1.
2 (ln x)2  3 ln x + 1 = 0
Attempting to factorize or using the quadratic formula
1
ln x = , ln x  1
2
x e , xe
(A1)
(M1)
A1A1
A1A1
N2
[6]
2.
METHOD 1
x2
x2
x2
 ln 2  ln 3 + …
ln x2 + ln
y
y
y
= ln x2 + (ln x2 – ln y) +(ln x2 – 2 ln y) + (ln x2 – 3 ln y) +…
n
35
S35 = (2u1 +(n – 1)d) =
(2 ln x2 – 34 ln y) = 35 ln x2 – 595 ln y
2
2
= ln x70 – ln y595
x 70
= ln 595 (Accept m = 70, n = 595)
y
METHOD 2
x2
x2
x2
x 2 x 2 ...x 2
 ln 2  ln 3  ...  ln
ln x2 + ln
y
y
y
1y...y 34
In the denominator, the sum of the powers of y is (0 + 34)
(M1)
(M1)(A1)
(A1)(A1) (N2)
(M1)(A1)
35
= 595
2
The required expression is
x 70
ln 595 (Accept m = 70, n = 595)
y
(A1)
(A1)(A1) (N2)
[5]
3.
Finding two equations
Correct equations ln (x + 3) = 1, ln (x + 3) =  1
ln (x + 3) = 1  x = e  3
M1
(A1)
(M1)A1
N2
1
e
(M1)A1
N2
ln (x + 3) = 1  x   3
Note:
 e
1

3
Award M1A0 for decimal answer.
[6]
4.
Taking logs, x ln 3 + (2x + 1) ln 4 = (x + 2) ln 6
x(ln 3 + 2 ln 4 – ln 6) = 21n6 – ln 4
2 ln 6 – ln 4
ln 3  2 ln 4 – ln 6
ln 9 
ln81

or equivalent 
=
 Accept
ln 8 
ln64

x=
(or a = 9, b = 8)
(M1)(A1)
(A1)
(A1)
(M1)(A1)
(C3)(C3)
[6]
1
5.
22x + 3  2x+1  3 = 0
Let p = 2x
8p2  2p  3 = 0
(2p + 1) (4p  3) = 0
1
3
p   or p 
2
4
3
2x 
4
x = log2 3  log2 4
= 2 + log2 3
(a = 2, b = 3 )
Note: Award no marks if candidates take
log2 of each term, even though this
gives the correct answer.
(M1)
A1
A1
A1
A1A1
N0
[6]
2