Mathematics · Oct / Nov 2023 · Paper 21 · Question 5
YMA01
As printed in the paper
The question as text
5. In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
5(i)[2]
a
3
=
70
log3
=
log70
a
log70
a
=
=
3.867
log3
5(ii)[4]
3
or
log 5
b
=
log 5
+
log
b
B1
4
=
log 81
or
3log
b
=
log
b
3
3
3
3
3
3
3
4
+
3log
b
=
log 5
b
log 81
b
=
log 5
b
3
3
3
3
3
2
81b
=
5b
=
...
b
5
=
b
81
This question demands working from the candidates so answers cannot appear without appropriate methods
(i)
M1: Takes (the same) log of both sides and uses the power law.
log 70
Alternatively writes
a =
with the correct base. But
a =
log70
is M0
3
A1: CSO. 3.867 following a suitable intermediate line. This is not awrt but isw after sight of 3.867
Cannot be achieved via Trial and Improvement
(ii) Note that this was marked M1 A1 dM1 A1. It is now being marked B1 M1 dM1 A1
B1: One correct log law used on a term of the initial equation.
Can be achieved without the bases being present
3
log 5
log 5
log
Look for
4
=
log 81
or
3log
=
log
or
b
=
+
b
b
b
3
3
3
3
3
3
M1: Correctly combines two terms of the initial equation to form a log equation of the correct form
log ...
=
log ...
or
log ...
= ±
4
.
3
3
3
3
b
Possible answers are;
log 81
3
=
log 5
,
log 81
b =
2
log 5
,
log
= −
4
b
b
3
3
3
3
3
5
b
For this mark you can condone a slip on the sign of the 4.
3
b
3
E.g.
4
+
3log
b
=
log 5
b
log
b
−
log 5
b
=
4
log
=
4
3
3
3
3
3
5
b
dM1: Removes the logs correctly and proceeds to a value for b using a correct method.
log ...
=
log ...
log ...
= ±
4
Must follow M1 and an equation of the form
or
3
3
3
1
5
A1:
b =
or exact equivalent such as
b =
5
. The negative (and 0) solution, if found, must be discarded.
81
9
...............................................................................................................................................................................
Correct answer from incorrect working: Watch for solutions like the following
log 5
b
5
b
5
4
+
3log
=
log 5
4
=
log 5
−
log
3
4
=
3
81
=
=
b
b
b
b
b
3
3
3
3
81
log
3
3
b
b
3
Score B1 M0 dM0 A0
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