CAIE P3 2020 November — Question 4 6 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2020
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLaws of Logarithms
TypeSolve log equation reducing to quadratic
DifficultyStandard +0.3 This is a straightforward logarithm equation requiring application of log laws (power rule) and converting to exponential form, leading to a standard quadratic. The steps are routine: simplify using 2log(x+1) = log(x+1)², rearrange to get log(2x+1) - log(x+1)² = -1, apply quotient rule, convert to exponential form, expand and solve the resulting quadratic. Slightly easier than average due to being a standard textbook exercise with well-practiced techniques.
Spec1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b

4 Solve the equation $$\log _ { 10 } ( 2 x + 1 ) = 2 \log _ { 10 } ( x + 1 ) - 1$$ Give your answers correct to 3 decimal places.

Question 4:
AnswerMarks Guidance
AnswerMarks Guidance
State or imply \(\log_{10} 10 = 1\)B1 \(\log_{10} 10^{-1} = -1\)
Use law of the logarithm of a power, product or quotientM1
Obtain a correct equation in any form, free of logsA1 e.g. \((2x+1)/(x+1)^2 = 10^{-1}\) or \(10(2x+1)/(x+1)^2 = 10^0\) or \(x^2 + 2x + 1 = 20x + 10\)
Reduce to \(x^2 - 18x - 9 = 0\), or equivalentA1
Solve a 3-term quadraticM1
Obtain final answers \(x = 18.487\) and \(x = -0.487\)A1 Must be 3 d.p. Do not allow rejection
## Question 4:

| Answer | Marks | Guidance |
|--------|-------|----------|
| State or imply $\log_{10} 10 = 1$ | B1 | $\log_{10} 10^{-1} = -1$ |
| Use law of the logarithm of a power, product or quotient | M1 | |
| Obtain a correct equation in any form, free of logs | A1 | e.g. $(2x+1)/(x+1)^2 = 10^{-1}$ or $10(2x+1)/(x+1)^2 = 10^0$ or $x^2 + 2x + 1 = 20x + 10$ |
| Reduce to $x^2 - 18x - 9 = 0$, or equivalent | A1 | |
| Solve a 3-term quadratic | M1 | |
| Obtain final answers $x = 18.487$ and $x = -0.487$ | A1 | Must be 3 d.p. Do not allow rejection |

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4 Solve the equation

$$\log _ { 10 } ( 2 x + 1 ) = 2 \log _ { 10 } ( x + 1 ) - 1$$

Give your answers correct to 3 decimal places.\\

\hfill \mbox{\textit{CAIE P3 2020 Q4 [6]}}