| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2020 |
| Session | October |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | Quadratic in exponential form |
| Difficulty | Standard +0.3 This requires setting the equations equal and recognizing that 2^(x+1) = 2ยท2^x, leading to a quadratic in 2^x. The substitution and algebraic manipulation are straightforward, and solving the resulting quadratic is routine. Slightly above average difficulty due to the exponential manipulation required, but still a standard textbook exercise. |
| Spec | 1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(15 - 2^{x+1} = 3\times 2^x\) | B1 | Correct use of index law to rewrite equation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(15 - 2 \times 2^x = 3 \times 2^x \Rightarrow 2^x = 3\) or e.g. \(\frac{15}{2^x} - 2 = 3 \Rightarrow 2^x = 3\) | M1 | Combines equations to reach \(15 - 2^{x+1} = 3 \times 2^x\) or equivalent |
| \(2^x = 3 \Rightarrow x = \ldots\) | dM1 | Uses logs correctly; depends on first M1 |
| \(x = \log_2 3\) | A1cso | Also accept \(\frac{\log 3}{\log 2}\) or \(\frac{\ln 3}{\ln 2}\); ignore attempts to find \(y\)-coordinate |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = 3 \times 2^x \Rightarrow 2^x = \frac{y}{3} \Rightarrow y = 15 - 2 \times \frac{y}{3}\) | B1 | Correct equation in \(y\) |
| \(3y + 2y = 45 \Rightarrow y = 9 \Rightarrow 3 \times 2^x = 9 \Rightarrow 2^x = 3\) | M1 | Solves for \(y\), makes \(2^x\) subject |
| \(2^x = 3 \Rightarrow x = \ldots\) | dM1 | Uses logs correctly; depends on M1 |
| \(x = \log_2 3\) | A1cso | Also accept \(\frac{\log 3}{\log 2}\) or \(\frac{\ln 3}{\ln 2}\) |
# Question 5:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $15 - 2^{x+1} = 3\times 2^x$ | B1 | Correct use of index law to rewrite equation |
# Question (Previous - solving simultaneous equations):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $15 - 2 \times 2^x = 3 \times 2^x \Rightarrow 2^x = 3$ or e.g. $\frac{15}{2^x} - 2 = 3 \Rightarrow 2^x = 3$ | M1 | Combines equations to reach $15 - 2^{x+1} = 3 \times 2^x$ or equivalent |
| $2^x = 3 \Rightarrow x = \ldots$ | dM1 | Uses logs correctly; depends on first M1 |
| $x = \log_2 3$ | A1cso | Also accept $\frac{\log 3}{\log 2}$ or $\frac{\ln 3}{\ln 2}$; ignore attempts to find $y$-coordinate |
**Alternative:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 3 \times 2^x \Rightarrow 2^x = \frac{y}{3} \Rightarrow y = 15 - 2 \times \frac{y}{3}$ | B1 | Correct equation in $y$ |
| $3y + 2y = 45 \Rightarrow y = 9 \Rightarrow 3 \times 2^x = 9 \Rightarrow 2^x = 3$ | M1 | Solves for $y$, makes $2^x$ subject |
| $2^x = 3 \Rightarrow x = \ldots$ | dM1 | Uses logs correctly; depends on M1 |
| $x = \log_2 3$ | A1cso | Also accept $\frac{\log 3}{\log 2}$ or $\frac{\ln 3}{\ln 2}$ |
---\begin{enumerate}
\item The curve with equation $y = 3 \times 2 ^ { x }$ meets the curve with equation $y = 15 - 2 ^ { x + 1 }$ at the point $P$. Find, using algebra, the exact $x$ coordinate of $P$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 2 2020 Q5 [4]}}