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10 Solve the equation (i) lg(2x) – lg(x – 3) = 1, [3] (ii) log3 y + 4logy 3 = 4. 9 [4] Solve (i) log4 2 ! log9 (2x ! 5) # log8 64, [4] (ii) 9y ! 5(3y 0 10) # 0. [4] Solve 8 (i) log3(2x ! 1) # 2 ! log3(3x 0 11), [4] (ii) log4 y ! log2 y # 9. [4] (a) Variables l and t are related by the equation l = l0 (1 + α)t where l0 and α are constants. 7 Given that l0 = 0.64 and α = 2.5 × 10 –3, find the value of t for which l = 0.66. (b) Solve the equation 1 + lg(8 – x) = lg(3x + 2). 9 [3] [4] (a) Given that u = log4 x, find, in simplest form in terms of u, (i) x, 16 (ii) log4 –– x , (iii) logx 8. [5] (b) Solve the equation (log3 y)2 + log3(y2) = 8. 5 (i) Express [4] as a power of 2. [1] (ii) Express (64) as a power of 2. [1] 32 1 x 1 x 8 (64) 1 (iii) Hence solve the equation . = x 32 2 [3] (i) Given that log9 x = a log3 x, find a. [1] (ii) Given that log27 y = b log3 y, find b. [1] (iii) Hence solve, for x and y, the simultaneous equations 6log9 x + 3 log27 y = 8, log3 x + 2 log9 y = 2. [4] 8 Solve the equation (i) 22x + 1 = 20, (ii) 5 5 4 y−1 25 y = [3] 125 y+3 25 2− y . [4] (a) Solve the equation 92x – 1 = 27x. (b) Given that 1 2 3 2 – 3 a– 2 b3 ab = a pbq, find the value of p and of q. [3] [2] 10 (a) Given that logp X = 6 and logp Y = 4, find the value of X2 (i) logp ––– , Y [2] (ii) logY X. [2] (b) Find the value of 2z, where z = 5 + log23. [3] (c) Express 512 as a power of 4. [2] 7 Given that logp X = 9 and logpY = 6, find (i) logp X , (ii) logp 6 () 1 , X [1] (iii) logp (XY ), [2] (iv) logY X. [2] Given that log8 p = x and log8 q = y, express in terms of x and/or y (i) log8 p + log8 q2, [2] q (ii) log8 – , 8 [2] (iii) log2 (64p). [3] 8 [1] (a) Solve the equation (23 – 4x) (4x + 4) = 2. [3] –– , giving your answer in the form k 3 , where k is an integer. (b) (i) Simplify 108 – 12 3 5 + 3 , giving your answer in the form a 5 + b, where a and b are integers. (ii) Simplify ––––– 5– 2 [2] [3] () n! . where n is a positive integer and r = (n – r)!r! 0606/1/M/J/02