Mathematics · Oct / Nov 2020 · Paper 21 · Question 9
YMA01
As printed in the paper
Show the answer and mark scheme
9(i)[4]
4
=
log 81
4
=
log 3
4
or
3
3
x
+
5
x
+
5
(
)
May be implied by e.g.
log
=
4
⇒
=
3
4
or 81
3
2
x
−
1
2
x
−
1
Examples:
x
+
5
(
)
log
x
+
5
−
log 81
=
log
3
3
3
81
or
x
+
5
(
)
(
)
log
x
+
5
−
log
2
x
−
1
=
log 2
3
3
3
x
−
1
or
log
(
2
x
1
)
log 81
log 81 2
(
x
1
)
−
+
=
−
3
3
3
This mark is for combining 2 log terms correctly and can be awarded following an incorrect
rearrangement e.g.
log
(
x
5
)
4
log
(
2
x
1
)
log
(
x
5
)
log
(
2
x
1
)
4
+
−
=
−
⇒
+
+
−
=
3
3
3
3
(
)(
)
⇒
log
2
x
−
1
x
+
5
=
...
3
Obtains this equation in any form e.g.
x
5
+
5
=
2
x
−
1
x
+
=
3
4
81
2
x
−
1
86
x =
Cao
161
Condone the omission of the base throughout
Alternative for first 3 marks:
log
(
x
5
)
4
log
(
2
x
1
)
3
log
(
x
+
5
)
−
4
3
log
(
2
x
−
1
)
+
−
=
−
⇒
3
=
3
3
3
5
x
+
(
)
⇒
3
log
x
+
5
×
3
−
4
=
2
x
−⇒
1
=
2
x
−
1
3
81
Score B1 for sight of 34 and M1 for applying 3
a b
±
=
3
a
×
3
±
b
and A1 as above
(a) Special Case
log
(
5
)
x
+
x
+
5
86
(
)
(
)
log
x
+
5
−
log
2
x
−
1
=
4
⇒
3
=
4
⇒
=
81
⇒
x
=
3
3
log
(
2
x
1
)
2
x
1
161
−
−
3
Scores B1(implied) M0 A0 A1
+ =
9(ii)(a)[4]
3
y
3
3
y
×
3
3
or
One correct index law seen or implied
2
2
anywhere in their working
2
1 2
−
y
=
2 2
×
−
2
y
or
or 2 4
×
−
y
or
2
2
y
4
y
Applies both correct index laws to the lhs
3
y
+
3
×
2
1 2
−
y
=
27 3
×
y
× ×
2 2
−
2
y
=
...
of the equation
108
or 3
y
108
3
y
108
3
y
2
−
2
y
y =
×
=
or
y =
27 2
×
4
27 2
×
2
2
54
Isolates the terms in y (as powers of 3 and 2(or 4)) on the lhs and the constants on the rhs.
3
y
×
3
3
=
27
y
There must be no incorrect work to combine terms e.g.
etc.
Cso. Reaches the given answer with no
y =
(
)
0.75
2
errors and all steps shown with 22y
*
appearing as 4y at some point.
Alternative 1 for (ii)(a) using logs:
(
)
log 3
y
+
3
×
2
1 2
−
y
=
log3
y
+
3
+
log 2
1 2
−
y
One correct log law seen or implied
+ =
(
)
Or
log3
y
3
y
+
3 log3
anywhere in their working. No bracketing
errors allowed for this mark.
(
)
Or
log 2
1 2
−
y
=
1 2
−
y
log 2
3
1 2
(
)
(
)
log3
y
+
+
log 2
−
y
=
y
+
3 log3
+
1 2
−
y
log 2
Applies the correct log laws to the lhs.
You can condone missing brackets around the y + 3 and/or 1 – 2y
(
)
(
)
y
+
3 log3
+
1 2
−
y
log 2
=
log108
⇒
log3
y
−
log 2
2
y
=
log108
−
3log3
−
log 2
3
y
108
log
log
⇒
=
2
3
2
y
3
×
2
Proceeds to isolate the terms in y on the lhs and combines the constants on the rhs
or e.g.
108
(
)
(
)
(
)
y
+
3 log3
+
1
−
2
y
log 2
=
log108
⇒
y
log3
−
2log 2
=
log 3
3
2
×
Proceeds to isolate the terms in y on the lhs and combines the constants on the rhs
y =
Cso. With e.g. 2log 2 seen as log 4 or log 22
(
)
0.75
2
*
or implied at some point.
Alternative 2 for (ii)(a) using factors of 108:
One correct index law seen or implied
3
1 2
2
3
3
y
+
×
2
−
y
=
108
=
2
×
3
3
y
3
+
=
3
y
3
y
+
3
×
2
1 2
−
y
anywhere in their working e.g.
or
3
3
⇒
=
...
2
2
×
3
3
2
1 2
−
y
3
y
2
−−
1 2
y
...
=
2
−−
1 2
y
⇒
×
=
2
2
Applies both correct index laws to the lhs of
3
y
2
−−
1 2
y
...
⇒
×
=
the equation
Proceeds to isolate the terms in y (as powers
3
y
2
−
2
y
2
of 3 and 2(or 4)) on the lhs and the
⇒
×
=
constants on the rhs
Cso. Reaches the given answer with no
y =
(
)
0.75
2
*
errors and all steps shown with 22y
appearing as 4y at some point.
9(ii)(b)[2]
log 2
(
0.75
)
y
2
y
=
⇒
=
log0.75
Correct processing to obtain a value for y
or
May be implied by awrt – 2.4
(
)
y
0.75
=
2
⇒
y
=
log
2
0.75
Awrt 2.409
−
y = −
2.409
A correct answer implies both marks
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