Mathematics · May / June 2023 · Paper 31 · Question 2
YMA01
As printed in the paper
The question as text
2. log6 T (0, 4) O log6 x (2, 0) Figure 1 Figure 1 shows the linear relationship between log6 T and log6 x The line passes through the points (0, 4) and (2, 0) as shown.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
2(a)[3]
6
6
E.g.
log
T
=
4
−
2log 216
lo
g
T
=
4
−
2
×
3
= −
2
T
=
.
.
.
(ii)
6
6
6
1
−
2
T
=
6
=
36
4 2log
−
x
log
T
=
4
−
2log
x
T
=
6
6
6
6
−
2
4
log
x
T
=
6
×
6
6
1296
T
=
2
x
Notes
Mark the question as a whole. Do not be concerned about part labelling.
(a)(i)
B1
Correct linear equation
log
T
=
4
−
2log
x
(oe) The 4 may be written as
log 1296
6
6
6
(ii)
M1
Substitutes
x =
216
into an equation linking T and x arising from a linear equation in the
logarithms and proceeds to make T the subject. They may have answered (b) first. Do not be
concerned about the process for this mark. May be implied by awrt 0.028 following a correct
equation.
1
A1 Correct value
T =
. Do not accept
6−.
2
36
(b)
M1
Makes a first step towards achieving an answer. Use of a correct log rule or law applied at some
stage in their attempt to eliminate logs from the equation.
As a rule of thumb this can be awarded for e.g.
''2''
''4''
4
application of a power rule−
''2''log
x
= −
log
x
or
''4''
=
log 6
or
4
→
6
(note that e.g.
6
6
6
−
2
4
log
T
= −
2log
x
+
4
→
x
+
6
implies this mark)
6
6
''4'' ''2''log
−
x
an attempt to make T the subject. E.g.log
T
=
''4'' ''2''log
−
x
T
=
6
6
6
6
dM1
Full and complete method in proceeding from an equation of form
log
T
=
a
+
b
log
6
±
n
to an equation of form
T
=
k
×
x
or equivalent. All log work must be correct but allow slips on
coefficients.
1296
2
A1
Achieves
T
=
or equivalent such as
Tx =
1296
and isw after a correct answer. Allow 64 for
x
2
1296.
Note: Allow the M marks if a different letter than T is used, e.g. y. But must be correct in terms of T and x
for the A mark.
)
d
(
1
2(b)[3]
6
6
−
2
4
log
x
T
=
6
×
6
6
1296
T
=
2
x
Notes
Mark the question as a whole. Do not be concerned about part labelling.
(a)(i)
B1
Correct linear equation
log
T
=
4
−
2log
x
(oe) The 4 may be written as
log 1296
6
6
6
(ii)
M1
Substitutes
x =
216
into an equation linking T and x arising from a linear equation in the
logarithms and proceeds to make T the subject. They may have answered (b) first. Do not be
concerned about the process for this mark. May be implied by awrt 0.028 following a correct
equation.
1
A1 Correct value
T =
. Do not accept
6−.
2
36
(b)
M1
Makes a first step towards achieving an answer. Use of a correct log rule or law applied at some
stage in their attempt to eliminate logs from the equation.
As a rule of thumb this can be awarded for e.g.
''2''
''4''
4
application of a power rule−
''2''log
x
= −
log
x
or
''4''
=
log 6
or
4
→
6
(note that e.g.
6
6
6
−
2
4
log
T
= −
2log
x
+
4
→
x
+
6
implies this mark)
6
6
''4'' ''2''log
−
x
an attempt to make T the subject. E.g.log
T
=
''4'' ''2''log
−
x
T
=
6
6
6
6
dM1
Full and complete method in proceeding from an equation of form
log
T
=
a
+
b
log
6
±
n
to an equation of form
T
=
k
×
x
or equivalent. All log work must be correct but allow slips on
coefficients.
1296
2
A1
Achieves
T
=
or equivalent such as
Tx =
1296
and isw after a correct answer. Allow 64 for
x
2
1296.
Note: Allow the M marks if a different letter than T is used, e.g. y. But must be correct in terms of T and x
for the A mark.
)
d
(
1
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