Edexcel C3 — Question 3 7 marks

Exam BoardEdexcel
ModuleC3 (Core Mathematics 3)
Marks7
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Mark schemeDownload PDF ↗
TopicExponential Equations & Modelling
TypeNatural logarithm equation solving
DifficultyStandard +0.3 Part (a) is routine algebraic factorization and simplification. Part (b) applies standard logarithm laws (ln A - ln B = ln(A/B)) to reduce to a simple exponential equation, but requires recognizing the connection to part (a) and careful algebraic manipulation. This is slightly above average difficulty for C3 due to the multi-step nature and need to connect parts, but remains a standard textbook-style question.
Spec1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b
3. (a) Simplify $$\frac { 2 x ^ { 2 } + 3 x - 9 } { 2 x ^ { 2 } - 7 x + 6 }$$ (b) Solve the equation $$\ln \left( 2 x ^ { 2 } + 3 x - 9 \right) = 2 + \ln \left( 2 x ^ { 2 } - 7 x + 6 \right)$$ giving your answer in terms of e.
AnswerMarks
(a) \(= \frac{(2x-3)(x+3)}{(2x-3)(x-2)} = \frac{x+3}{x-2}\)M1 A2
(b) \(\ln(2x^2+3x-9) - \ln(2x^2-7x+6) = 2\), \(\ln\frac{2x^2+3x-9}{2x^2-7x+6} = 2\)M1
\(\ln\frac{x+3}{x-2} = 2\), \(\frac{x+3}{x-2} = e^2\)A1
\(x+3 = e^2(x-2)\)
AnswerMarks Guidance
\(3+2e^2 = x(e^2-1)\)M1
\(x = \frac{2e^2+3}{e^2-1}\)A1 (7) 
**(a)** $= \frac{(2x-3)(x+3)}{(2x-3)(x-2)} = \frac{x+3}{x-2}$ | M1 A2

**(b)** $\ln(2x^2+3x-9) - \ln(2x^2-7x+6) = 2$, $\ln\frac{2x^2+3x-9}{2x^2-7x+6} = 2$ | M1

$\ln\frac{x+3}{x-2} = 2$, $\frac{x+3}{x-2} = e^2$ | A1

$x+3 = e^2(x-2)$

$3+2e^2 = x(e^2-1)$ | M1

$x = \frac{2e^2+3}{e^2-1}$ | A1 | **(7)**
3. (a) Simplify

$$\frac { 2 x ^ { 2 } + 3 x - 9 } { 2 x ^ { 2 } - 7 x + 6 }$$

(b) Solve the equation

$$\ln \left( 2 x ^ { 2 } + 3 x - 9 \right) = 2 + \ln \left( 2 x ^ { 2 } - 7 x + 6 \right)$$

giving your answer in terms of e.\\

\hfill \mbox{\textit{Edexcel C3  Q3 [7]}}
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