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5 indices & logarithms

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Here are the key steps to solve equations in indices that involve logarithms: 1. Isolate the term with the index (e.g. ax) on one side of the equation. 2. Take the logarithm (to an appropriate base) of both sides. 3. Use the property that loga(bx) = loga(b) + xloga(a) to split up logarithms of products/quotients. 4. Simplify the resulting equation so it is in the form of x = value. 5. Solve for x by isolating and evaluating the logarithm. Some examples: 1) 32x = 8 Take log base

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5. INDICES AND LOGARITHMS IMPORTANT NOTES : UNIT 5.1 Law of Indices I. am x an = am + n II. am ÷ an = am – n III (am)n = amn 1 Other Results : 0 a = 1, a −m = , (ab)m = am bm am 1. 1000 = 3– 2 = (3p)2 = 5.1.1 “BACK TO BASIC” BIL am × an = am + n am ÷ an = am – n (am)n = amn 1. a3 × a2 = a3 + 2 = a5 a4 ÷ a = a5 – 1 = a 4 (a3)2 = a3x2 = a6 23 × 24 = 23 + 4 3 5 3–5 □ a ÷a = a = a 2. □ 2 4 2x4 □ = 2 (3 ) = 3 = 3 1 = a2 p3 × p – 4 = p3 + ( – 4) p – 4 ÷ p5 = p – 4 – 5 □ □ –5 2 – 5 x2 □ = p = p (p ) = p = p 3. 1 = = = ( ) p 2k3 × (2k)3 (4a)2 ÷ 2a5 = (42a2) ÷ (2a 5) (3x2)3 = 33 × x2x3 = 2k3 × 23 × k3 2 16a 4. = 2a 5 = □ = ( )k = 5 Indices & Logarithms 1
UNIT 5.2 SIMPLE EQUATIONS INVOLVING INDICES Suggested Steps: S1 : Use the laws of indices to simplify expression (if necessary) S2 : Make sure the base is the SAME S3 : Form a linear equation by equating the indices S4 : Solve the linear equation No Example Exercise 1 Exercise 2 x x x 1. 3 = 81 2 = 32 4 = 64 3x = 34 x= 4 x = x= 2. 8x = 16 4x = 32 27x = 9 (23)x = 24 23x = 24 3x = 4 4 x = 3 x = x= 3. 8x = 16x – 3 4x+2 = 32x - 1 271 – x = 92x (23)x = 24(x – 3) 23x = 24x – 12 3x = 4x – 12 x = 12 x = x= 4. 2 × 82x = 16x + 3 16 × 42x – 3 = 322 – x 251 – 3x = 5 × 125x 21 × (23)x = 24 (x + 3) 21+3x = 24x + 12 1 + 3x = 4x + 12 1 – 12 = 4x – 3x x = x= x = – 11 5 Indices & Logarithms 2
UNIT 5.2 LOGARITMS Do YOU know that .... If a number N can be expressed in the form N = ax , then the logarithm of N to the base a is x? N = ax ⇔ loga N = x 100 = 102 ⇔ log10 100 = 2 64 = 43 ⇔ log4 64 = 3 0.001 = 10-3 ⇔ log10 0.001 = – 3 5 Indices & Logarithms 3
Unit 5.2.1To convert numbers in index form to logaritmic form and vice-versa. No. Index Form Logaritmic Form 1. 102 = 100 log10 100 = 2 2. 23 = 8 log2 8 = 3 3. pq = r logp r = q 4. 104 = 10000 5. a3 = b 6. 81 = 34 7. logp m = k 8. 2x = y 9. V = 10x 10. log3 x = y 11. loga y = 2 12. 25 = 32 13. log3 (xy) = 2 14. 10x = y3 15. log10 100y = p 5 Indices & Logarithms 4
UNIT 5.2.2 To find the value of a given Logaritm IMPORTANT : x loga a = x No. Logaritmic Form Notes 103 = 1000 1. log10 1000 = 3 dan log10 103 = 3 25 = 32 2. log2 32 = 5 dan log2 25 = 5 10 = 0.01 3. log10 0.01 = dan log10 = 4 = 64 4. log 4 64 = dan log4 = 5. log p p = (REINFORCEMENT) 6. log p p8 = log a a2 = 1 7. log m m-1 = logm m2 = 8. log a a⅓ = log p p-5 = 1 9. loga a4 = log b bk = 10. log p (p×p2) = log p p p = 5 Indices & Logarithms 5
UNIT 5.3 Laws of Logaritm I. loga (xy) = loga x + loga y x II. loga    y = loga x – loga y   III loga xm = m loga x Other Results : loga 1 = 0 (since 1 = a0) loga a = 1 (since a = a1) NO. Examples Exercises 1. loga 3pr = loga 3 + loga p + loga r (a)loga 2mn = (b) loga 3aq = (c) log10 10yz = (d) log10 1000xy = (e) log2 4mn = 2. p p loga = loga p – loga q (a) loga = loga p – loga 2r q 2r = loga p – (loga 2 + loga r) = (c) log10   = 4 10 (b) log2 =   m  kx  (d) log10   = xy 3a   (e) loga =  100  m 5 Indices & Logarithms 6
UNIT 5.3.2 Aplication of the law : loga xn = n loga x 3. Example : 1 –2 (a) loga = loga x 3 x2 loga x = 3 loga x = (b) log2 ( xy 4 ) = log2 x + log2 y 4 (c) log2 ( 4 y 4 ) = = log2 x + = y4 y4 (d) log2 = (e) log2 = x 8 Reinforcement exercises (Laws of Logaritm) 1. Example : (a) log10 10000x 5 = 3 log10 100x = log10 100 + log10x3 = = log10 102 + 3 log10 x = = 2 + 3 log10 x xy 4 (c) logp ( 8 p 5 ) = (b) log2 ( ) = 8 = = k2 y3 (d) log2 = (e) log4 = 4x3 64 5 Indices & Logarithms 7
UNIT 5.4 EQUATIONS IN INDICES (Which involves the use of LOGARITM) I. Equation in the form ax = b Steps to be followed: S1 : Take logaritm (to base 10) on both sides. S2 : Use the law log10 ax = x log10 a. S3 : Solve the linear equation with the help of a calculator. No. Example Exercise 1 Exercise 2 x x 1. 3x = 18 2 = 9 7 = 20 log10 3x = log10 18 x log10 3 = log10 18 log10 18 x = log10 3 x = 2.631 x = x= 2. 5x+2 = 16 4x+1 = 28 3x-2 = 8 log10 5x+2 = log10 16 (x+2) log10 3 = log10 18 log10 16 x+2 = log10 5 x+2 = x = x = x= 3. 2x+3 = 200 71-x = 2.8 63x-2 = 66 log10 2x+3 = log10 200 x+3 = x = x = x= 5 Indices & Logarithms 8
UNIT 5.5 Change of Base of Logarithms log b x Formula : loga x = log b a No. Example Exercise 1 Exercise 2 1. log 2 8 log 2 32 (b) log16 8 = log4 8 = (a) log4 32 = log 2 4 log 2 4 = 3 = 2 (c) log8 2 = (d) log9 27 = (e) log81 9 = = = = (With a calculator) – Change to base 10 1. log10 9 log10 20 (b) log4 0.8 = log4 9 = (a) log5 20 = log10 4 log10 5 = 1.585 = (c) log7 2 = (d) log9 77 = (e) log3 9.6 = (f) log6 2.5 = (g) log5 2000 = (h) log12 6 = Ans : 0.5114 Ans : 4.723 Ans : 0.7211 5 Indices & Logarithms 9
UNIT 5.6 Aplication of the Laws of Logaritm To Solve Simple Equations involvong logaritms EXAMPLE EVERCISE C1. Solve the equation log2 (x+1) = 3. L1. Solve the equation log2 (x – 3 ) = 2. Answers: log2 (x+1) = 3 Jawapan: x + 1 = 23 x+1= 8 x = 7 Ans : x = 7 C2. Solve the equation log10 (3x – 2) = – 1 . L2. Solve the equation log5 (4x – 1 ) = – 1 . Jawapan: 3x – 2 = 10-1 3x – 2 = 0.1 3x = 2.1 x = 0.7 Ans : x = 0.3 L3. Solve the equation log3 (x – 6) = 2. L4. Solve the equation log10 (1+ 3x) = 2 Ans : x = 15 Ans : x = 33 L5. Solve the equation log3 (2x – 1) + log2 4 = 5 L6. Solve the equation . log4 (x – 2) + 3log2 8 = 10. Ans : x = 6 Ans : x = 14 L7. Solve the equation L8. Solve the equation log2 (x + 5) = log2 (x – 2) + 3. log5 (4x – 7) = log5 (x – 2) + 1. Ans : x = 3 Ans : x = 3 L9. Solve log3 3(2x + 3) = 4 L10 . Solve log2 8(7 – 3x) = 5 Ans : x = 12 Ans : x = 1 5 Indices & Logarithms 10
UNIT 5.6.1 To Determine the value of a logarithm without using calculator. EXAMPLE EXERCISE C1. Given log2 3 = 1.585, log2 5 = 2.322. Without L1. Given log3 5 = 1.465 , log3 7 = 1.771 . using a calculator find the value of Withouf using calculator, evaluate (a) log2 15 = log2 (3 × 5) (a) log3 35 = = log2 3 + log2 5 = = 1.585 + 2.322 = = = (b) log2 25 = log2 (5 × 5) (b) log3 49 = = = = = = = 3 (c) log3 1.4 = (c) log2 0.6 = log2 ( ) 5 = log2 3 – log2 5 = = = = = (d) log2 10 = log2 (2 × 5) (d) log3 21 = = log2 2 + log2 5 = = = = = log 2 5 (e) log4 5 = (e) log9 21 = log 2 4 = 2.322 = = 2 = = log 2 2 (f) log5 3 = (f) log5 2 = log 2 5 1 1 = = ( ) ( ) = = 5 Indices & Logarithms 11
Enrichment Exercises (SPM Format Questions) EXERCISE EXCERCISE L1 Given log3 x = m and log2 x = n. L2. Given log3 x = p and log2 x = q. Find logx 24 in terms of m and n. Find logx 36 in terms of m and n. [SPM 2001] [4] [4] (Ans : 3/n + 1/m ) (Ans: 2/p + 2/q ) L3. Given log3 x = p and log9 y = q. L4. Given log3 x = p and log9 y = q. Find log3 xy2 in terms of p and q. Find log3 x2y3 in terms of p and q. [SPM 1998] [4] [4] (Ans: p + 4q ) (Ans: 2p + 6q ) L5 Given log5 2 = m and log5 7 = p, express L6. Given log5 2 = m and log5 7 = p, express log5 4.9 in terms of m and p. [4] log5 2.82 in terms of m and p. [SPM 2004] [4] (Ans: 2p – m – 1 ) (Ans: 2(p + m – 1 ) L7 Given log 2 T - log4 V = 3, express T in terms of L8. Given log 4 T + log 2 V = 2, express V. [4] T in terms of V. [4] [SPM 2003] (Ans: T = 8V ½ ) (Ans: 16V-2 ) 2x – 1 x L9 Solve 4 = 7. [4] L10. Solve 42x – 1 = 9x. [4] ( Ans: x = 1.677 ) ( Ans: x = 2.409 ) 5 Indices & Logarithms 12
INDICES & LOGARITMS Enrichment Exercises (Past Year SPM Questions) : PAPER 1 EXERCISE EXCERCISE 1 1 2. Given log2 xy = 2 + 3log2 x – log2 y, Solve the equation 82 x −3 = . express y in terms of x. 4x+2 [SPM 2006 P1,Q 7] [4] [SPM 2006 P1,Q6] [3] (Ans : x = 1 ) (Ans: y = 4x ) 3 Solve the equation 2 + log 3 ( x − 1) = log 3 x . 4. Solve the equation 2 x + 4 − 2 x +3 = 1 . [SPM 2006 P1,Q 8] [3] [SPM 2005 P1,Q7] [3] (Ans : x= 9/8 ) (Ans : x = -3 ) 5 Solve the equation log 3 4 x − log 3 (2 x − 1) = 1 . 6. Given that log m 2 = p and log m 3 = r , [SPM 2005 P1,Q 8] [3]  27 m  express log m   in terms of p and r.  4  [SPM 2005 P1,Q 9] [4] (Ans: x = 3/2 ) (Ans: 3r – 2p + 1 ) 7 Solve the equation 8 2 x −3 8 x +6 = 4 . 8. Given that log 5 2 = m and log 5 7 = p , [SPM 2004 P1,Q7] [3] express log 5 4.9 in terms of m and p. [SPM 2004 P1,Q 8] [4] (Ans : x=3 ) (Ans: 2p – m - 1 ) 5 Indices & Logarithms 13
INDICES & LOGARITMS Enrichment Exercises (Past Year SPM Questions) : PAPER 1 (Cont...) EXERCISE EXCERCISE 1 Solve the equation 162 x −3 = 84 x . 2. Given log4 x = log2 3, find the value of x. [SPM 2008 P1,Q7] [3] [SPM 2008 P1,Q 8] [4] (Ans : x = -3 ) (Ans: x = 9 ) 3 Given that log 2 T – log 4 V = 3, express T in 4. Solve the equation 42 x −1 = 7 x . terms of V [SPM 2003 P1,Q6] [3] [SPM 2006 P1,Q 8] [3] (Ans : ) (Ans : ) 5 Solve the equation log 3 9 x − log 3 (2 x + 1) = 1 . 6. Given that log m 2 = p and log m 3 = r , [3]  27m 2  express log m   in terms of p and r.  16  [4] (Ans: x = 1 ) (Ans: 3r – 4p +2 ) 7 Solve the equation 85 x −3 = 32 x +6 . 8. Given that log 5 2 = m and log 5 3 = p , [3] express log 5 2.7 in terms of m and p. [4] (Ans : x = 3.9 ) (Ans: 3p – m – 1 ) 5 Indices & Logarithms 14

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5 indices & logarithms

  • 1. 5. INDICES AND LOGARITHMS IMPORTANT NOTES : UNIT 5.1 Law of Indices I. am x an = am + n II. am ÷ an = am – n III (am)n = amn 1 Other Results : 0 a = 1, a −m = , (ab)m = am bm am 1. 1000 = 3– 2 = (3p)2 = 5.1.1 “BACK TO BASIC” BIL am × an = am + n am ÷ an = am – n (am)n = amn 1. a3 × a2 = a3 + 2 = a5 a4 ÷ a = a5 – 1 = a 4 (a3)2 = a3x2 = a6 23 × 24 = 23 + 4 3 5 3–5 □ a ÷a = a = a 2. □ 2 4 2x4 □ = 2 (3 ) = 3 = 3 1 = a2 p3 × p – 4 = p3 + ( – 4) p – 4 ÷ p5 = p – 4 – 5 □ □ –5 2 – 5 x2 □ = p = p (p ) = p = p 3. 1 = = = ( ) p 2k3 × (2k)3 (4a)2 ÷ 2a5 = (42a2) ÷ (2a 5) (3x2)3 = 33 × x2x3 = 2k3 × 23 × k3 2 16a 4. = 2a 5 = □ = ( )k = 5 Indices & Logarithms 1
  • 2. UNIT 5.2 SIMPLE EQUATIONS INVOLVING INDICES Suggested Steps: S1 : Use the laws of indices to simplify expression (if necessary) S2 : Make sure the base is the SAME S3 : Form a linear equation by equating the indices S4 : Solve the linear equation No Example Exercise 1 Exercise 2 x x x 1. 3 = 81 2 = 32 4 = 64 3x = 34 x= 4 x = x= 2. 8x = 16 4x = 32 27x = 9 (23)x = 24 23x = 24 3x = 4 4 x = 3 x = x= 3. 8x = 16x – 3 4x+2 = 32x - 1 271 – x = 92x (23)x = 24(x – 3) 23x = 24x – 12 3x = 4x – 12 x = 12 x = x= 4. 2 × 82x = 16x + 3 16 × 42x – 3 = 322 – x 251 – 3x = 5 × 125x 21 × (23)x = 24 (x + 3) 21+3x = 24x + 12 1 + 3x = 4x + 12 1 – 12 = 4x – 3x x = x= x = – 11 5 Indices & Logarithms 2
  • 3. UNIT 5.2 LOGARITMS Do YOU know that .... If a number N can be expressed in the form N = ax , then the logarithm of N to the base a is x? N = ax ⇔ loga N = x 100 = 102 ⇔ log10 100 = 2 64 = 43 ⇔ log4 64 = 3 0.001 = 10-3 ⇔ log10 0.001 = – 3 5 Indices & Logarithms 3
  • 4. Unit 5.2.1To convert numbers in index form to logaritmic form and vice-versa. No. Index Form Logaritmic Form 1. 102 = 100 log10 100 = 2 2. 23 = 8 log2 8 = 3 3. pq = r logp r = q 4. 104 = 10000 5. a3 = b 6. 81 = 34 7. logp m = k 8. 2x = y 9. V = 10x 10. log3 x = y 11. loga y = 2 12. 25 = 32 13. log3 (xy) = 2 14. 10x = y3 15. log10 100y = p 5 Indices & Logarithms 4
  • 5. UNIT 5.2.2 To find the value of a given Logaritm IMPORTANT : x loga a = x No. Logaritmic Form Notes 103 = 1000 1. log10 1000 = 3 dan log10 103 = 3 25 = 32 2. log2 32 = 5 dan log2 25 = 5 10 = 0.01 3. log10 0.01 = dan log10 = 4 = 64 4. log 4 64 = dan log4 = 5. log p p = (REINFORCEMENT) 6. log p p8 = log a a2 = 1 7. log m m-1 = logm m2 = 8. log a a⅓ = log p p-5 = 1 9. loga a4 = log b bk = 10. log p (p×p2) = log p p p = 5 Indices & Logarithms 5
  • 6. UNIT 5.3 Laws of Logaritm I. loga (xy) = loga x + loga y x II. loga    y = loga x – loga y   III loga xm = m loga x Other Results : loga 1 = 0 (since 1 = a0) loga a = 1 (since a = a1) NO. Examples Exercises 1. loga 3pr = loga 3 + loga p + loga r (a)loga 2mn = (b) loga 3aq = (c) log10 10yz = (d) log10 1000xy = (e) log2 4mn = 2. p p loga = loga p – loga q (a) loga = loga p – loga 2r q 2r = loga p – (loga 2 + loga r) = (c) log10   = 4 10 (b) log2 =   m  kx  (d) log10   = xy 3a   (e) loga =  100  m 5 Indices & Logarithms 6
  • 7. UNIT 5.3.2 Aplication of the law : loga xn = n loga x 3. Example : 1 –2 (a) loga = loga x 3 x2 loga x = 3 loga x = (b) log2 ( xy 4 ) = log2 x + log2 y 4 (c) log2 ( 4 y 4 ) = = log2 x + = y4 y4 (d) log2 = (e) log2 = x 8 Reinforcement exercises (Laws of Logaritm) 1. Example : (a) log10 10000x 5 = 3 log10 100x = log10 100 + log10x3 = = log10 102 + 3 log10 x = = 2 + 3 log10 x xy 4 (c) logp ( 8 p 5 ) = (b) log2 ( ) = 8 = = k2 y3 (d) log2 = (e) log4 = 4x3 64 5 Indices & Logarithms 7
  • 8. UNIT 5.4 EQUATIONS IN INDICES (Which involves the use of LOGARITM) I. Equation in the form ax = b Steps to be followed: S1 : Take logaritm (to base 10) on both sides. S2 : Use the law log10 ax = x log10 a. S3 : Solve the linear equation with the help of a calculator. No. Example Exercise 1 Exercise 2 x x 1. 3x = 18 2 = 9 7 = 20 log10 3x = log10 18 x log10 3 = log10 18 log10 18 x = log10 3 x = 2.631 x = x= 2. 5x+2 = 16 4x+1 = 28 3x-2 = 8 log10 5x+2 = log10 16 (x+2) log10 3 = log10 18 log10 16 x+2 = log10 5 x+2 = x = x = x= 3. 2x+3 = 200 71-x = 2.8 63x-2 = 66 log10 2x+3 = log10 200 x+3 = x = x = x= 5 Indices & Logarithms 8
  • 9. UNIT 5.5 Change of Base of Logarithms log b x Formula : loga x = log b a No. Example Exercise 1 Exercise 2 1. log 2 8 log 2 32 (b) log16 8 = log4 8 = (a) log4 32 = log 2 4 log 2 4 = 3 = 2 (c) log8 2 = (d) log9 27 = (e) log81 9 = = = = (With a calculator) – Change to base 10 1. log10 9 log10 20 (b) log4 0.8 = log4 9 = (a) log5 20 = log10 4 log10 5 = 1.585 = (c) log7 2 = (d) log9 77 = (e) log3 9.6 = (f) log6 2.5 = (g) log5 2000 = (h) log12 6 = Ans : 0.5114 Ans : 4.723 Ans : 0.7211 5 Indices & Logarithms 9
  • 10. UNIT 5.6 Aplication of the Laws of Logaritm To Solve Simple Equations involvong logaritms EXAMPLE EVERCISE C1. Solve the equation log2 (x+1) = 3. L1. Solve the equation log2 (x – 3 ) = 2. Answers: log2 (x+1) = 3 Jawapan: x + 1 = 23 x+1= 8 x = 7 Ans : x = 7 C2. Solve the equation log10 (3x – 2) = – 1 . L2. Solve the equation log5 (4x – 1 ) = – 1 . Jawapan: 3x – 2 = 10-1 3x – 2 = 0.1 3x = 2.1 x = 0.7 Ans : x = 0.3 L3. Solve the equation log3 (x – 6) = 2. L4. Solve the equation log10 (1+ 3x) = 2 Ans : x = 15 Ans : x = 33 L5. Solve the equation log3 (2x – 1) + log2 4 = 5 L6. Solve the equation . log4 (x – 2) + 3log2 8 = 10. Ans : x = 6 Ans : x = 14 L7. Solve the equation L8. Solve the equation log2 (x + 5) = log2 (x – 2) + 3. log5 (4x – 7) = log5 (x – 2) + 1. Ans : x = 3 Ans : x = 3 L9. Solve log3 3(2x + 3) = 4 L10 . Solve log2 8(7 – 3x) = 5 Ans : x = 12 Ans : x = 1 5 Indices & Logarithms 10
  • 11. UNIT 5.6.1 To Determine the value of a logarithm without using calculator. EXAMPLE EXERCISE C1. Given log2 3 = 1.585, log2 5 = 2.322. Without L1. Given log3 5 = 1.465 , log3 7 = 1.771 . using a calculator find the value of Withouf using calculator, evaluate (a) log2 15 = log2 (3 × 5) (a) log3 35 = = log2 3 + log2 5 = = 1.585 + 2.322 = = = (b) log2 25 = log2 (5 × 5) (b) log3 49 = = = = = = = 3 (c) log3 1.4 = (c) log2 0.6 = log2 ( ) 5 = log2 3 – log2 5 = = = = = (d) log2 10 = log2 (2 × 5) (d) log3 21 = = log2 2 + log2 5 = = = = = log 2 5 (e) log4 5 = (e) log9 21 = log 2 4 = 2.322 = = 2 = = log 2 2 (f) log5 3 = (f) log5 2 = log 2 5 1 1 = = ( ) ( ) = = 5 Indices & Logarithms 11
  • 12. Enrichment Exercises (SPM Format Questions) EXERCISE EXCERCISE L1 Given log3 x = m and log2 x = n. L2. Given log3 x = p and log2 x = q. Find logx 24 in terms of m and n. Find logx 36 in terms of m and n. [SPM 2001] [4] [4] (Ans : 3/n + 1/m ) (Ans: 2/p + 2/q ) L3. Given log3 x = p and log9 y = q. L4. Given log3 x = p and log9 y = q. Find log3 xy2 in terms of p and q. Find log3 x2y3 in terms of p and q. [SPM 1998] [4] [4] (Ans: p + 4q ) (Ans: 2p + 6q ) L5 Given log5 2 = m and log5 7 = p, express L6. Given log5 2 = m and log5 7 = p, express log5 4.9 in terms of m and p. [4] log5 2.82 in terms of m and p. [SPM 2004] [4] (Ans: 2p – m – 1 ) (Ans: 2(p + m – 1 ) L7 Given log 2 T - log4 V = 3, express T in terms of L8. Given log 4 T + log 2 V = 2, express V. [4] T in terms of V. [4] [SPM 2003] (Ans: T = 8V ½ ) (Ans: 16V-2 ) 2x – 1 x L9 Solve 4 = 7. [4] L10. Solve 42x – 1 = 9x. [4] ( Ans: x = 1.677 ) ( Ans: x = 2.409 ) 5 Indices & Logarithms 12
  • 13. INDICES & LOGARITMS Enrichment Exercises (Past Year SPM Questions) : PAPER 1 EXERCISE EXCERCISE 1 1 2. Given log2 xy = 2 + 3log2 x – log2 y, Solve the equation 82 x −3 = . express y in terms of x. 4x+2 [SPM 2006 P1,Q 7] [4] [SPM 2006 P1,Q6] [3] (Ans : x = 1 ) (Ans: y = 4x ) 3 Solve the equation 2 + log 3 ( x − 1) = log 3 x . 4. Solve the equation 2 x + 4 − 2 x +3 = 1 . [SPM 2006 P1,Q 8] [3] [SPM 2005 P1,Q7] [3] (Ans : x= 9/8 ) (Ans : x = -3 ) 5 Solve the equation log 3 4 x − log 3 (2 x − 1) = 1 . 6. Given that log m 2 = p and log m 3 = r , [SPM 2005 P1,Q 8] [3]  27 m  express log m   in terms of p and r.  4  [SPM 2005 P1,Q 9] [4] (Ans: x = 3/2 ) (Ans: 3r – 2p + 1 ) 7 Solve the equation 8 2 x −3 8 x +6 = 4 . 8. Given that log 5 2 = m and log 5 7 = p , [SPM 2004 P1,Q7] [3] express log 5 4.9 in terms of m and p. [SPM 2004 P1,Q 8] [4] (Ans : x=3 ) (Ans: 2p – m - 1 ) 5 Indices & Logarithms 13
  • 14. INDICES & LOGARITMS Enrichment Exercises (Past Year SPM Questions) : PAPER 1 (Cont...) EXERCISE EXCERCISE 1 Solve the equation 162 x −3 = 84 x . 2. Given log4 x = log2 3, find the value of x. [SPM 2008 P1,Q7] [3] [SPM 2008 P1,Q 8] [4] (Ans : x = -3 ) (Ans: x = 9 ) 3 Given that log 2 T – log 4 V = 3, express T in 4. Solve the equation 42 x −1 = 7 x . terms of V [SPM 2003 P1,Q6] [3] [SPM 2006 P1,Q 8] [3] (Ans : ) (Ans : ) 5 Solve the equation log 3 9 x − log 3 (2 x + 1) = 1 . 6. Given that log m 2 = p and log m 3 = r , [3]  27m 2  express log m   in terms of p and r.  16  [4] (Ans: x = 1 ) (Ans: 3r – 4p +2 ) 7 Solve the equation 85 x −3 = 32 x +6 . 8. Given that log 5 2 = m and log 5 3 = p , [3] express log 5 2.7 in terms of m and p. [4] (Ans : x = 3.9 ) (Ans: 3p – m – 1 ) 5 Indices & Logarithms 14