SPS SPS SM 2020 October — Question 6 5 marks

Exam BoardSPS
ModuleSPS SM (SPS SM)
Year2020
SessionOctober
Marks5
TopicExponential Functions
TypeSolve exponential equation by substitution
DifficultyModerate -0.8 Part (i) requires spotting two algebraic errors in worked solutions (misapplying index laws: 2^{2x+4} ≠ 2^{2x} + 2^4, and an arithmetic error in the quadratic constant term). Part (ii) is a standard exponential equation solved by substitution, leading to a straightforward quadratic. Both parts test routine A-level techniques with minimal problem-solving demand, making this easier than average.
Spec1.06g Equations with exponentials: solve a^x = b

  1. A student was asked to solve the equation \(2^{2x+4} - 9(2^x) = 0\). The student's attempt is written out below. $$2^{2x+4} - 9(2^x) = 0$$ $$2^{2x} + 2^4 - 9(2^x) = 0$$ $$\text{Let } y = 2^x$$ $$y^2 - 9y + 8 = 0$$ $$(y - 8)(y - 1) = 0$$ $$y = 8 \text{ or } y = 1$$ $$\text{So } x = 3 \text{ or } x = 0$$ Identify the two mistakes that the student has made. [2]
  2. Solve the equation \(2^{2x+4} - 9(2^x) = 0\), giving your answer in exact form. [3]

\begin{enumerate}[label=(\roman*)]
\item A student was asked to solve the equation $2^{2x+4} - 9(2^x) = 0$.
The student's attempt is written out below.

$$2^{2x+4} - 9(2^x) = 0$$
$$2^{2x} + 2^4 - 9(2^x) = 0$$
$$\text{Let } y = 2^x$$
$$y^2 - 9y + 8 = 0$$
$$(y - 8)(y - 1) = 0$$
$$y = 8 \text{ or } y = 1$$
$$\text{So } x = 3 \text{ or } x = 0$$

Identify the two mistakes that the student has made. [2]

\item Solve the equation $2^{2x+4} - 9(2^x) = 0$, giving your answer in exact form. [3]
\end{enumerate}

\hfill \mbox{\textit{SPS SPS SM 2020 Q6 [5]}}