| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Laws of Logarithms |
| Type | Express log in terms of given variables |
| Difficulty | Moderate -0.3 This is a straightforward application of logarithm laws (power rule, product rule) with some algebraic manipulation. Part (a) is immediate using the power rule. Part (b) requires factoring then applying product rule. Part (c) needs factoring out a common term before applying log laws. All parts are standard textbook exercises requiring only routine manipulation, making it slightly easier than average. |
| Spec | 1.06f Laws of logarithms: addition, subtraction, power rules |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2}a\) | B1 | \(\frac{1}{2}a\) or \(\frac{a}{2}\) or \(0.5a\) isw. Condone omission of base 2 in all parts. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\log_2 x(x+8) \Rightarrow \log_2 x + \log_2(x+8)\) | M1 | Takes factor of \(x\) out of bracket and applies addition law of logs. Condone missing brackets or omission of base 2. \(\log_2 x \times \log_2(x+8)\) on its own is M0 but allow if they proceed to \(a+b\) |
| \(= a + b\) | A1 | \(a+b\) or simplified equivalent. Note \(\log_2 x \times \log_2(x+8) = a+b\) is M1A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(8 + \frac{64}{x} = \frac{8x+64}{x}\) | B1 | Writes \(8 + \frac{64}{x}\) as single fraction e.g. \(\frac{8x+64}{x}\) or \(\frac{8}{x}(x+8)\) or \(8x^{-1}(x+8)\) or \(8\left(\frac{x^2+8x}{x^2}\right)\). May be implied by later work. |
| \(\log_2 \frac{8}{x}(x+8) = 3 - \log_2 x + \log_2(x+8)\) | M1 | Attempts to apply laws of logs, uses \(\log_2 8 = 3\) and proceeds to \(3 \pm \log_2 x \pm \log_2(x+8)\). May be implied by \(3 \pm b \pm a\). Note if they write \(\log_2(x+8)\) as \(\log_2 x + \log_2 8\) this is M0 |
| \(3 + b - a\) | A1 | \(3+b-a\) or simplified equivalent. Note \(\log_2\frac{8}{x}(x+8) = 3 \div \log_2 x \times \log_2(x+8) \Rightarrow 3-a+b\) is M1A0 |
## Question 6:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}a$ | B1 | $\frac{1}{2}a$ or $\frac{a}{2}$ or $0.5a$ isw. Condone omission of base 2 in all parts. |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\log_2 x(x+8) \Rightarrow \log_2 x + \log_2(x+8)$ | M1 | Takes factor of $x$ out of bracket and applies addition law of logs. Condone missing brackets or omission of base 2. $\log_2 x \times \log_2(x+8)$ on its own is M0 but allow if they proceed to $a+b$ |
| $= a + b$ | A1 | $a+b$ or simplified equivalent. Note $\log_2 x \times \log_2(x+8) = a+b$ is M1A0 |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $8 + \frac{64}{x} = \frac{8x+64}{x}$ | B1 | Writes $8 + \frac{64}{x}$ as single fraction e.g. $\frac{8x+64}{x}$ or $\frac{8}{x}(x+8)$ or $8x^{-1}(x+8)$ or $8\left(\frac{x^2+8x}{x^2}\right)$. May be implied by later work. |
| $\log_2 \frac{8}{x}(x+8) = 3 - \log_2 x + \log_2(x+8)$ | M1 | Attempts to apply laws of logs, uses $\log_2 8 = 3$ and proceeds to $3 \pm \log_2 x \pm \log_2(x+8)$. May be implied by $3 \pm b \pm a$. Note if they write $\log_2(x+8)$ as $\log_2 x + \log_2 8$ this is M0 |
| $3 + b - a$ | A1 | $3+b-a$ or simplified equivalent. Note $\log_2\frac{8}{x}(x+8) = 3 \div \log_2 x \times \log_2(x+8) \Rightarrow 3-a+b$ is M1A0 |
---6.
$$a = \log _ { 2 } x \quad b = \log _ { 2 } ( x + 8 )$$
Express in terms of $a$ and/or $b$
\begin{enumerate}[label=(\alph*)]
\item $\log _ { 2 } \sqrt { x }$
\item $\log _ { 2 } \left( x ^ { 2 } + 8 x \right)$
\item $\log _ { 2 } \left( 8 + \frac { 64 } { x } \right)$
Give your answer in simplest form.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 1 2023 Q6 [6]}}