| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2016 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Laws of Logarithms |
| Type | Model y=ab^x: linearise and find constants from graph/data |
| Difficulty | Moderate -0.3 This is a standard logarithmic linearization question requiring routine manipulation of logarithms and straight-line equations. Part (a) involves taking log₁₀ of both sides (a common textbook exercise), while parts (b)(i-ii) require finding gradient from two points and converting back to exponential form—all mechanical procedures with no novel insight required. Slightly easier than average due to its formulaic nature. |
| Spec | 1.06f Laws of logarithms: addition, subtraction, power rules1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\log_{10} y = \log_{10} a + \log_{10} b^x\) | M1 | \(\log ab^x = \log a + \log b^x\) seen or used |
| \(\log_{10} y = \log_{10} a + x\log_{10} b\) | A1 | \(Y = \log_{10} a + x\log_{10} b\) (is a linear relationship between \(Y\) and \(x\)) |
| Answer | Marks | Guidance |
|---|---|---|
| (b)(i) (gradient of line =) \(-0.4\) | B1 | Correct value for gradient |
| Answer | Marks | Guidance |
|---|---|---|
| (b)(ii) \(\log_{10} a = 2.5\), \(a = 10^{2.5}\) | M1 | \(\log_{(10)} a = 2.5\) OE PI by \(a = 316\) |
| \(a = 316\) (to 3 sf) | A1 | CAO \(a = 316\) |
| \(\log_{10} b =\) gradient of line \(= -0.4\) | M1 | \(\log_{(10)} b = -0.4\) OE ft c's (b)(i) answer |
| \(b = 0.398\) to 3 sf | A1 | CAO \(b = 0.398\) |
**(a)** $\log_{10} y = \log_{10} a + \log_{10} b^x$ | M1 | $\log ab^x = \log a + \log b^x$ seen or used
$\log_{10} y = \log_{10} a + x\log_{10} b$ | A1 | $Y = \log_{10} a + x\log_{10} b$ (is a linear relationship between $Y$ and $x$)
**Total for (a): 2 marks**
**(b)(i)** (gradient of line =) $-0.4$ | B1 | Correct value for gradient
**Total for (b)(i): 1 mark**
**(b)(ii)** $\log_{10} a = 2.5$, $a = 10^{2.5}$ | M1 | $\log_{(10)} a = 2.5$ OE PI by $a = 316$
$a = 316$ (to 3 sf) | A1 | CAO $a = 316$
$\log_{10} b =$ gradient of line $= -0.4$ | M1 | $\log_{(10)} b = -0.4$ OE ft c's (b)(i) answer
$b = 0.398$ to 3 sf | A1 | CAO $b = 0.398$
**Total for (b)(ii): 4 marks**
**Overall Total: 7 marks**
**Additional Notes:**
- (a) If base 10 is missing or if $\log_{10} y$ has not been replaced by $Y$ then A0
---The variables $y$ and $x$ are related by an equation of the form
$$y = a(b^x)$$
where $a$ and $b$ are positive constants.
Let $Y = \log_{10} y$.
\begin{enumerate}[label=(\alph*)]
\item Show that there is a linear relationship between $Y$ and $x$.
[2 marks]
\item The graph of $Y$ against $x$, shown below, passes through the points $(0, 2.5)$ and $(5, 0.5)$.
\includegraphics{figure_3}
\begin{enumerate}[label=(\roman*)]
\item Find the gradient of the line.
[1 mark]
\item Find the value of $a$ and the value of $b$, giving each answer to three significant figures.
[4 marks]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2016 Q3 [7]}}