SPS SPS SM Pure 2024 September — Question 10 6 marks

Exam BoardSPS
ModuleSPS SM Pure (SPS SM Pure)
Year2024
SessionSeptember
Marks6
TopicLaws of Logarithms
TypeSolve by showing reduces to polynomial
DifficultyStandard +0.3 This is a standard logarithm manipulation question requiring systematic application of log laws (power rule, addition rule) to reduce to a polynomial, followed by routine factorization given a root. The steps are predictable and well-practiced, making it slightly easier than average despite requiring careful algebraic manipulation.
Spec1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b

  1. In this question you must show detailed reasoning.
Solutions relying entirely on calculator technology are not acceptable.
  1. Given that $$2 \log _ { 4 } ( x + 3 ) + \log _ { 4 } x = \log _ { 4 } ( 4 x + 2 ) + \frac { 1 } { 2 }$$ show that $$x ^ { 3 } + 6 x ^ { 2 } + x - 4 = 0$$
  2. Given also that - 1 is a root of the equation $$x ^ { 3 } + 6 x ^ { 2 } + x - 4 = 0$$ solve $$2 \log _ { 4 } ( x + 3 ) + \log _ { 4 } x = \log _ { 4 } ( 4 x + 2 ) + \frac { 1 } { 2 }$$ \section*{(Total for Question 10 is 6 marks)}

\begin{enumerate}
  \item In this question you must show detailed reasoning.
\end{enumerate}

Solutions relying entirely on calculator technology are not acceptable.\\
(a) Given that

$$2 \log _ { 4 } ( x + 3 ) + \log _ { 4 } x = \log _ { 4 } ( 4 x + 2 ) + \frac { 1 } { 2 }$$

show that

$$x ^ { 3 } + 6 x ^ { 2 } + x - 4 = 0$$

(b) Given also that - 1 is a root of the equation

$$x ^ { 3 } + 6 x ^ { 2 } + x - 4 = 0$$

solve

$$2 \log _ { 4 } ( x + 3 ) + \log _ { 4 } x = \log _ { 4 } ( 4 x + 2 ) + \frac { 1 } { 2 }$$

\section*{(Total for Question 10 is 6 marks)}

\hfill \mbox{\textit{SPS SPS SM Pure 2024 Q10 [6]}}