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15.4, 5 Solving Logarithmic Equations

15.4, 5 Solving Logarithmic Equations. OBJ:  To solve a logarithmic equation. DEF:  Property of equality for logarithms. If log b x = log b y, then x = y Note: Each apparent solution must be checked since log b x is defined only when x > 0.

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15.4, 5 Solving Logarithmic Equations

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  1. 15.4, 5 Solving Logarithmic Equations OBJ: To solve a logarithmic equation

  2. DEF: Property of equality for logarithms If log b x = log b y, then x = y • Note: Each apparent solution must be checked since log b x is defined only when x > 0

  3. HW 8 p399 ( 16 – 22 even) P 398 Solve EX: 5 log52t + log5 (t – 4) = log548 – log5 2 log 5 2t(t – 4) = log 5 (48 / 2) 2t(t – 4) = (48 / 2) 2t2 – 8t =24 2t2 – 8t – 24 = 0 2(t2 – 4t – 12) = 0 2(t – 6)(t + 2) = 0 t = -2, t = 6 now 

  4. Check log 5 2t(t – 4) = log 5 (48 / 2) log 5 2(6)(6 – 4) = log 5 48/2 log 5 12(2) = log 5 48/2  x = 6 log 5 2(-2) + log 5 (-2 – 4) = log 5 48 – log5 2 x ≠ -2 because cannot have log 5 -4

  5. Solve and check EX: log b 8t + log b 2 = log b 48 log b 8t (2) = log b 48 log b 16t = log b 48 16t = 48 t = 3 • log b 8t (2) = log b 48 log b 8(3)(2) = log b 48  t = 3

  6. Solve and Check EX: log 5 x + log 5 12 = log 5 ( x – 4) log 5 x + log 5 12 = log 5 ( x – 4) log 5 x (12) = log 5 ( x – 4) 12x = x – 4 11x = -4 x = -4/11 Ø

  7. HW 8 P402 (16, 20, 24)P 401 Solve EX:5 2 log 3 t = log 3 2 + log 3 (t + 12 ) log 3 t2 = log 3 2 (t + 12 ) t2 = 2 (t + 12 ) t2 – 2 t – 24 = 0 (t – 6) ( t + 4) = 0 t = 6, t = -4 now 

  8. Check log 3 62 = log 3 2(6 + 12 ) log 3 36 = log 3 2(18 )  x = 6 2 log 3 -4 = log 3 2 + log 3 (-4 + 12 ) x ≠ -4 because cannot have 2 log 3 -4

  9. HW 8 P 402 (18, 22)P 401Solve EX:6½ log b a + ½ log b (a – 6) = log b 4 log b√a + log b√(a – 6) = log b 4 log b√a(a – 6) = log b 4 √a(a – 6) = 4 (√a(a – 6))2 = 42 a(a – 6) = 42 a2 – 6a – 16 = 0 (a – 8) (a + 2) = 0 a = 8, -2 now 

  10. Check log b√8 (8 – 6) = log b 4 √8 (2) = 4 √16 = 4 x = 8 ½ log b -2 + ½ log b (-2 – 6) = log b 4 x ≠ -2 because cannot have ½ log b -2

  11. Solve EX: 2 log b x = log b 3 + log b ( x + 6 ) log b x2 = log b 3( x + 6 ) x2 = 3( x + 6 ) x2 = 3x + 18 x2 – 3x– 18 = 0 (x – 6) (x + 3) = 0 x = 6, x = -3 now 

  12. Check log b 62 = log b 3( 6 + 6 ) log b 36 = log b 3(12)  x = 6 2 log b -3 = log b 3 + log b ( -3 + 6 ) x ≠-3 because cannot have 2 log b -3

  13. Solve EX:  ½ log 4 x + ½ log 4 (x – 8) = log 4 3 ½ log 4 x + ½ log 4 (x – 8) = log 4 3 log 4√x(x – 8) = log 4 3 √x(x – 8) = 3 (√x(x – 8))2 = 32 x(x – 8) = 32 x2 – 8x– 9 = 0 (x– 9) (x + 1) = 0 x = 9, x = -1 now 

  14. Check log 4√9(9 – 8) = log 4 3 log 4√9(1) = log 4 3  x = 9 ½ log 4 -1 + ½ log 4 (-1 – 8) = log 4 3 x ≠ -1 because cannot have ½ log 4-1

  15. NOTE: Remember if log b x =y, then b y = x EX: log 4 (3y + 1) + log 4 (y–1) = 3 log 4 (3y + 1) (y–1) = 3 43 = (3y + 1) (y–1) 64 = 3y2 – 3y + 1y – 1 3y2 – 2y– 65 = 0 (3y + 13) (y – 5) = 0 y = -13/3, y = 5  43 = (3· 5 + 1) (5 –1) 64 = (16) (4)

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