Edexcel P2 2021 January — Question 3 8 marks

Exam BoardEdexcel
ModuleP2 (Pure Mathematics 2)
Year2021
SessionJanuary
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicExponential Equations & Modelling
TypeLogarithmic equation solving
DifficultyModerate -0.3 Part (i) is straightforward application of logarithms to isolate x, requiring only basic log manipulation. Part (ii) involves combining logarithms and solving a quadratic, which is standard P2 fare but requires careful algebraic manipulation and checking solutions are valid. Overall slightly easier than average due to being routine textbook-style exercises with no novel insight required.
Spec1.06c Logarithm definition: log_a(x) as inverse of a^x1.06f Laws of logarithms: addition, subtraction, power rules
3. (i) Solve $$7 ^ { x + 2 } = 3$$ giving your answer in the form \(x = \log _ { 7 } a\) where \(a\) is a rational number in its simplest form.
(ii) Using the laws of logarithms, solve $$1 + \log _ { 2 } y + \log _ { 2 } ( y + 4 ) = \log _ { 2 } ( 5 - y )$$
Question 3:
Part (i):
AnswerMarks Guidance
Working/AnswerMark Guidance
\(7^{x+2} = 3 \Rightarrow 7^2 \times 7^x = 3\)M1 Applies addition law to write in form \(7^n \times 7^x = 3\); or takes logs both sides and applies power law
\(x = \log_7 \frac{3}{7^2}\)A1 Any correct exact expression, e.g. \(\log_7 3 - \log_7 49\), \(\log_7 3 - 2\), \(\frac{\log 3}{\log 7} - 2\)
\(x = \log_7 \frac{3}{49}\)A1 Correct answer only
Part (ii):
AnswerMarks Guidance
Working/AnswerMark Guidance
\(1 + \log_2 y + \log_2(y+4) = \log_2(5-y)\); uses \(1 \rightarrow \log_2 2\) or combines two log termsM1 Correctly applies addition/subtraction law at least once, or correctly replaces 1 with \(\log_2 2\)
\(\log_2\!\left(\frac{y(y+4)}{5-y}\right) = -1 \Rightarrow \frac{y(y+4)}{5-y} = \frac{1}{2}\) or \(\log_2(2y(y+4)) = \log_2(5-y)\)dM1 Rearranges to \(\log_2(\ldots) = C\) or \(\log_2(\ldots) = \log_2(\ldots)\) with no incorrect log work; removes logs correctly; dependent on M1
\(2y^2 + 9y - 5 = 0\)A1 oe, 3TQ
\((2y-1)(y+5) = 0 \Rightarrow y = \ldots\)ddM1 Attempts to solve quadratic; dependent on both previous method marks
\(y = \frac{1}{2}\)A1 Only (the \(-5\) must have been rejected if seen) 
# Question 3:

## Part (i):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $7^{x+2} = 3 \Rightarrow 7^2 \times 7^x = 3$ | M1 | Applies addition law to write in form $7^n \times 7^x = 3$; or takes logs both sides and applies power law |
| $x = \log_7 \frac{3}{7^2}$ | A1 | Any correct exact expression, e.g. $\log_7 3 - \log_7 49$, $\log_7 3 - 2$, $\frac{\log 3}{\log 7} - 2$ |
| $x = \log_7 \frac{3}{49}$ | A1 | Correct answer only |

## Part (ii):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $1 + \log_2 y + \log_2(y+4) = \log_2(5-y)$; uses $1 \rightarrow \log_2 2$ or combines two log terms | M1 | Correctly applies addition/subtraction law at least once, or correctly replaces 1 with $\log_2 2$ |
| $\log_2\!\left(\frac{y(y+4)}{5-y}\right) = -1 \Rightarrow \frac{y(y+4)}{5-y} = \frac{1}{2}$ or $\log_2(2y(y+4)) = \log_2(5-y)$ | dM1 | Rearranges to $\log_2(\ldots) = C$ or $\log_2(\ldots) = \log_2(\ldots)$ with no incorrect log work; removes logs correctly; dependent on M1 |
| $2y^2 + 9y - 5 = 0$ | A1 | oe, 3TQ |
| $(2y-1)(y+5) = 0 \Rightarrow y = \ldots$ | ddM1 | Attempts to solve quadratic; dependent on both previous method marks |
| $y = \frac{1}{2}$ | A1 | Only (the $-5$ must have been rejected if seen) |
3. (i) Solve

$$7 ^ { x + 2 } = 3$$

giving your answer in the form $x = \log _ { 7 } a$ where $a$ is a rational number in its simplest form.\\
(ii) Using the laws of logarithms, solve

$$1 + \log _ { 2 } y + \log _ { 2 } ( y + 4 ) = \log _ { 2 } ( 5 - y )$$

\hfill \mbox{\textit{Edexcel P2 2021 Q3 [8]}}
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