| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2015 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Solve exponential equation by substitution |
| Difficulty | Moderate -0.8 This is a standard C1 substitution question with clear scaffolding. Part (a) guides students to recognize 4^x = (2^2)^x = y^2, then part (b) becomes a routine quadratic equation. The technique is commonly practiced and requires only basic index laws and solving a quadratic—no problem-solving insight needed. |
| Spec | 1.02a Indices: laws of indices for rational exponents1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((4^x =)\, y^2\) | B1 | Allow \(y^2\) or \(y \times y\) or "y squared". "\(4^x=\)" not required. Must be seen in part (a). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(8y^2 - 9y + 1 = (8y-1)(y-1) = 0 \Rightarrow y = \ldots\) or \((8(2^x)-1)((2^x)-1) = 0 \Rightarrow 2^x = \ldots\) | M1 | Attempt to solve as 3 term quadratic in \(y\) or \(2^x\). Apply usual rules for solving quadratic. Allow \(x\) (or any other letter) instead of \(y\). |
| \(2^x \text{(or } y) = \frac{1}{8},\ 1\) | A1 | Both correct answers of \(\frac{1}{8}\) (oe) and 1. Not \(x\) unless \(2^x\) (or \(y\)) is implied later. |
| \(x = -3 \quad x = 0\) | M1A1 | M1: Correct attempt to find one numerical value of \(x\) from their \(2^x\) (or \(y\)), must come from 3 term quadratic. If logs used they must be evaluated. A1: Both \(x=-3\) and/or \(x=0\), e.g. \(2^{-3}=\frac{1}{8}\) and \(2^0=1\), and no extra values. |
## Question 7:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(4^x =)\, y^2$ | B1 | Allow $y^2$ or $y \times y$ or "y squared". "$4^x=$" not required. Must be seen in part (a). |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $8y^2 - 9y + 1 = (8y-1)(y-1) = 0 \Rightarrow y = \ldots$ or $(8(2^x)-1)((2^x)-1) = 0 \Rightarrow 2^x = \ldots$ | M1 | Attempt to solve as 3 term quadratic in $y$ or $2^x$. Apply usual rules for solving quadratic. Allow $x$ (or any other letter) instead of $y$. |
| $2^x \text{(or } y) = \frac{1}{8},\ 1$ | A1 | Both correct answers of $\frac{1}{8}$ (oe) and 1. Not $x$ unless $2^x$ (or $y$) is implied later. |
| $x = -3 \quad x = 0$ | M1A1 | M1: Correct attempt to find one numerical value of $x$ from their $2^x$ (or $y$), must come from 3 term quadratic. If logs used they must be evaluated. A1: Both $x=-3$ and/or $x=0$, e.g. $2^{-3}=\frac{1}{8}$ and $2^0=1$, and no extra values. |
---\begin{enumerate}
\item Given that $y = 2 ^ { x }$,\\
(a) express $4 ^ { x }$ in terms of $y$.\\
(b) Hence, or otherwise, solve
\end{enumerate}
$$8 \left( 4 ^ { x } \right) - 9 \left( 2 ^ { x } \right) + 1 = 0$$
\hfill \mbox{\textit{Edexcel C1 2015 Q7 [5]}}