| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2020 |
| Session | October |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Laws of Logarithms |
| Type | Solve log equation with domain restrictions |
| Difficulty | Moderate -0.3 This is a straightforward logarithm equation requiring standard manipulation (using power law and equating arguments), followed by factorizing a simple quadratic and checking domain restrictions. The multi-part structure guides students through each step, making it slightly easier than average but still requiring understanding of logarithm domains. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.06c Logarithm definition: log_a(x) as inverse of a^x1.06f Laws of logarithms: addition, subtraction, power rules |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2\log(4-x) = \log(4-x)^2\) | B1 | States or uses \(2\log(4-x)=\log(4-x)^2\) |
| \((4-x)^2 = (x+8)\) or \(\frac{(4-x)^2}{(x+8)}=1\) | M1 | Correct attempt at eliminating logs to form quadratic; may be implied by \(\log\frac{(4-x)^2}{(x+8)}=0 \Rightarrow (4-x)^2=x+8\) |
| \(16-8x+x^2 = x+8 \Rightarrow x^2-9x+8=0\)* | A1* | Proceeds to given answer; must be no errors; at least one line where \((4-x)^2\) has been multiplied out |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (i) \(x = 1, 8\) | B1 | Writes down both values |
| (ii) \(8\) is not a solution as \(\log(4-8)\) cannot be found | B1 | Chooses \(8\) and gives reason referring to logs; must reference \(\log(4-x)\) or \(\log(-4)\); acceptable reasons: \(\log(-4)\) can't be found/is undefined/\(= \) n/a; no contradictory statements |
# Question 3(a) (Logarithms):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2\log(4-x) = \log(4-x)^2$ | B1 | States or uses $2\log(4-x)=\log(4-x)^2$ |
| $(4-x)^2 = (x+8)$ or $\frac{(4-x)^2}{(x+8)}=1$ | M1 | Correct attempt at eliminating logs to form quadratic; may be implied by $\log\frac{(4-x)^2}{(x+8)}=0 \Rightarrow (4-x)^2=x+8$ |
| $16-8x+x^2 = x+8 \Rightarrow x^2-9x+8=0$* | A1* | Proceeds to given answer; must be no errors; at least one line where $(4-x)^2$ has been multiplied out |
# Question 3(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| (i) $x = 1, 8$ | B1 | Writes down both values |
| (ii) $8$ is **not** a solution as $\log(4-8)$ cannot be found | B1 | Chooses $8$ and gives reason referring to logs; must reference $\log(4-x)$ or $\log(-4)$; acceptable reasons: $\log(-4)$ can't be found/is undefined/$= $ n/a; no contradictory statements |
---\begin{enumerate}
\item (a) Given that
\end{enumerate}
$$2 \log ( 4 - x ) = \log ( x + 8 )$$
show that
$$x ^ { 2 } - 9 x + 8 = 0$$
(b) (i) Write down the roots of the equation
$$x ^ { 2 } - 9 x + 8 = 0$$
(ii) State which of the roots in (b)(i) is not a solution of
$$2 \log ( 4 - x ) = \log ( x + 8 )$$
giving a reason for your answer.
\hfill \mbox{\textit{Edexcel Paper 2 2020 Q3 [5]}}