = amn Pre-Knowledge For any b, c, u, v +, and b≠ 1, c ≠ 1, there exists some x, y , such that u = bx, v = by By the previous section knowledge, as long as taking x = logbu, y = logbv 4. Power Property logbuk = k logbu Proof logbuk = logb(bx)k= logbb kx = kx = k logbu

7. 5. Change-of-Base Formula Proof Note that bx = u, logbu = x Taking the logarithm with base c at both sides: logcbx = logcu or x logcb = logcu

8. 6. Reciprocal Formula Proof

9. 7. Raise Power Formula Proof

10. Example 1 Assume that log95 = a, log911 = b, evaluate • log9 (5/11) • log955 • log9125 • log9(121/45) • log9275

11. Example 2 Expanding the expression • ln(3y4/x3) ln(3y4/x3) = ln(3y4)– lnx3 = ln3 + lny4 – lnx3 = ln3 + 4 ln|y| – 3 lnx b) log3125/6x9 log3125/6x9 = log3125/6 + log3x9 = 5/6 log312 + 9 log3x = 5/6 log3(3· 22) + 9 log3x = 5/6 (log33 + log322) + 9 log3x = 5/6 ( 1 + 2 log32) + 9 log3x

12. Example 3 Condensing the expression a) 3 ( ln3 – lnx ) + ( lnx – ln9 ) 3 ( ln3 – lnx ) + ( lnx – ln9 ) = 3 ln3 – 3 lnx + lnx – 2 ln3 = ln3 – 2 lnx = ln(3/x2) b) 2 log37 – 5 log3 x + 6 log9 y2 2 log37 – 5 log3 x + 6 log9 y2 = log349 – log3 x5 + 6 ( log3 y2/ log39) = log3(49/x5) + 3 log3 y2 = log3(49y6/x5)

13. Practice A) P. 199 Q 7 – 18 • P. 199 Q 19 – 20 How do you change to make it to be true? True or Falselog ay = – log 1/ay c) P. 200 Q 7 – 27 (odd)

14. Example 4 Calculate log48 and log615 using common and natural logarithms. a) log48 log48 = log8 / log4 = 3 log2 / (2 log2) = 3/2 log48 = ln8 / ln4 = 3 ln2 / (2 ln2) = 3/2 b) log615 = log15 / log6 = 1.511

15. in terms of Example 5. Express

16. More on Expand/Condense logarithmic expressions Example 6 Expand

17. Example 7 Expand in terms of sums and differences of logarithms More on Expand/Condense logarithmic expressions

18. More on Expand/Condense logarithmic expressions Example 8 Expand to express all powers as factors

19. Example 9 Condense to a single logarithm. More on Expand/Condense logarithmic expressions

20. Assignment: 5.6 P. 196 #36 – 44 (even) P. 200 #2 – 22 (even), 21 – 33 (odd), 41, 43, 45

21. Solving Exponential Equations • One way to solve exponential equations is to use the property that if 2 powers with the same base are equal, then their exponents are equal. • For b > 0 and b≠1 if bx = by, then x = y

22. Solve by Equating Exponents Example 10: Solve 43x = 8x+1 (22)3x = (23)x+1 rewrite with same base 26x = 23x+3 6x = 3x + 3 x = 1 Check → 43*1 = 81+1 64 = 64

23. Your turn! Solve: 16x = 32x–1 24x = 32x–1 24x = (25)x–1 4x = 5x – 5 x = 5 Be sure to check your answer!!!

24. When you can’t rewrite using the same base, you can solve by taking a log of both sides or use the definition of log Example 11: Solve 2x = 7 log22x = log27 x = log27 x = ≈ 2.807

25. Example 12: Solve 102x – 3 + 4 = 21 When you can’t rewrite using the same base, you can solve by taking a log of both sides or use the definition of log 102x – 3 = 17 log10102x – 3 = log1017 2x – 3 = log 17 2x = 3 + log17 x = ½(3 + log17) ≈ 2.115

26. Your turn! Solve: 5x + 2 + 3 = 25 5x+2 = 22 log55x+2 = log522 x + 2 = log522 x = (log522) – 2 = (log22/log5) – 2 ≈ –0.079

27. More on Solving Exponential Equations Example 13: Solve ex – 3e-x = 2 [Answer] Multiply ex at both sides of the equation: ex(ex – 3e-x ) = 2ex e2x–3 = 2ex e2x–2ex –3 = 0 (ex)2–2(ex)–3 = 0 Denote ex= u, then (u)2–2(u)–3 = 0 (u)2–2(u)–3 = 0 (u – 3)(u+ 1) = 0 u = 3, or u= –1 ex= 3, or ex = –1 (discard) x = ln3

28. Solving Logarithmic Equations 1. To solve use the property for logs with the same base: • b, x, y+ and b  1 • If logb x = logb y, then x = y 2. When you can’t rewrite both sides as logs with the same base exponentiate each side • b, x+ and b  1 • if logb x = y, then x = by • This can get the expression in the log out of the log simply.

29. Solving Logarithmic Equations Example 14: Newton’s Law of Cooling • The temperature T of a cooling substance at time t (in minutes) is: • T = (T0 – TR) e-rt + TR • T0= initial temperature • TR= room temperature • r = constant cooling rate of the substance

30. Solving Logarithmic Equations Example 14: You’re cooking stew. When you take it off the stove the temp. is 212°F. The room temp. is 70°F and the cooling rate of the stew is r = 0.046. How long will it take to cool the stew to a serving temp. of 100°? T = (T0 – TR) e -rt + TR T0 = 212, TR = 70, T = 100 r = 0.046 So solve: 100 = (212 – 70)e -0.046t + 70

31. Solving Logarithmic Equations 30 = 142e -0.046t(subtract 70) 15/71 = e -0.046t(divide by 142) • How do you get the variable out of the exponent? ln(15/71) = lne-.046t (take the ln of both sides) ln(15/71) = – 0.046t ln(15/71)/(– 0.046) = t t =(ln15 – ln71)/(– 0.046) = t ≈ – 1.556/(– 0.046) t ≈ 33.8 about 34 minutes to cool!

32. Example 15: Solve log3(5x – 1) = log3(x + 7) 5x – 1 = x + 7 4x = 8 x = 2 and check log3(52 – 1) = log3(2 + 7) log39 = log39 Because the domain of log functions doesn’t include all reals, you should check for extraneous solutions. Solving Logarithmic Equations

33. Example 16: Solve log5(3x + 1) = –2 3x + 1 = 5-2 3x + 1 = 1/25 x = –8/25 and check Because the domain of log functions doesn’t include all reals, you should check for extraneous solutions. Solving Logarithmic Equations

34. More on Solving Logarithmic Equations Example 16: Solve log5x + log(x + 1) = 2 log[5x(x + 1)]= 2 (product property) log (5x2 + 5x) = 2 5x2 + 5x = 100 x2 + x – 20 = 0 (subtract 100 and divide by 5) (x + 5)(x – 4) = 0 x = – 5, or x = 4 check and you’ll see x = 4 is the only solution.

35. Your Turn! Solve log2x + log2(x – 7) = 3 log2 [x(x – 7)]= 3 log2 (x2 – 7x) = 3 x2 – 7x = 23 x2 – 7x = 8 x2 – 7x – 8 = 0 (x – 8)(x + 1) = 0 x = 8 x = –1 Check log28 + log2(8 – 7) =3 3 + 0 = 3

36. One More!Solve log6(x – 2) + log6(x + 3) = 2 log6(x – 2) + log6(x + 3) = 2 log6 [(x – 2)(x + 3)] = 2 log6 (x2 + x – 6) = 2 x2 + x – 6 = 36 x2 + x – 42 = 0 (x – 6)(x + 7)=0 x = 6 x = –7 Check log64 + log69 =2 log636 = 2 Check log64 + log69 =2 log636 = 2

37. Challenge!Example 17: Solve log2x + log4(x2 – 4x + 4) = 3 log4 x2 + log4(x – 2)2 = 3 (Raise Power Formula) log4 [x2(x – 2)2] = 3 x2(x – 2)2 = 43 [x(x – 2)]2 = 64 A2B2 = (AB)2 x(x – 2) = ±8 A2 = 64, A = ±8 x2 – 2x – 8 = 0 or x2 – 2x + 8 =0 x = 4 x = –2 No real solution

38. Challenge!Solve log2x + ½ log2(x – 1) = 1 2log2x + log2(x – 1) = 2 log2x2 + log2(x - 1) = 2 log2 [x2(x – 1)] = 2 x2(x – 1) = 4 x3– x2 – 4 = 0 (Rational Zero Theorem) (x – 2)(x2 + x + 2) = 0 x – 2 = 0 or x2 + x + 2 = 0 x = 2 No real solution Check log22 + ½ log21 =1 2 + 0 = 2

39. Challenge Simplify (No calculator) 1) 2) 3) 4) 5) Proof

40. Assignment: 5.7 P. 201 #24 – 34 (even), 42, 44 P. 179 #49 – 52