QA #6 Factors and Multiples

Questions and Answers Sheet 5 Factors and Multiples Question #1 A polynomial P is given. (a) Factor P into linear and irreducible quadratic factors with real coefficients. (b) Factor P completely into linear factors with complex coefficients. P(𝑥 ) = 𝑥 4 + 8𝑥 2 − 9 Answer: a) To find: The factorization of the polynomial P(x) in linear and irreducible quadratic factors with real coefficients. Given information: The polynomial P(x) is, 𝑃(𝑥 ) = 𝑥 4 + 8𝑥 2 − 9 Concept used: Linear and Quadratic Factors Theorem: Any polynomial that has real coefficients can be factored into the product of linear and irreducible quadratic factors. Calculation: The given polynomial P(x) is, 𝑃(𝑥 ) = 𝑥 4 + 8𝑥 2 − 9 Rewrite the above polynomial as, 𝑃(𝑥 ) = 𝑥 4 + 9𝑥 2 − 𝑥 2 − 9 Now, factor the polynomial P(x) as, 𝑃 (𝑥 ) = 𝑥 2 (𝑥 2 + 9) − 1(𝑥 2 + 9) = (𝑥 2 − 1)(𝑥 2 + 9) = (𝑥 − 1)(𝑥 + 1)(𝑥 2 + 9) The factor (𝑥 − 1) and (𝑥 + 1) are linear factors with real zeros. The factor (𝑥 2 + 9) is irreducible, since it has no real zeros. Conclusion: Thus, the factored form of the polynomial P(x) that has linear and irreducible quadratic factors is (𝑥 − 1)(𝑥 + 1)(𝑥 2 + 9) b) To find: The factors of the polynomial P(x) that has linear factors with complex coefficients. Given: The polynomial P(x) is, 𝑃(𝑥 ) = 𝑥 4 + 8𝑥 2 − 9 Calculation: From part (a) the factored form of the polynomial P(x) is, 𝑃(𝑥 ) = (𝑥 − 1)(𝑥 + 1)(𝑥 2 + 9) Now, factor the remaining quadratic factor to obtain the complete factorization as, 𝑃(𝑥 ) = (𝑥 − 1)(𝑥 + 1)(𝑥 2 + 9) = (𝑥 − 1)(𝑥 + 1)(𝑥 2 − (3𝑖 )2 ) = (𝑥 − 1)(𝑥 + 1)(𝑥 − 3𝑖 )(𝑥 + 3𝑖 ) The above factors are linear factors with complex coefficients. Conclusion: Thus, the factored form of the polynomial P(x) that has linear factors is (𝑥 − 1)(𝑥 + 1)(𝑥 − 3𝑖 )(𝑥 + 3𝑖 ) . Question #2 A polynomial P is given (a) Factor P into linear and irreducible quadratic factors with real coefficients. (b) Factor P completely into linear factors with complex coefficients. 𝑃(𝑥 ) = 𝑥 3 − 2𝑥 − 4 Answer: a) To find: The factorization of the polynomial P(x) in linear and irreducible quadratic factors with real coefficients. Consider, 𝑃(𝑥 ) = 𝑥 3 − 2𝑥 − 4 Using the Rational Zeros Theorem, possible rational zeros are ±1, ±2, ±4. Substitute 𝑥 = 2 𝑖𝑛 𝑃 (𝑥) = 𝑥 3 − 2𝑥 − 4 , we get 𝑃 (2) = 23 − 2(2) − 4 ⇒ 𝑃 (2) = 8 − 4 − 4 ⇒ 𝑃 (2) = 0 Therefore, 𝑥 = 2 is a factor of 𝑃 (𝑥) = 𝑥 3 − 2𝑥 − 4 . Factorize the polynomial, 𝑃(𝑥 ) = 𝑥 3 − 2𝑥 − 4 . \Rightarrow 𝑃 (𝑥 ) = (𝑥 − 2)(𝑥 2 + 2𝑥 + 2) Therefore, 𝑃 (𝑥 ) = (𝑥 − 2)(𝑥 2 + 2𝑥 + 2) . b) To find: The factor completely into linear factors with complex coefficients. Consider, 𝑃(𝑥 ) = 𝑥 3 − 2𝑥 − 4 Using the Rational Zeros Theorem, possible rational zeros are ±1, ±2, ±4 . Substitute 𝑥 = 2 in P(𝑥 ) = 𝑥 3 − 2𝑥 − 4 , we get 𝑃 (2) = 23 − 2(2) − 4 ⇒ 𝑃 (2) = 8 − 4 − 4 ⇒ 𝑃 (2) = 0 Therefore, 𝑥 = 2 is a factor of 𝑃 (𝑥) = 𝑥 3 − 2𝑥 − 4 . Factorize the polynomial, 𝑃(𝑥 ) = 𝑥 3 − 2𝑥 − 4 ⇒ 𝑃(𝑥 ) = (𝑥 − 2)(𝑥 2 + 2𝑥 + 2) ⇒ 𝑃(𝑥 ) = (𝑥 − 2)(𝑥 − (−1 + 𝑖 ))(𝑥 − (−1 − 𝑖 )) Therefore, 𝑃 (𝑥 ) = (𝑥 − 2)(𝑥 − (−1 + 𝑖 ))(𝑥 − (−1 − 𝑖 )) . Question #3 Taj says the expression 3 − 2𝑠 cannot be factored using the GCF. Is he correct? Explain why or why not. Select the correct answer below and, if necessary, fill in the answer box to complete your choice. A. Taj is incorrect because all expressions with two terms can be factored. B. Taj is correct because 3 and −2𝑠 have no common factors. C. Taj is correct because the greatest common factor of 3 and −2𝑠 is 1 and factoring out 1 does not change the expression. D. Taj is incorrect because the greatest common factor of 3 and −2𝑠 is ?. The factored expression is ?. Answer: Step 1 factorisation is Process of splitting an expression in the product of some other expressions. 𝐺𝐶𝐷 ⇒ greatest common factor. Step 2 Taj is correct because the greatest common factor of 3 and -2s is 1 and factoring out 1 does not change the expression. Question #4 add or subtract as indicated, and express answers in lowest terms. 1 6 1 +8. Answer: Step 1 The given expression is, 1 6 1 +8 The objective is to find the sum of two fractions with different denominator. Step 2 Since, the denominators of the fractions are not same, take LCM of the denominators to make same. The multiples of 6 are 6,12,18,24,30,36,42,48,... The multiples of 8 are 8,16,24,32,40,48,... The least common multiplier of 6 and 8 is 24. Now, simplify the expression as follows. 1 6 1 1×4 1×3 + 8 = 6×4 + 8×3 4 3 = 24 + 24 = 4+3 24 7 = 24 Thus, the value of the expression is, 7 24 . Question #5 Solving Basic Trigonometric Equations by Factoring Solve the given equation. 3 sin2 𝜃 − 7 sin 𝜃 + 2 = 0 Answer: Approach: The domain of the trigonometry function of sin is lies between [−1,1]. No solution exists beyond this domain. Sine has period 2 𝜋 , we find solution in any interval of length 2 𝜋 . SIne function is positive in first and second quadrant. Calculation: Consider the trigonometry equation. 3 sin2 𝜃 − 7 sin 𝜃 + 2 = 0 The factor of the given equation is obtained by, 3 sin2 𝜃 − 7 sin 𝜃 + 2 = 0 (sin 𝜃 − 2)(3 sin 𝜃 − 1) = 0 The factors of above equation are, sin 𝜃 − 2 = 0 … … (1) 3 sin 𝜃 − 1 = 0 … … (2) The solution obtained for the factor in which sine function involved so we will get the solution in the interval of [0,2𝜋]. Consider equation (1). sin 𝜃 − 2 = 0 The above equation has no solution because value lies outside the domain. Add 1 both sides in equation (2). 3 sin 𝜃 − 1 = 0 1 𝜃 = sin1 (3) 𝜃 = 0.34, 2.80 Here, the angles are in radian. The sine has period, 2 𝜋 So we get all solutions of the equation by adding integer multiples of 2 𝜋 to these solutions: S 𝜃 = 0.34 + 2𝑘 𝜋 𝜃 = 2.80 + 2𝑘 𝜋 Therefore, the solutions of the trigonometry equation 3 sin2 𝜃 − 7 sin 𝜃 + 2 = 0 are 𝜃 = 0.34 + 2𝑘 𝜋 and 𝜃 = 2.80 + 2𝑘 𝜋 . Concusion: Thus, the solutions of the trigonometry equation 3 sin2 𝜃 − 7 sin 𝜃 + 2 = 0 are 𝜃 = 0.34 + 2𝑘 𝜋 and 𝜃 = 2.80 + 2𝑘 𝜋 . Question #6 3 2 Evaluate (14) − (21) Answer: Step 1 The given expression is, 3 2 (14) − (21) The objective is to evaluate the expression Step 2 Consider the expression. 3 2 (14) − (21) Note that, the denominator of the two fractions are not same. To make it same, find the LCM of 14 and 21. The multiples of 14 are, 14, 28,42,56, 70, 84, 98,... The multiples of 21 are, 21, 42,63,84,... Thus, the LCM of 14 and 21 is 42. Now, evaluate the expression as follows. 3 2 3 3 2 2 (14) − (21) = 14 ⋅ 3 − 21 ⋅ 2 9 4 = 42 − 42 = 9−4 42 5 = 42 Thus, the value of the expression is, 5 42 Question #7 A ball is tossed upward from the ground. Its height in feet above ground after t seconds is given by the function ℎ(𝑡) = −16𝑡 2 + 24task. Find the maximum height of the ball and the number of seconds it took for the ball to reach the maximum height. Answer: 1) Concept: The maximum height of the ball is the maximum value of h(t). The degree of the function ℎ(𝑡) = −16𝑡 2 + 24task is 2 and hence it is a quadratic function. The graphical representation of the quadratic function is a parabola. The coefficient of the 𝑡 2 is -16. Therefore, it is a parabola that opens downward. So, the h(t) or the y coordinate of the vertex gives the maximum height and the x coordinate of the vertex gives the number of seconds it took for the ball to reach the maximum height. Step 1: Compare h(t) with the general quadratic equation 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 . Find the values of a and b. −𝑏 −𝑏 Step 2: Find the vertex using the formula ( 2𝑎 , ℎ ( 2𝑎 )) . Step 3: The x coordinate of the vertex is the numbers of seconds the ball takes to reach the maximum height and the y coordinate of the vertex is the maximum height of the ball. 2) Calculation: Comparing ℎ(𝑡) = −16𝑡 2 + 24task with 𝑎𝑥 2 + 𝑏𝑥 + 𝑐, we have 𝑎 = −16; 𝑏 = 24; Now, to find the x-coordinate of the vertex, substitute the values of a and b in −𝑏 2𝑎 −𝑏 2𝑎 . −24 = 2(−16) 24 and 16 are multiples of 2. So,we can write them as factors of 2 and the common factors can be cancelled out. Since both the numerator and the denominator has negative sign, the fraction becomes positive. −24 2(−16) 2×2×2×3 3 = 2×2×2×2×2 = 4 3 3 3 3 2 3 Now, find ℎ (4). Substitute 4 for t in the function ℎ(𝑡) = −16𝑡 2 + 24 task. ℎ (4) = −16 (4) + 24 (4) Again, write the numbers as product of their factors. 3 3×3 3 ℎ (4) = −2 × 2 × 2 × 2 × (4×4) + 2 × 2 × 2 × 3 (4) Cancelling out the common factors in the numerator and the denominator, we have, 3 3×3 3 ℎ (2) = −2 × 2 × 2 × 2 (2×2×2×2) + 2 × 2 × 2 × 3 (2×2) Evaluate using order of operations. 3 ℎ (2) = −3 × 3 + 2 × 3 × 3 3 ℎ (2) = −9 + 18 = 9 Now, we have, −𝑏 2𝑎 3 −𝑏 = 4 and ℎ ( 2𝑎 ) = 9 . 3 Therefore, the vertex is (4 , 9) So, it took 0.75 seconds for the ball to reach the ground. Conclusion: The maximum height of the ball is 9 feet and it took 3 4 or 0.75 seconds for the ball to reach the ground. Question #8 Use a Double or Half-Angled Formula to solve the equation in the interval [0,2 \pi) . 1 sin 𝜃 − cos 𝜃 = 2 Answer: Approach: The range of the trigonometric functions of sin and cos are lie between [-1,1]. No solution exists beyond this range. Simplify the equation. Obtain the factors of the equation. Use sine Double-Angled formulas, sin 2 𝑥 = 2 sin 𝑥 cos 𝑥 Cosine and sine functions has period of 2 𝜋 , thus find the solution in any interval of length 2 𝜋. Cosine function is positive in first and fourth quadrant. SIne function is positive in first and second quadrant. Calculation: Consider the equation. 1 sin 𝜃 − cos 𝜃 = 2 Squaring both sides in above equation, 1 2 (sin\ 𝑡ℎ𝑒𝑡 − cos 𝜃 )2 = ( ) 2 1 cos 2 𝜃 + sin2 𝜃 − 2 cos 𝜃 sin 𝜃 = 4 1 − 2 cos 𝜃 sin 𝜃 = 1 4 Use doubled-Angled the equation, 3 sin 2 𝜃 = 4 Taking sine inverse both sides, 3 sin−1 sin 2 𝜃 = sin−1 (4) 3 2𝜃 = sin−1 ( ) 4 2 𝜃 ≈ 0.85 𝜃 ≈ 0.43 The solution of the equation is obtained by adding in the integer multiples of 𝜋 , 𝑃𝑆𝐾 𝜃 ≈ 0.43 + 𝑘 𝜋 Substitute 𝑘 = 0,1,2 𝜃 ≈ 0.43 + 0 = 0.43 𝜃 ≈ 0.43 + 𝜋 = 3.56 𝜃 ≈ 0.43 + 2 𝜋 = 1.15 From the above obtained solutions only 1.15 and is 3.56 satisfying the equation, so these are the only solutions of the equation. 1 Therefore, the solution of the trigonometry equation sin 𝜃 − cos 𝜃 = 2 in the interval [0,2𝜋) is 𝜃 = 1.15, 3.56 . Conclusion: 1 Hence, the solution of the trigonometry equation sin 𝜃 − cos 𝜃 = 2 in the interval [0,2𝜋) is 𝜃 = 1.15,3.56. Question #9 The binomials (𝑥 − 2) and (2𝑥 + 3) are the factors of which polynomial? A. 2𝑥 2 − 6 B. 2𝑥 2 − 𝑥 − 6 C. 2𝑥 2 + 𝑥 − 6 D. 2𝑥 2 + 7𝑥 − 6 Answer: Step 1 NSL We have to find the polynomial which the binomials (𝑥 − 2) and (2𝑥 + 3) are the factors. Then, The polynomial is Step 2 = (𝑥 − 2)(2𝑥 + 3) = 2𝑥 2 + 3𝑥 − 4𝑥 − 6 = 2𝑥 2 − 𝑥 − 6 Hence , Polynomial is 2𝑥 2 − 𝑥 − 6 Answer = (B) Question #10 To solve a rational inequality, we factor the numerator and the denominator into irreducible factors. The cut points are the real __________ of the numerator and the real ________ denominator. Then we find the intervals determined by the _________ ________, and we use test points to find the sign of the rational function on each interval. Let (𝑥+2)(𝑥−1) 𝑟(𝑥 ) = (𝑥−3)(𝑥+4) Fill in the diagram below to find the intervals on which 𝑟(𝑥) ≥ 0. Sign of x+2 x −1 x −3 x+4 ( x + 2)( x − 1) ( x − 3)( x + 4) −4 −2 1 3 From the diagram we see that 𝑟(𝑥 ) ≥ 0 on the intervals _______, _______, and _______. Answer: To solve a we factor the numerator and the denominator into irreducible factors. The cut points are the real zeros of the numerator and the real zeros denominator. Then we find the intervals determined by the cut points, and we use test points to find the sign of the rational function on each interval. (𝑥+2)(𝑥−1) Let 𝑟(𝑥 ) = 𝑟(𝑥 ) = (𝑥−3)(𝑥+4) Set the numerator equal to zero and solve for x. (𝑥 + 2)(𝑥 − 1) = 0 ⇒ 𝑥 + 2 = 0 and 𝑥 − 1 = 0 ⇒ 𝑥 = −2 and 𝑥 = 1 Set the denominator equal to zero and solve for x. (𝑥 − 3)(𝑥 + 4) = 0 ⇒ 𝑥 − 3 = 0 and 𝑥 + 4 = 0 ⇒ 𝑥 = 3 and 𝑥 = −4 Therefore, the cut points are 𝑥 = −4, 𝑥 = −2, 𝑥 = 1 and 𝑥 = 3 . Sign of x+2 x −1 x −3 x+4 ( x + 2)( x − 1) ( x − 3)( x + 4) (−, −4) (−4, −2) (−2,1) (1,3) (3, ) − − + + + − − − + + − − − − + − + + + + + − + − + From the diagram, we see that 𝑟(𝑥 ) ≥ 0 on the intervals (−∞, −4), (−2,1) and (3, ∞). Question #11 Let M,N and K be matrices of type 3 × 2, 2 × 3 and 3 × 3 respectively, such that det (MN) = 2 and 𝑑𝑒𝑡(𝐾 ) = 6 . Then which of the following is the value of the det [(−3𝑀𝑁)𝑇 𝐾 −1 ]−1 ? −1 a) −32 . 2 b) 33 .2 c) 2 32 d) −32 4 e) −1 32 Answer: Step 1 Determinant properties help to find the determinant of the given matrix. Determinant property for the multiples helps to do the required, which is defined as |𝑘 ⋅ 𝐴|%𝑛 = 𝑘 𝑛 ⋅ |𝐴|𝑛 . Here k is the real number whereas A is the square matrix. The determinant of the matrix is equal to the determinant of its transpose. Step 2 Apply the determinant properties for the multiples to find the required answer. Apply the determinant properties for the product of matrices, which is defined as |𝐴 ⋅ 𝐵| = |𝐴| ⋅ |𝐵| . Put the values of the determinants in equation (1) and solve. 𝑇 𝑑𝑒𝑡[(−3𝑀𝑁)𝑇 𝐾 −1 ]−1 = (−3)−1 (𝑑𝑒𝑡(𝑀𝑁)) − 1𝑑𝑒𝑡(𝐾 )−1 − 1 = −13(𝑑𝑒𝑡(𝑀𝑁)𝑇) − 1𝑑𝑒𝑡(𝐾 ) … … … … … … … . . (1) = −13(2) − 1(6) = −13126 = −1 Question #12 Suppose 𝜆 is an eigenvalue of the 𝑛 × 𝑛 matrix A. If u and v are eigenvectors of A corresponding to 𝜆 , which of the following statement MUST be false? 1) 𝜆 = 0 2) u and v are scalars multiple of each other. 3) nullity (𝐴 − 𝜆𝐼 ) = 0 4) 7𝑣 − 3𝑢 is also an eigenvector of A. Answer: Step 1 In the question we have to find falsa statement. Step 2 Result - option d Given that u and v are eigen vectors of 𝐴(𝑛 × 𝑛) matrix. a) 𝜆 = 0 - true eigen values can be zero or non zero or positive or negative too. b) u and v are scalar multiple of each other - true because eigen vactor are written as [1,3]t or any quantity which implies eigen vectors are scalar multiples. c) nullity 𝐴 − 𝜆𝐼 = 0 - true as eigen values are found by putting eigen values in this form. d) 7𝑣 − 3𝑢 is also a eigen vector of matrix A - false because eigen vectors can not be calculated by taking sum, difference or any calculus. Question #13 To find: The partial fraction decomposition for the rational expression 11−2𝑥 𝑥 2 −8𝑥+16 . Answer: Steps to solve: Partial fraction decomposition of 𝑓(𝑥) 𝑔(𝑥) : To form a partial fraction decomposition of a rational expression, follow these steps. 𝑓(𝑥) 1. If 𝑔(𝑥) is not a proper fraction (a fraction with the numerator of lesser degree than the denominator), divide f(x) by g(x). 2. Factor the denominator g(x) completely into factors of the form (𝑎𝑥 + 𝑏)𝑚 or (𝑐𝑥 2 + 𝑑𝑥 + 𝑒)𝑛 , where 𝑐𝑥 2 + 𝑑𝑥 + 𝑒 is irreducible and m and n are positive integers. 3. For each repeated linear factor (𝑎𝑥 + 𝑏)𝑚 , the decomposition must include the terms 𝐴1 𝑎𝑥+𝑏 𝐴 2 + (𝑎𝑥+𝑏) 2+ 𝐴 𝑚 ⋯ + (𝑎𝑥+𝑏) 𝑚 . 4. Use algebraic techniques to solve for the constants in the numerators or the decomposition. Consider the rational expression. 11−2𝑥 𝑥2−8𝑥+16 The factor of 𝑥 2 − 8𝑥 + 16 = (𝑥 − 4)(𝑥 − 4) = (𝑥 − 4)2 As the denominator is a repeated linear factor (𝑥 − 4)2 , the decomposition must include the terms 𝐴 (𝑥−4) 𝐵 + (𝑥−4)2 . The rational expression becomes 11−2𝑥 𝑥 2 −8𝑥+16 𝐴 𝐵 = (𝑥−4) + (𝑥−4)2 ⇒ (1) Taking least common multiples on both the sides of the denominator, 11−2𝑥 𝑥 2 −8𝑥+16 = (𝑥−4)𝐴+𝐵 (𝑥−4)2 11 − 2𝑥 = (𝑥 − 4)𝐴 + 𝐵 ⇒ (2) Put 𝑥 = 4 to get the value of B from equation(2). 11 − 2(4) = (4 − 4)𝐴 + 𝐵 3=𝐵 Put 𝑥 = 0 to get the value of A from equation (2) 11 − 2(0) = (0 − 4)𝐴 + 𝐵 11 = −4𝐴 + 𝐵 ⇒ (3) Now substitute the value of B in equation (3) 11 = −4𝐴 + (3) −4𝐴 = 8 𝐴 = −2 Substituting the values of A and B in equation (1), 11−2𝑥 𝑥 2 −8𝑥+16 −2 3 = (𝑥−4) + (𝑥−4)2 11−2𝑥 Therefore, the partial fraction decomposition of the rational expression 𝑥 2−8𝑥+16 is 3 (𝑥−4)2 2 − (𝑥−4). Question #14 Perform the indicated operation, and simplify if possible. Assume any factors you cancel are not zero. 5 8 + 𝑥 𝑥2 3 5 − 𝑥2 𝑥 Answer: = 5𝑥+8 ) 𝑥2 3−5𝑥 𝑥 2( 2 ) 𝑥 𝑥 2( { multiplying num of deno by x 2 } {∴ 𝑥 ≠ 0} 5𝑥+8 = 3−5𝑥 Question #15 Write expression as a sum and/or difference or logarithms, express powers as factors: log 3(𝑥 4 /2 − 𝑥 ) Answer: Step 1 Use the properties of logarithms to write the expression as a sum/or difference of logarithms and express the powers as factors as follows. 𝑥4 log 3 (2−𝑥) = log 3 𝑚 𝑥 4 − log 3(2 − 𝑥 ) (∵ log 𝑎 ( 𝑛 ) = log𝑎 𝑚 − log 𝑎 𝑛) Step 2 Answer: The expression can be rewritten as 4 log 3 𝑥 − log 3(2 − 𝑥 ) . Question #16 Find all the zeros, real and nonreal, of the polynomial and use that information to express p(x) as a product of linear factors. 𝑝(𝑥 ) = 𝑥 3 + 11𝑥 Answer: Step 1 𝑝(𝑥 ) = 𝑥 3 + 11𝑥 Step 2 Find zeroes, 𝑝(𝑥 ) = 0 ⇒ 𝑥 3 + 11𝑥 = 0 ⇒ 𝑥(𝑥 2 + 11) = 0 ⇒ 𝑥 = 0, 𝑥 2 + 11 = 0 ⇒ 𝑥 = 0, 𝑥 = ±√11𝑖 linear factors are x, (𝑥 − √11𝑖 )(𝑥 + √11𝑖 ) ⇒ 𝑝(𝑥) = 𝑥(𝑥 − √11𝑖)(𝑥 + √11𝑖 ) Question #17 What are the factors of 𝑚2 − 12𝑚 + 20 ? 1) 𝑚 − 14 and 𝑚 − 6 2) 𝑚 − 10 and 𝑚 − 2 3) 𝑚 − 9 and 𝑚 − 3 4) 𝑚 − 5 and 𝑚 − 4 Answer: 1) Given: log5 (25𝑥) since |𝑙𝑜𝑔𝑐 (𝑎𝑏) = log 𝑐 (𝑎) + log 𝑐 (𝑏)) therefore log 5(25𝑥) = log 5(25) + log5 (25) and we have log 5 (25) = 2 so, 𝑙𝑜𝑔5 (25𝑥) = 2 + 𝑙𝑜𝑔5 𝑥 𝑥 2) Given: log3 9 since we have log𝑐 𝑥 therefore log 3 9 𝑎 𝑏 = log𝑐 (𝑎) − log𝑐 (𝑏) = log 3(𝑥 ) − log3 (9) log 3(9) = 2 log 3 𝑥 9 = log 3(𝑥 ) − 2 3) Given: log2 𝑧 3 we have 𝑙𝑜𝑔𝑐 𝑚𝑛 = 𝑛log 𝑐 𝑚 then log 2 𝑧 3 = 3 log2 𝑧 Question #18 Write each expression as a sum and/or difference of logarithms. Express powers as factors. log 5(25𝑥) Answer: Given: log 5(25𝑥 ) since |𝑙𝑜𝑔𝑐 (𝑎𝑏) = log 𝑐 (𝑎) + log 𝑐 (𝑏)) therefore log 5(25𝑥) = log 5(25) + log5 (25) and we have log 5(25) = 2 so, 𝑙𝑜𝑔5 (25𝑥) = 2 + 𝑙𝑜𝑔5 𝑥 Question #19 Write each expression as a sum and/or difference of logarithms. Express powers as factors. log 3 𝑥 9 Answer: Given: log 3 𝑥 9 since we have log𝑐 therefore log 3 𝑥 9 𝑎 𝑏 = log𝑐 (𝑎) − log𝑐 (𝑏) = log 3 (𝑥) − log 3(9) log 3(9) = 2 log 3 𝑥 9 = log 3(𝑥 ) − 2 Question #20 Write each expression as a sum and/or difference of logarithms. Express powers as factors. log 2 𝑧 3 Answer: Given: log 2 𝑧 3 we have log 𝑐 𝑚𝑛 = 𝑛 log 𝑐 𝑚 then log 2 𝑧 3 = 3 log2 𝑧