| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2021 |
| Session | November |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Indices and Surds |
| Type | Solve exponential equations |
| Difficulty | Moderate -0.8 This is a straightforward exponential equation requiring only the standard technique of expressing all terms as powers of the same base (3), then equating exponents. It's a routine AS-level question with a single clear method and no problem-solving insight required, making it easier than average. |
| Spec | 1.02a Indices: laws of indices for rational exponents |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\frac{9^{x-1}}{3^{y+2}} = 81 \Rightarrow \frac{3^{2x-2}}{3^{y+2}} = 3^4\) or \(\frac{9^{x-1}}{3^{y+2}} = 81 \Rightarrow \frac{9^{x-1}}{9^{\frac{1}{2}(y+2)}} = 9^2\) | M1 | Attempts to set \(9^{x-1}\) and 81 as powers of 3. Condone \(9^{x-1} = 3^{2x-1}\) and \(9^{x-1} = 3^{3x-3}\). Alternatively attempts to write each term as a logarithm of base 3 or 9 — base must be written |
| \(\Rightarrow 2x - 2 - y - 2 = 4 \Rightarrow y =\) or \(\Rightarrow x - 1 - \frac{1}{2}y - 1 = 2 \Rightarrow y =\) | dM1 | Attempts to use the addition (or subtraction) index law, or laws of logarithms, correctly and rearranges to reach \(y\) in terms of \(x\). Condone slips in rearrangement |
| \(y = 2x - 8\) | A1 | cao |
| Answer | Marks |
|---|---|
| \(\log_3\!\left(\frac{9^{x-1}}{3^{y+2}}\right) = \log_3 81\) | M1 |
| \((x-1)\log_3(9^{x-1}) - (y+2)\log_3(3^{y+2}) = 4 \Rightarrow 2(x-1) - y - 2 = 4 \Rightarrow y =\) | dM1 |
| \(y = 2x - 8\) | A1 |
## Question 2:
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\frac{9^{x-1}}{3^{y+2}} = 81 \Rightarrow \frac{3^{2x-2}}{3^{y+2}} = 3^4$ or $\frac{9^{x-1}}{3^{y+2}} = 81 \Rightarrow \frac{9^{x-1}}{9^{\frac{1}{2}(y+2)}} = 9^2$ | M1 | Attempts to set $9^{x-1}$ and 81 as powers of 3. Condone $9^{x-1} = 3^{2x-1}$ and $9^{x-1} = 3^{3x-3}$. Alternatively attempts to write each term as a logarithm of base 3 or 9 — base must be written |
| $\Rightarrow 2x - 2 - y - 2 = 4 \Rightarrow y =$ or $\Rightarrow x - 1 - \frac{1}{2}y - 1 = 2 \Rightarrow y =$ | dM1 | Attempts to use the addition (or subtraction) index law, or laws of logarithms, correctly and rearranges to reach $y$ in terms of $x$. Condone slips in rearrangement |
| $y = 2x - 8$ | A1 | cao |
**Alt method:**
| $\log_3\!\left(\frac{9^{x-1}}{3^{y+2}}\right) = \log_3 81$ | M1 | |
| $(x-1)\log_3(9^{x-1}) - (y+2)\log_3(3^{y+2}) = 4 \Rightarrow 2(x-1) - y - 2 = 4 \Rightarrow y =$ | dM1 | |
| $y = 2x - 8$ | A1 | |\begin{enumerate}
\item In this question you should show all stages of your working.
\end{enumerate}
Solutions relying on calculator technology are not acceptable.\\
Given
$$\frac { 9 ^ { x - 1 } } { 3 ^ { y + 2 } } = 81$$
express $y$ in terms of $x$, writing your answer in simplest form.
\hfill \mbox{\textit{Edexcel AS Paper 1 2021 Q2 [3]}}