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

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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

HW 7 p399 ( 16 – 22 even) P 398 Solve EX: 5 log 5 2t + log 5 (t – 4) = log 5 48 – log 5 2 log 5 2t(t – 4) = log 5 (48 / 2) 2t(t – 4) = (48 / 2) 2t 2 – 8t =24 2t 2 – 8t – 24 = 0 2(t 2 – 4t – 12) = 0 2(t – 6)(t + 2) = 0 t = -2, t = 6 now 

Check log 5 2t + log 5 (t – 4) = log 5 48 – log 5 2 log 5 2(6) + log 5 (6 – 4) = log 5 48 – log 5 2 log 5 2(6)(6 – 4) = log 5 48/2 log 5 12(2) = log 5 48/2 log 5 24 = log 5 24  x = 6 log 5 2(-2) + log 5 (-2 – 4) = log 5 48 – log 5 2 x ≠ -2 because cannot have log 5 -4

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 8(3) + log b 2 = log b 48 log b 8(3)(2) = log b 48 log b 48 = log b 48  t = 3

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

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

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

HW 7 P 402 (18, 22) P 401Solve EX: 6 ½ log b a + ½ log b (a – 6) = log b 4 ½ (log b a(a – 6)) = log b 4 log b √a(a – 6) = log b 4 √a(a – 6) = 4 (√a(a – 6)) 2 = 4 2 a(a – 6) = 4 2 a 2 – 6a – 16 = 0 (a – 8) (a + 2) = 0 a = 8, -2 now 

Check ½ log b a + ½ log b (a – 6) = log b 4 ½ log b 8 + ½ log b (8 – 6) = log b 4 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

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

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

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 = 3 2 x(x – 8) = 3 2 x 2 – 8x– 9 = 0 (x– 9) (x + 1) = 0 x = 9, x = -1 now 

Check ½ log 4 x + ½ log 4 (x – 8) = log 4 3 ½ log 4 9 + ½ log 4 (9 – 8) = log 4 3 log 4 √9(9 – 8) = log 4 3 log 4 √9(1) = log 4 3 log 4 3 = 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

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 4 3 = (3y + 1) (y–1) 64 = 3y 2 – 3y + 1y – 1 3y 2 – 2y– 65 = 0 (3y + 13) (y – 5) = 0 y = -13/3, y = 5  4 3 = (3· 5 + 1) (5 –1) 64 = (16) (4)

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