Edexcel PURE 2024 October — Question 6

Exam BoardEdexcel
ModulePURE
Year2024
SessionOctober
PaperDownload PDF ↗
TopicLaws of Logarithms
TypeSolve log equation reducing to quadratic
DifficultyStandard +0.3 This is a straightforward logarithm equation requiring application of standard log laws (power rule, combining logs) and solving a resulting quadratic. While it requires multiple steps and careful algebraic manipulation, it follows a standard template that students practice extensively. The 'show all working' requirement and rejection of calculator methods is typical but doesn't significantly increase difficulty beyond ensuring proper technique.
Spec1.06c Logarithm definition: log_a(x) as inverse of a^x1.06f Laws of logarithms: addition, subtraction, power rules
  1. In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
Using the laws of logarithms, solve $$\log _ { 4 } ( 12 - 2 x ) = 2 + 2 \log _ { 4 } ( x + 1 )$$
Question 6:
AnswerMarks
62 l o g ( x + 1 ) = l o g ( x + 1 ) 2
4 4B1
12−2x
e.g. log ( 12−2x)−log (x+1 )2 =log
AnswerMarks
4 4 4 (x+1 )2M1
1 2 − 2 x
= 1 6
AnswerMarks
( x + 1 2 )A1
 8 x 2 + 1 7 x + 2 = 0  x = . . .M1
1
x = − oe e.g. −0.125
AnswerMarks
8A1
(5)

Total 5

Alternative:
log
( 12−2x)=2+2log (x+1 )
4 4
 4 lo g 4 (1 2 − 2 x ) = 4 2 + 2 lo g 4 ( x + 1 ) = 4 2 + lo g 4 ( x + 1 2)
=424log
4
(x+1)2
 1 2 − 2 x = 1 6 ( x + 1 ) 2
Special Case: Beware incorrect log work which leads to the correct answer:
2log
(x+1 )=log (x+1 )2
4 4
log
( 12−2x)
log ( 12−2x)−log (x+1 )2 = 4
4 4
log
(x+1 )2
4
12−2x
=16
(x+1 )2
etc.
Score as:
B1: 2log (x+1 )=log (x+1 )2
4 4
M1: Uses correct index law e.g. 4a+b =4a4b
Then follow main scheme.
 8 x 2 + 1 7 x + 2 = 0  x = . . .
x = −
1
8
This scores B1M0A0M1A0
Allow this SC if no combination of logs is shown at all, but the correct quadratic is
produced.
e.g. l o g
4
( 1 2 − 2 x ) = 2 + 2 l o g
4
( x + 1 ) 
1
(
2
x
−
+
2
1
x
2 )
= 1 6 etc.
or e.g.
l o g
4
( 1 2 − 2 x ) = l o g
4
1 6 + l o g
4
( x + 1 ) 2  1 2 − 2 x = 1 6 ( x + 1 ) 2 etc.
Could both score B1M0A0M1A0
Special Case: Beware incorrect work removing logs which leads to the correct
answer:
2 l o g
4
( x + 1 ) = l o g
4
( x + 1 ) 2
l o g
4
( 1 2 − 2 x ) − l o g
4
( x + 1 ) 2 = l o g
4
1
(
2
x
−
+
2
1
x
2 )
= 2
1
(
2
x
−
+
2
1
x
2 )
= 2 4 = 1 6
 8 x 2 + 1 7 x + 2 = 0  x = . . .
x = −
1
8
PMT
This scores B1M1A0M1A0
Question 6:
6 | 2 l o g ( x + 1 ) = l o g ( x + 1 ) 2
4 4 | B1
12−2x
e.g. log ( 12−2x)−log (x+1 )2 =log
4 4 4 (x+1 )2 | M1
1 2 − 2 x
= 1 6
( x + 1 2 ) | A1
 8 x 2 + 1 7 x + 2 = 0  x = . . . | M1
1
x = − oe e.g. −0.125
8 | A1
(5)
Total 5
Alternative:
log
( 12−2x)=2+2log (x+1 )
4 4
 4 lo g 4 (1 2 − 2 x ) = 4 2 + 2 lo g 4 ( x + 1 ) = 4 2 + lo g 4 ( x + 1 2)
=424log
4
(x+1)2
 1 2 − 2 x = 1 6 ( x + 1 ) 2
Special Case: Beware incorrect log work which leads to the correct answer:
2log
(x+1 )=log (x+1 )2
4 4
log
( 12−2x)
log ( 12−2x)−log (x+1 )2 = 4
4 4
log
(x+1 )2
4
12−2x
=16
(x+1 )2
etc.
Score as:
B1: 2log (x+1 )=log (x+1 )2
4 4
M1: Uses correct index law e.g. 4a+b =4a4b
Then follow main scheme.
 8 x 2 + 1 7 x + 2 = 0  x = . . .
x = −
1
8
This scores B1M0A0M1A0
Allow this SC if no combination of logs is shown at all, but the correct quadratic is
produced.
e.g. l o g
4
( 1 2 − 2 x ) = 2 + 2 l o g
4
( x + 1 ) 
1
(
2
x
−
+
2
1
x
2 )
= 1 6 etc.
or e.g.
l o g
4
( 1 2 − 2 x ) = l o g
4
1 6 + l o g
4
( x + 1 ) 2  1 2 − 2 x = 1 6 ( x + 1 ) 2 etc.
Could both score B1M0A0M1A0
Special Case: Beware incorrect work removing logs which leads to the correct
answer:
2 l o g
4
( x + 1 ) = l o g
4
( x + 1 ) 2
l o g
4
( 1 2 − 2 x ) − l o g
4
( x + 1 ) 2 = l o g
4
1
(
2
x
−
+
2
1
x
2 )
= 2
1
(
2
x
−
+
2
1
x
2 )
= 2 4 = 1 6
 8 x 2 + 1 7 x + 2 = 0  x = . . .
x = −
1
8
PMT
This scores B1M1A0M1A0
\begin{enumerate}
  \item In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
\end{enumerate}

Using the laws of logarithms, solve

$$\log _ { 4 } ( 12 - 2 x ) = 2 + 2 \log _ { 4 } ( x + 1 )$$

\hfill \mbox{\textit{Edexcel PURE 2024 Q6}}
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