OCR MEI C3 — Question 5 4 marks

Exam BoardOCR MEI
ModuleC3 (Core Mathematics 3)
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicExponential Equations & Modelling
TypeInverse function with exponentials
DifficultyModerate -0.8 This is a straightforward algebraic manipulation requiring students to rearrange an exponential equation and apply logarithm rules. The steps are routine: divide both sides by x₀, take natural log of both sides, and simplify using log laws. It's easier than average as it's purely procedural with no problem-solving or conceptual insight required, though it does test basic fluency with exponentials and logarithms.
Spec1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b
5 Given that \(x\) and \(t\) are related by the formula \(x = x _ { 0 } \mathrm { e } ^ { - 3 t }\), show that \(t = \ln \left( \frac { a } { x } \right) ^ { b }\) where \(a\) and \(b\) are to be determined.
AnswerMarks Guidance
\(x = x_0 e^{-3t} \Rightarrow e^{3t} = \frac{x_0}{x}\)M1 A1 Take logs
\(\Rightarrow 3t = \ln\left(\frac{x_0}{x}\right) \Rightarrow t = \frac{1}{3}\ln\left(\frac{x_0}{x}\right)\)A1
\(\Rightarrow t = \ln\left(\frac{x_0}{x}\right)^{\frac{1}{3}}\)A1 A1 or any equivalent method
i.e. \(a = x_0\), \(b = \frac{1}{3}\)
4 
$x = x_0 e^{-3t} \Rightarrow e^{3t} = \frac{x_0}{x}$ | M1 A1 | Take logs

$\Rightarrow 3t = \ln\left(\frac{x_0}{x}\right) \Rightarrow t = \frac{1}{3}\ln\left(\frac{x_0}{x}\right)$ | A1 | 

$\Rightarrow t = \ln\left(\frac{x_0}{x}\right)^{\frac{1}{3}}$ | A1 A1 | or any equivalent method

i.e. $a = x_0$, $b = \frac{1}{3}$ | 
| 4 |
5 Given that $x$ and $t$ are related by the formula $x = x _ { 0 } \mathrm { e } ^ { - 3 t }$, show that $t = \ln \left( \frac { a } { x } \right) ^ { b }$ where $a$ and $b$ are to be determined.

\hfill \mbox{\textit{OCR MEI C3  Q5 [4]}}
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