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# test review ex log (1)

advertisement ```test review, ex, log [74 marks]
1.
Solve (ln x)2 − (ln 2) (ln x) < 2(ln 2)2.
[6 marks]
Markscheme
(ln x)2 − (ln 2) (ln x) − 2(ln 2)2 (= 0)
EITHER
ln x =
=
ln 2± √(ln 2)2+8(ln 2)2
M1
2
ln 2± 3 ln 2
2
A1
OR
(ln x − 2 ln 2) (ln x + 2 ln 2) (= 0)
M1A1
THEN
ln x = 2 ln 2 or −ln 2
⇒ x = 4 or x =
1
2
A1
(M1)A1
Note: (M1) is for an appropriate use of a log law in either case, dependent on the previous M1 being awarded, A1 for both correct
answers.
solution is
1
2
<x<4
A1
[6 marks]
1
+
2a. Show that logr2x = 2 logr x where r, x ∈ R .
Markscheme
METHOD 1
logr2x =
=
logr x
logr r2
logr x
(=
logr x
2 logr r
AG
2
[2 marks]
METHOD 2
logr2x =
1
logx r2
=
1
2 logx r
A1
=
logr x
2
[2 marks]
AG
M1
)
M1A1
[2 marks]
It is given that log2 y + log4 x + log4 2x = 0.
q
2b. Express y in terms of x. Give your answer in the form y = px , where p , q are constants.
[5 marks]
Markscheme
METHOD 1
log2 y + log4 x + log4 2x = 0
log2 y + log4 2x2 = 0
M1
log2 y + 1 log2 2x2 = 0
M1
2
log2 y = − 1 log2 2x2
2
log2 y = log2 (
y=
1 −1
x
√2
1
)
√2x
M1A1
A1
Note: For the final A mark, y must be expressed in the form pxq.
[5 marks]
METHOD 2
log2 y + log4 x + log4 2x = 0
log2 y + 1 log2 x + 1 log2 2x = 0
2
2
1
2
1
2
log2 y + log2 x + log2 (2x) = 0
log2 ( √2xy) = 0
√2xy = 1
y=
1 −1
x
√2
M1
M1
M1
A1
A1
Note: For the final A mark, y must be expressed in the form pxq.
[5 marks]
2c.
The region R, is bounded by the graph of the function found in part (b), the x-axis, and the lines x = 1 and x = α where α > 1.
The area of R is √2.
Find the value of α.
Markscheme
α
∫
the area of R is 1
=[
=
α
1
ln x]
√2
1
1
ln α
√2
1
ln α
√2
α = e2
1 −1
x dx
√2
M1
A1
A1
= √2
M1
A1
Note: Only follow through from part (b) if y is in the form y = pxq
[5 marks]
[5 marks]
3.
Solve the equation log2(x + 3) + log2(x − 3) = 4.
Markscheme
log2(x + 3) + log2(x − 3) = 4
log2(x2 − 9) = 4
x2 − 9
x2
=
24
(= 16)
= 25
x = ±5
x=5
(A1)
A1
[5 marks]
(M1)
M1A1
[5 marks]
4.
Find the solution of log2x − log25 = 2 + log23.
Markscheme
log2x − log25 = 2 + log23
collecting at least two log terms
eg
log2 x5
= 2 + log23 or
log2 15x
(M1)
=2
obtaining a correct equation without logs
eg
x
5
= 12 OR
x = 60
x
15
=
22
(A1)
A1
[4 marks]
(
)
(M1)
[4 marks]
5.
Given that log10 (
1
(p + 2q))
2√2
= 1 (log10p + log10q) , p > 0, q > 0, find p in terms of q.
2
Markscheme
log10
1
(p + 2q)
2√2
= 12 (log10p + log10q)
log10
1
(p + 2q)
2√2
= 12 log10pq
log10
1
(p + 2q)
2√2
1
(p + 2q)
2√2
= log10(pq)
= (pq)
1
2
(A1)
(p + 2q)2 = 8pq
p2 + 4pq + 4q 2 = 8pq
p2 − 4pq + 4q 2 = 0
(p − 2q)2 = 0
M1
hence p = 2q
A1
[5 marks]
+
(M1)
1
2
(M1)
[5 marks]
6.
Solve the equation 4x + 2x+ 2 = 3.
[5 marks]
Markscheme
attempt to form a quadratic in 2x
(2x)2 + 4 ∙ 2x − 3
2 =
x
− 4± √16+ 12
2
=0
M1
A1
(= −2 ± √7)
M1
2x = −2 + √7 (as − 2 − √7 < 0)
x = log2 (−2 + √7) (x =
R1
ln(− 2+ √7)
ln2
)
A1
Note: Award R0 A1 if final answer is x = log2 (−2 + √7) .
[5 marks]
7.
Consider
a = log23 × log34 × log45 × … × log3132. Given that
a ∈ Z, find the value of a.
[5 marks]
Markscheme
log3
log 2
log 4
log3
×
×…×
=
log32
log 2
A1
=
5log 2
log 2
(M1)
=5
log32
log31
M1A1
A1
hence
a=5
Note:
log2x.
Accept the above if done in a specific base eg
[5 marks]
8.
Solve the equation
8x− 1 = 63x. Express your answer in terms of
ln 2 and
ln 3.
[5 marks]
Markscheme
METHOD 1
23(x− 1) = (2 × 3)3x
Note:
M1
Award M1 for writing in terms of 2 and 3.
23x × 2−3 = 23x × 33x
2−3 = 33x
A1
−3
ln(2 ) = ln(33x)
(M1)
−3 ln 2 = 3x ln 3
x=
− ln2
ln 3
A1
A1
METHOD 2
ln 8x− 1 = ln 63x
(M1)
3
(x − 1) ln 2 = 3x ln(2 × 3)
M1A1
3x ln 2 − 3 ln 2 = 3x ln 2 + 3x ln 3
x=
− ln2
ln 3
A1
A1
METHOD 3
ln 8x− 1 = ln 63x
(M1)
(x − 1) ln 8 = 3x ln 6
x=
x=
ln 8
ln 8−3 ln 6
3 ln2
ln( 23 )
3
A1
A1
M1
6
x = − ln23
ln
A1
[5 marks]
9.
The first terms of an arithmetic sequence are
1
, 1 , 1 , 1 ,
log2x log8x log32x log128x
…
Find x if the sum of the first 20 terms of the sequence is equal to 100.
[6 marks]
Markscheme
METHOD 1
d=
=
1
log8x
log28
log2x
−
−
1
log2x
1
log2x
(M1)
(M1)
Note: Award this M1 for a correct change of base anywhere in the question.
=
2
log2x
(A1)
20
(2 × log1 x
2
2
=
400
log2x
100 =
2
)
log2x
+ 19 ×
M1
(A1)
400
log2x
log2x = 4 ⇒ x = 24 = 16
A1
METHOD 2
20th term =
1
log239x
100 =
20
( log1 x
2
2
+
100 =
20
( log1 x
2
2
+
A1
1
)
log239x
log2239
log2x
M1
)
M1(A1)
Note: Award this M1 for a correct change of base anywhere in the question.
100 =
400
log2x
(A1)
log2x = 4 ⇒ x = 24 = 16
A1
METHOD 3
1
log2x
1
log2x
+
1
log8x
+
log28
log2x
+
1
log32x
+
log232
log2x
+
1
log128x
+…
+
log2128
+…
log2x
(M1)(A1)
Note: Award this M1 for a correct change of base anywhere in the question.
=
1
(1 + 3 + 5 + …)
log2x
=
1
( 202 (2 + 38))
log2x
100 =
A1
(M1)(A1)
400
log2x
log2x = 4 ⇒ x = 24 = 16
A1
[6 marks]
10. Solve the equation
2 − log3(x + 7) = log 1 2x .
3
[5 marks]
Markscheme
9
log3 ( x+7
) = log3 21x
M1M1A1
Note: Award M1 for changing to single base, M1 for incorporating the 2 into a log and A1 for a correct equation with maximum one log
expression each side.
x + 7 = 18x
x=
7
17
M1
A1
[5 marks]
11.
Let
f(x) = ln x . The graph of f is transformed into the graph of the function g by a translation of
3
(
), followed by a reflection in the x-axis. Find an expression for
−2
g(x), giving your answer as a single logarithm.
[5 marks]
Markscheme
h(x) = f(x − 3) − 2 = ln(x − 3) − 2
g(x) = −h(x) = 2 − ln(x − 3)
(M1)(A1)
M1
Note: Award M1 only if it is clear the effect of the reflection in the x-axis:
the expression is correct OR
there is a change of signs of the previous expression OR
there’s a graph or an explanation making it explicit
= ln e2 − ln(x − 3)
= ln( x−3 )
e2
M1
A1
[5 marks]
12.
Solve the following system of equations.
[6 marks]
logx+ 1y = 2
logy+ 1x =
1
4
Markscheme
logx+ 1y = 2
logy+ 1x =
1
4
so
(x + 1)2 = y
A1
1
4
(y + 1) = x
A1
EITHER
x4 − 1 = (x + 1)2
M1
x = −1, not possible
R1
x = 1.70,
y = 7.27 A1A1
OR
1
1
(x2 + 2x + 2) 4 − x = 0
M1
attempt to solve or graph of LHS
M1
x = 1.70,
y = 7.27 A1A1
[6 marks]
13. Solve the equation
4x− 1 = 2x + 8.
[5 marks]
Markscheme
22x− 2 = 2x + 8
1 2x
2
4
= 2 +8
x
(M1)
(A1)
22x − 4 × 2x − 32 = 0
A1
(2x − 8)(2x + 4) = 0
(M1)
2 =8⇒x=3
x
A1
Notes: Do not award final A1 if more than 1 solution is given.
[5 marks]
© International Baccalaureate Organization 2018
International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®
Printed for Bahrain Bayan School
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