CAIE P2 2020 November — Question 2 5 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2020
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicExponential Equations & Modelling
TypeRational exponential equation
DifficultyModerate -0.3 This is a straightforward algebraic manipulation question requiring substitution (let y = 2^(3x)), solving a linear equation, then applying logarithms. While it involves multiple steps, each step is routine and the substitution strategy is clearly signposted by the question structure. Slightly easier than average due to the guided approach and standard techniques.
Spec1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b

2 Given that \(\frac { 2 ^ { 3 x + 2 } + 8 } { 2 ^ { 3 x } - 7 } = 5\), find the value of \(2 ^ { 3 x }\) and hence, using logarithms, find the value of \(x\) correct to 4 significant figures.

Question 2:
AnswerMarks Guidance
AnswerMarks Guidance
Use \(2^{3x+2} = 4 \times 2^{3x}\)B1 OE
Solve equation for \(2^{3x}\)M1
Obtain \(2^{3x} = 43\)A1
Apply logarithms and use power law for \(2^{3x} = k\) where \(k > 0\)M1
Obtain \(1.809\)A1 AWRT
Total5
## Question 2:

| Answer | Marks | Guidance |
|--------|-------|----------|
| Use $2^{3x+2} = 4 \times 2^{3x}$ | B1 | OE |
| Solve equation for $2^{3x}$ | M1 | |
| Obtain $2^{3x} = 43$ | A1 | |
| Apply logarithms and use power law for $2^{3x} = k$ where $k > 0$ | M1 | |
| Obtain $1.809$ | A1 | AWRT |
| **Total** | **5** | |

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2 Given that $\frac { 2 ^ { 3 x + 2 } + 8 } { 2 ^ { 3 x } - 7 } = 5$, find the value of $2 ^ { 3 x }$ and hence, using logarithms, find the value of $x$ correct to 4 significant figures.\\

\hfill \mbox{\textit{CAIE P2 2020 Q2 [5]}}