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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