CAIE P3 2009 June — Question 1 4 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2009
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicExponential Equations & Modelling
TypeNatural logarithm equation solving
DifficultyModerate -0.8 This is a straightforward single-step logarithm equation requiring only basic manipulation: exponentiate both sides to get 2 + e^(-x) = e^2, rearrange to isolate e^(-x), then take ln again. It's a routine procedural question with no conceptual challenges, making it easier than average but not trivial since it requires careful algebraic manipulation of exponentials.
Spec1.06g Equations with exponentials: solve a^x = b
1 Solve the equation \(\ln \left( 2 + \mathrm { e } ^ { - x } \right) = 2\), giving your answer correct to 2 decimal places.
Question 1:
AnswerMarks Guidance
Answer/WorkingMark Guidance
State or imply \(2 + e^{-x} = e^2\)B1
Carry out method for finding \(\pm x\) from \(e^{\pm x} = k\), where \(k > 0\), following sound ln or exp workM1 SR: M1 available for attempts starting with \(2 + e^{-x} = 10^2\)
Obtain \(x = -\ln(e^2 - 2)\), or equivalent expression for \(x\)A1
Obtain answer \(x = -1.68\)A1 Answer must be given to 2 decimal places 
## Question 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| State or imply $2 + e^{-x} = e^2$ | B1 | |
| Carry out method for finding $\pm x$ from $e^{\pm x} = k$, where $k > 0$, following sound ln or exp work | M1 | SR: M1 available for attempts starting with $2 + e^{-x} = 10^2$ |
| Obtain $x = -\ln(e^2 - 2)$, or equivalent expression for $x$ | A1 | |
| Obtain answer $x = -1.68$ | A1 | Answer must be given to 2 decimal places |

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1 Solve the equation $\ln \left( 2 + \mathrm { e } ^ { - x } \right) = 2$, giving your answer correct to 2 decimal places.

\hfill \mbox{\textit{CAIE P3 2009 Q1 [4]}}
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