| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2019 |
| Session | June |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | Inverse function with exponentials |
| Difficulty | Moderate -0.8 This question requires manipulating exponential expressions using index laws (converting 4 to 2², converting 1/(2√2) to 2^(-3/2)) and then equating powers to solve for y in terms of x. While it involves multiple steps, these are standard A-level techniques with no problem-solving insight required—it's a straightforward algebraic manipulation exercise that's slightly easier than average. |
| Spec | 1.02a Indices: laws of indices for rational exponents1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2^x \times 2^{2y} = 2^{-\frac{3}{2}}\) | B1 | Writes a correct equation in powers of 2 only |
| \(2^{x+2y} = 2^{-\frac{3}{2}} \Rightarrow x + 2y = -\frac{3}{2} \Rightarrow y = \ldots\) | M1 | Complete process using correct index laws to obtain \(y\) as a function of \(x\) |
| e.g. \(y = -\frac{1}{2}x - \frac{3}{4}\) or \(y = -\frac{1}{4}(2x+3)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\log(2^x \times 4^y) = \log\!\left(\frac{1}{2\sqrt{2}}\right)\) | B1 | Writes a correct equation involving logarithms |
| \(x\log 2 + y\log 4 = \log 1 - \log(2\sqrt{2}) \Rightarrow y = \ldots\) | M1 | Complete process using correct log laws to obtain \(y\) as a function of \(x\) |
| \(y = \dfrac{-\log(2\sqrt{2}) - x\log 2}{\log 4}\) \(\left\{\Rightarrow y = -\frac{1}{2}x - \frac{3}{4}\right\}\) | A1 | Also accept \(y = \dfrac{-\ln(2\sqrt{2})-x\ln 2}{\ln 4}\) or \(y = \dfrac{\log\!\left(\frac{1}{2\sqrt{2}}\right)-\log(2^x)}{\log 4}\) |
# Question 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2^x \times 2^{2y} = 2^{-\frac{3}{2}}$ | B1 | Writes a correct equation in powers of 2 only |
| $2^{x+2y} = 2^{-\frac{3}{2}} \Rightarrow x + 2y = -\frac{3}{2} \Rightarrow y = \ldots$ | M1 | Complete process using correct index laws to obtain $y$ as a function of $x$ |
| e.g. $y = -\frac{1}{2}x - \frac{3}{4}$ or $y = -\frac{1}{4}(2x+3)$ | A1 | |
**Way 2/3/4 (logarithm methods):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\log(2^x \times 4^y) = \log\!\left(\frac{1}{2\sqrt{2}}\right)$ | B1 | Writes a correct equation involving logarithms |
| $x\log 2 + y\log 4 = \log 1 - \log(2\sqrt{2}) \Rightarrow y = \ldots$ | M1 | Complete process using correct log laws to obtain $y$ as a function of $x$ |
| $y = \dfrac{-\log(2\sqrt{2}) - x\log 2}{\log 4}$ $\left\{\Rightarrow y = -\frac{1}{2}x - \frac{3}{4}\right\}$ | A1 | Also accept $y = \dfrac{-\ln(2\sqrt{2})-x\ln 2}{\ln 4}$ or $y = \dfrac{\log\!\left(\frac{1}{2\sqrt{2}}\right)-\log(2^x)}{\log 4}$ |
---\begin{enumerate}
\item Given
\end{enumerate}
$$2 ^ { x } \times 4 ^ { y } = \frac { 1 } { 2 \sqrt { 2 } }$$
express $y$ as a function of $x$.
\hfill \mbox{\textit{Edexcel Paper 2 2019 Q1 [3]}}