10 (i) x –3)=1, [3] - WELCOME IGCSE | Dr. Tayeb's Websiteiii) Hence solve, for x and y, the...
Transcript of 10 (i) x –3)=1, [3] - WELCOME IGCSE | Dr. Tayeb's Websiteiii) Hence solve, for x and y, the...
10 Solve the equation
(i) lg(2x) – lg(x – 3) = 1, [3]
(ii) log3 y + 4logy 3 = 4. [4]
9 Solve
(i) log�4 2 ! log�9 (2x ! 5) # log�8 64, [4]
(ii) 9�y ! 5(3�y 0 10) # 0. [4]
8 Solve
(i) log�3(2x ! 1) # 2 ! log�3(3x 0 11), [4]
(ii) log�4 y ! log�2 y # 9. [4]
7 (a) Variables l and t are related by the equation l = l0(1 + α)t where l0 and α are constants.
Given that l0 = 0.64 and α = 2.5 × 10–3, find the value of t for which l = 0.66. [3]
(b) Solve the equation 1 + lg(8 – x) = lg(3x + 2). [4]
9 (a) Given that u = log4 x, find, in simplest form in terms of u,
(i) x,
(ii) log4� �,(iii) logx 8.
[5]
(b) Solve the equation (log3 y)2 + log3(y2) = 8. [4]
16––x
5 (i) Express 32
as a power of 2. [1]
(ii) Express ( )641x as a power of 2. [1]
(iii) Hence solve the equation ( )64
2
132
1x
x = . [3]
8 (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, [3]
(ii) 5
25
125
25
4 1 3
2
y
y
y
y
− +
−= . [4]
5 (a) Solve the equation 92x – 1 = 27x. [3]
(b) Given that
a b– 12
23
a b3 – 23
= a pbq, find the value of p and of q. [2]
10 (a) Given that logp X = 6 and logp Y = 4, find the value of
(i) logp � X2–––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 , [1]
(ii) logp ( 1X ) , [1]
(iii) logp (XY ), [2]
(iv) logY X. [2]
6 Given that log8 p = x and log8 q = y, express in terms of x and/or y
(i) log8 p + log8 q2, [2]
(ii) log8 �q–8�, [2]
(iii) log2 (64p). [3]
8 (a) Solve the equation (23 – 4x) (4x + 4) = 2. [3]
(b) (i) Simplify 108 – 12–– 3
, giving your answer in the form k 3 , where k is an integer. [2]
(ii) Simplify 5 + 3––––– 5 – 2
, giving your answer in the form a 5 + b, where a and b are integers. [3]
0606/1/M/J/02
where n is a positive integer and = .n!(n – r)!r!)r(