ࡱ> []Z]|r5@ 8bjbj22 -XX06666666JN!N!N!N!T"LJ<"821212121212121< < < < < < <$>=R?n-<62212122-<662121B<:::2621621<:2<:::66:21" `Y(/N!3:<X<0<:?7"?:JJ6666?6: 21"T1:l11212121-<-<JJ N!: JJN!Log Problems Assume log bases are positive but not 1. 1. (a) log2 1 = ? (b) log 1 = ? (c) loga 1 = ? 2. (a) log 10 = ? (b) log3 3 = ? (c) loga a = ? 3. (a) log5 0 = ? (b) log3 (1/9) = ? (c) log4 -2 = ? 4. Give a counterexample to the statement that:  EMBED Equation.DSMT4 . 5. log2 84 = ? 6.  EMBED Equation.DSMT4 =? 7. log37 log24 log727 = ? 8. log3 (1/81) = ? 9.  EMBED Equation.DSMT4 = ? 10.  EMBED Equation.DSMT4 = ? 11. (True/False)?  EMBED Equation.DSMT4  12. Solve for x: log927 = x 13. Solve for x: log3x = -3 14. Solve for x: logx81 = 4 15. Solve for x: log2x1/5 = 2 16. log4(x-3) + log4x = log428 17. log2(1+1/x) = 3 18.  EMBED Equation.DSMT4  19. logx(x+6) = 2 20. Under what condition(s) is it true that logaM > logaN ? 21.  EMBED Equation.DSMT4  22. Determine whether the following logs are positive or negative. (a) log1214 (b) log1002 (c) log3.1(.8) (d)  EMBED Equation.DSMT4  (e) log0.32.1 (f)  EMBED Equation.DSMT4  23. Find the log of  EMBED Equation.DSMT4  in terms of base a. 24. Evaluate:  EMBED Equation.DSMT4 = ? 25.  EMBED Equation.DSMT4  26.  EMBED Equation.DSMT4  27.  EMBED Equation.DSMT4  28. loga8 = 2, a = ? 29. loga 15 = -1, a = ? 30.  EMBED Equation.DSMT4  31.  EMBED Equation.DSMT4  Log Problems (continued) 32.  EMBED Equation.DSMT4  33.  EMBED Equation.DSMT4  34. log0.1 x = -2, x = ? 35. log121 x = , x = ? 36. log4 x = -0.5, x = ? 37.  EMBED Equation.DSMT4  38.  EMBED Equation.DSMT4  39.  EMBED Equation.DSMT4 = ? 40. Solve for x:  EMBED Equation.DSMT4  41. Solve for x:  EMBED Equation.DSMT4  42. Express as a logarithm to the base 2:  EMBED Equation.DSMT4  43. Evaluate:  EMBED Equation.DSMT4  44. Be able to use a calculator to approximate log23. 45. log2 x + log3 x = 1 46. Simplify:  EMBED Equation.DSMT4  47. Solve for x:  EMBED Equation.DSMT4  48.  EMBED Equation.DSMT4  49.  EMBED Equation.DSMT4  50. Complete the following statement regarding problem #49. If two power expressions are equal with equal exponents, then 51.  EMBED Equation.DSMT4  52. Solve the following system:  EMBED Equation.DSMT4  53. Solve the following system:  EMBED Equation.DSMT4  54. Solve the following system:  EMBED Equation.DSMT4  More Log Problems! 55.  EMBED Equation.DSMT4  56.  EMBED Equation.DSMT4  57.  EMBED Equation.DSMT4  58.  EMBED Equation.DSMT4  59.  EMBED Equation.DSMT4  = 0 60.  EMBED Equation.DSMT4  61.  EMBED Equation.DSMT4  62.  EMBED Equation.DSMT4  Log Answers 1. 0, 0, 0 2. 1, 1, 1 3. undefined, -2, undefined 4. Let x = (or any number such that 0 < x < 1). 5. 12 6. 7. 6 8. -4 9. -7/2 10.  EMBED Equation.DSMT4  11. True 12. 3/2 13. 1/27 14. x = 3 (not -3 since x was a log base) 15. 45 or 210 or 1024 16. 7 17. 1/7 18.  EMBED Equation.DSMT4  19. 3 20. If base, a > 1, then log is increasing, so the statement is true for M > N. If base, 0 < a < 1, the log is decreasing, so we need M < N. 21. -7/3 22. (a) positive (b) positive (c) negative (d) positive (e) negative (f) positive 23.  EMBED Equation.DSMT4  24.  EMBED Equation.DSMT4  25. 12 26. 1/5 27. 3/8 28.  EMBED Equation.DSMT4  29. 1/15 30. 1/144 31. .001 32.  EMBED Equation.DSMT4  33. 16 34. 100 35. 11 36. 37. 729 38. 729 Log Answers (continued) 39. 72 40.  EMBED Equation.DSMT4  41. 2/3 42.  EMBED Equation.DSMT4  43. 44. Use the Change of Base Thm to compute  EMBED Equation.DSMT4  EMBED Equation.DSMT4  *45.  EMBED Equation.DSMT4  [See below for solution process.] 46.  EMBED Equation.DSMT4  *47.  EMBED Equation.DSMT4  [See below for solution process.] *48. 1, 3 [See below for solution process.] *49. 1,2,3 [See below for solution process.] 50. their bases are equal (assuming positive bases). Exception if exponents are zero! *51. 1/3, 9 [See below for solution process.] *52. (2,4) [See below for solution process.] *53. (4,7),(1,4),(-1,2) [See below for solution process.] *54. (2,4) and (4,2) [See below for solution process.] *55.  EMBED Equation.DSMT4  [See below for solution process.] *56. (2,2), (8,1/8) [See below for solution process.] *57. 4 [See below for solution process.] *58. 2 [See below for solution process.] *59. 4 [See below for solution process.] *60. 1,2 [See below for solution process.] *61. (2,3),(1/2,3),(4,3/2),(1/4,3/2) [See below for solution process.] *62. (1/3,1/5) [See below for solution process.] Log Solutions (selected problems) 45. 3 different ways! Method 1 Method 2 Method 3 log2 x + log3 x = 1 or  EMBED Equation.DSMT4  or  EMBED Equation.DSMT4   EMBED Equation.DSMT4  or  EMBED Equation.DSMT4  or  EMBED Equation.DSMT4   EMBED Equation.DSMT4  or  EMBED Equation.DSMT4  or  EMBED Equation.DSMT4   EMBED Equation.DSMT4  or  EMBED Equation.DSMT4  or  EMBED Equation.DSMT4   EMBED Equation.DSMT4  or  EMBED Equation.DSMT4  or  EMBED Equation.DSMT4   EMBED Equation.DSMT4  or  EMBED Equation.DSMT4  or  EMBED Equation.DSMT4  =  EMBED Equation.DSMT4   EMBED Equation.DSMT4  or  EMBED Equation.DSMT4  or  EMBED Equation.DSMT4   EMBED Equation.DSMT4  =  EMBED Equation.DSMT4  =  EMBED Equation.DSMT4  *Suggesting the following theorem? Thm/  EMBED Equation.DSMT4   EMBED Equation.DSMT4  Use Down in front on the right side to get:  EMBED Equation.DSMT4  Then use Collapsing Log Theorem to get:  EMBED Equation.DSMT4 , done! 47.  EMBED Equation.DSMT4  (Exponentiate both sides, base 2)  EMBED Equation.DSMT4  EMBED Equation.DSMT4  48.  EMBED Equation.DSMT4  (This is going to be a quadratic. Let u =  EMBED Equation.DSMT4 .)  EMBED Equation.DSMT4   EMBED Equation.DSMT4  EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4  Log Solutions (continued) 49.  EMBED Equation.DSMT4  Since x > 0, equal exponents means equal bases (bases > 0) with the exception (see problem #50) that if the exponents are zero, the bases can be unequal. (i)  EMBED Equation.DSMT4  (ii) exponent equals zero case/  EMBED Equation.DSMT4  (i) & (ii)  EMBED Equation.DSMT4  51.  EMBED Equation.DSMT4  Here the exponent, 2, is not zero, so the bases must be equal (base > 0).  EMBED Equation.DSMT4  Let u = log3 x (x > 0 and since x is in the base position of a log, x is not 1.)  EMBED Equation.DSMT4  EMBED Equation.DSMT4   EMBED Equation.DSMT4  52.  EMBED Equation.DSMT4  x, y > 0 Well work with equation #1 and then substitute into #2.  EMBED Equation.DSMT4 (Substitute into equation #2.)  EMBED Equation.DSMT4  EMBED Equation.DSMT4  but not -11/3 since y > 0, our only y-solution is 4 EMBED Equation.DSMT4  53.  EMBED Equation.DSMT4  (Looking at equation #1, x cant be zero, but it might be negative.)  EMBED Equation.DSMT4  Now substitute for y in the equation #1  EMBED Equation.DSMT4  (much better!) Consider the following Thm/ au = 1 if (i) u = 0, a  EMBED Equation.DSMT4  (ii) a = 1 (iii) a = -1, u is even. So either (i) 2x 8 = 0 or (ii) x = 1 (iii) x = -1 if 2x-8 is even (Here, its -10.) (i) x = 4 and y = 7 (ii) x = 1 and y = 4 (iii) x = -1 and y = 2  EMBED Equation.DSMT4  54.  EMBED Equation.DSMT4   EMBED Equation.DSMT4  EMBED Equation.DSMT4  so 2log 8 = 3log x and log 64 = log x3, x = 4 and so y = 2, (4,2)  EMBED Equation.DSMT4  EMBED Equation.DSMT4  so 3log x = log 8 and x3 = 8, x = 2 and y = 4, (2,4)  EMBED Equation.DSMT4  Log Solutions (continued) 55.  EMBED Equation.DSMT4  y = 2  EMBED Equation.DSMT4  y = -3  EMBED Equation.DSMT4  56.  EMBED Equation.DSMT4  Equation #1:  EMBED Equation.DSMT4  x,y > 0, x,y EMBED Equation.DSMT4 ,  EMBED Equation.DSMT4  (or y = x-1) Now plug into Equation #2: (i) 2log2 x + log2(x) = 3 EMBED Equation.DSMT4   EMBED Equation.DSMT4  (ii) 2log2 x + log2(x-1) = 3 EMBED Equation.DSMT4 log2 x = 3  EMBED Equation.DSMT4 x = 23 = 8 and y = 1/8  EMBED Equation.DSMT4  57.  EMBED Equation.DSMT4  with x > 0 and x > -4  EMBED Equation.DSMT4  x > 0  EMBED Equation.DSMT4  EMBED Equation.DSMT4  EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4  58.  EMBED Equation.DSMT4  with x > 0 and x  EMBED Equation.DSMT4 , let u = logx (x+2). u + 1/u = 5/2  EMBED Equation.DSMT4  2u2 + 2 = 5u  EMBED Equation.DSMT4  2u2 5u + 2 = (2u-1)(u-2) = 0, (i) u = , 2 (i) u = logx (x+2) =  EMBED Equation.DSMT4  EMBED Equation.DSMT4  (ii) u = logx (x+2) = 2  EMBED Equation.DSMT4  x2 = x + 2  EMBED Equation.DSMT4  x2 x 2 = (x-2)(x+1) = 0, only take  EMBED Equation.DSMT4  59.  EMBED Equation.DSMT4  = 0 iff  EMBED Equation.DSMT4  = 1 iff  EMBED Equation.DSMT4 = 5, x > 0 So square both sides (moving one radical over if you prefer):  EMBED Equation.DSMT4   EMBED Equation.DSMT4  60.  EMBED Equation.DSMT4 , no restrictions on x for a change!  EMBED Equation.DSMT4  EMBED Equation.DSMT4  EMBED Equation.DSMT4  EMBED Equation.DSMT4  Bases are equal, so EMBED Equation.DSMT4  EMBED Equation.DSMT4  EMBED Equation.DSMT4   EMBED Equation.DSMT4 , multiplying by 9 to clear of fractions  EMBED Equation.DSMT4   EMBED Equation.DSMT4  Log Solutions (continued) 61.  EMBED Equation.DSMT4  or  EMBED Equation.DSMT4  Equation #2 is easier to work with, start there. Multiply by 8u to clear of fractions: 8u2 + 8 = 65u or 8u2 65u + 8 = (8u 1)(u 8) = 0 With 1/8 and 8, well go back to Equation #1: (i) u = 8 = xy  EMBED Equation.DSMT4  EMBED Equation.DSMT4 . Now clear of fractions to get: 64 + x4 = 20x2 x4 20x2 + 64 = (x2 16)(x2 4) = 0  EMBED Equation.DSMT4  x =  EMBED Equation.DSMT4  four solutions? (x=4) 4y = 8  EMBED Equation.DSMT4  22y = 23 EMBED Equation.DSMT4 y = 3/2  EMBED Equation.DSMT4  (x=-4) (-4)y = 8 undefined, try y=3/2 vs 6/4 (x=2) 2y = 8  EMBED Equation.DSMT4  y = 3  EMBED Equation.DSMT4  (x=-2) (-2)y = 8 undefined, try y=3 vs 6/2 (ii) u = 1/8 = xy  EMBED Equation.DSMT4  EMBED Equation.DSMT4  EMBED Equation.DSMT4 1 + 64x4 = 20x2  EMBED Equation.DSMT4 (16x2-1)(4x2-1) = 0 EMBED Equation.DSMT4 x =  EMBED Equation.DSMT4  (x=1/2) (1/2)y = 1/8  EMBED Equation.DSMT4  y=3  EMBED Equation.DSMT4  (x=-1/2) (-1/2)y = 1/8, same problem! (x=1/4) (1/4)y = 1/8  EMBED Equation.DSMT4  y=3/2  EMBED Equation.DSMT4  (x=-1/4) (-1/4)y = 1/8, no answer 62.  EMBED Equation.DSMT4   EMBED Equation.DSMT4  Lets just start cranking away by taking the log5 of both sides of equation #1  EMBED Equation.DSMT4 . Now exponentiate both sides base 10  EMBED Equation.DSMT4  (whew!) Now well use that theorem:  EMBED Equation.DSMT4  Why?!?! (I think I found this in a Russian book!) So lets start using equation #2  EMBED Equation.DSMT4  Now watch the switch around with xlog 3 into 3log x. Very nifty!  EMBED Equation.DSMT4  Now the simple Law of Exponents (1) 3log 3 + log x = 5log 5 + log y (Looking at the extra room below, I think Ill change font size!)  EMBED Equation.DSMT4  (Much better!) Now substitute that early result: x =  EMBED Equation.DSMT4 .  EMBED Equation.DSMT4  and take the log3 of both sides to work on the left a bit.  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