Mathematics · May / June 2023 · Paper 21 · Question 5
YMA01
As printed in the paper
The question as text
5. Use the laws of logarithms to solve log2 (16x) + log2 (x + 1) = 3 + log2 (x + 6) (5)
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
53
1
+
, but allow if some
3
of the terms of the binomial expansion are found and 243 is the constant term.
e.g. 243
+
405
+
...
Do not accept e.g. 5
×
243 (
×
)
0
for this mark.
px
C
px
0
(b)(i)
(
)
or (
)
M1
=
5 3
×
4
( 405 )
=
=
10 3
×
2
3
( 90
=
3
)
(simplified or unsimplified including
B
p
p
D
p
p
unprocessed indices).
The coefficients of B or D may be embedded within the binomial expansion such as
5 3
5 3
(
)
3
... 5 3
+ ×
4
px
+
... 10 3
+
×
2
px
. Also allow e.g.
4
or
2
3
or 5
13
4
p or 5
p
p
C
C
1
3
(
)
5 3
4
( 405 )
and (
)
10 3
2
3
( 90
3
)
A1
B
=
×
p
=
p
D
=
×
p
=
p
(simplified or unsimplified) Allow
this mark if the coefficients have not been isolated from x and
3x , but the brackets around
(
)
and
(
)
3
must be correctly removed. Binomial notation such as 5
or
5
C must have
px
px
C
1
3
been evaluated.
Uses "
"
18"
"
M1
B
=
D
to form a cubic equation in p and attempts to take out a factor of p or divide
by p leading to
p =
2
...
(or may be implied by their value of p). " "
=
"
"
is M0
B
D
Condone equations involving x and
3x as long as they are removed as they proceed to
Note that equations may have been simplified before they are formed by cancelling factors
5 3
×
2
e.g.
23 , 3, 5 or p so look carefully at equations which are acceptable such as
p
2
2
or
9
=
36p
p
=
...
(condone if negative)
3 2
scores M1A1M0A0 in (b)(i) (did not use "
"
=
18"
Note
405
p
=
90
p
3
p
= ±
B
2
They can proceed in (b)(ii) to score M1A0.
1
A1
p = −
only following correct coefficients for B and D (the positive root and
p =
0
2
1
be rejected if found / a clear choice of
p = −
) Watch
2
out for notation rejecting the positive root such as:
Can only be scored provided all previous marks have been awarded.
1
Note if the coefficients of B and D are incorrect leading to
p = −
then this is max
2
M0A0M1A0 in (b) and M1A0 in (c)
Alternative method (b)(i)
2
3
5 4
×
5 4 3
××
px
px
px
If the expansion is written as
then 243 is not required to
243 1 5
+
+
+
3
2!
3
3!
3
find p as this will cancel from both sides when setting "
B
"
=
18"
D
"
.
5 4 3
××
3
p
p
or (
)
M1
(
) (243 )5 3
=
(243 )
×
(unsimplified with or without the 243)
B
=
×
D
3!
3
5 4 3
××
3
p
p
and (
)
A1
(
) (243 )5 3
=
(243 )
×
(unsimplified either both with or both
B
=
×
D
3!
3
without the 243 included). Allow this mark if the coefficients have not been isolated from x and
3
px
px
3x , but the brackets around
and
must be correctly removed. Any binomial notation
3
3
would now have to be evaluated as in the main scheme notes.
M1
As above in main scheme notes
A1
As above in main scheme notes
(ii)
M1
Uses their value of
p , p or p and either attempts 5
2
3
3
2
to find a value for C, or attempts to
C
×
p
2
5
2 3
×
3
2
2
find
C
p x
and then proceeds to identify the value for C.
1
Condone values for p where x had not been removed from (i) e.g.
±
being substituted in
2 x
provided a value without x is identified/extracted in (ii)
Condone a slip in calculating 5
C and allow this mark if they substitute both values of p into the
2
expression and find two values for C.
The method mark may be implied by the value found using their p
A1 135
or 270
1
1
2 o.e. e.g. 67.5 or
67
1
4 following a correctly found
p = ±
or
p =
2
2
2
4
Condone where both values of p are used provided they both give the correct answer.
135
Do not accept e.g.
x unless the coefficient is clearly identified
2
2
5
log 16
x
+
log (
x
+
1)
=
3
+
log (
x
+
6)
2
2
2
log
2b
a
=
b
=
a
2
log 16 (
+
1)
=
log 8(
+
6)
x x
x
2
2
2
2
+
−
6
=
0
x
x
(2
−
3)(
+
2)
=
0
x
x
3
x =
only
2
Note a correct answer with no working is 0 marks
log 8
3
B1
Correct use of
log
a
=
b
2b
=
a
This will usually result from either
=
or
2
2
log 16
=
4
being stated or seen applied.
2
16 (
+
1)
16 (
+
1)
x x
=
x x
e.g.
log
3
=
8
2
x
+
6
x
+
6
(
1)
x x
+
= −
May be implied by e.g.
log 16 (
x x
+
1)
=
log 8(
x
+
6)
or
log
1
o.e.
2
2
2
+
6
x
M1
Applies the addition (or subtraction) laws of logs at least once correctly. Condone invisible
brackets. e.g.
log 16
2
16
x
+
x
2
This mark can be scored for expressions where the law has been applied correctly and does
+
6
x
not need to be a full equation e.g.
log
scores M1. Do not withhold this mark if they
2
x
+
1
subsequently apply a law incorrectly.
e.g.
still scores M1
log 16 (
x x
+
1)
=
3
+
log
x
+
log 6
2
2
2
Also allow the mark to be scored despite an error dealing with the 3 when writing it as a log
e.g.
3
+
log (
+
6)
=
log 6
+
log (
+
6)
=
log 6(
+
6)
x
x
x
2
2
2
2
A1
2
2
6
0
or equivalent 3TQ (quite often
16
2
8
48
0
) Terms must be collected but do
x
+
x
−
=
x
+
x
−
=
not need to be collected on the same side. Condone the omission of = 0 . Can only be scored from
correct working.
dM1
Attempts to solve their 3TQ via any valid method including via a calculator (you may need
to check this). Condone complex roots. It is dependent on the previous method mark.
A1cso 3
2 o.e. only ( 2
− must be rejected if found)
This mark can only be scored if a correct quadratic is achieved (following correct log work)
+
4
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