| Exam Board | AQA |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | Finding x from given y value |
| Difficulty | Moderate -0.8 This is a straightforward application question requiring basic logarithm evaluation (part a), solving a simple logarithmic equation (part b), and recognizing that ln(x) is undefined at x=0 and negative for 0<x<1 (part c). All parts are routine with no problem-solving insight needed, making it easier than average. |
| Spec | 1.06d Natural logarithm: ln(x) function and properties1.06e Logarithm as inverse: ln(x) inverse of e^x |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(16\ln 5 + 31 = 57\) years | B1 | AO 3.3 — obtains 57 (AWFW 56 to 57) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\ln x = \frac{40-31}{16}\), so \(x = e^{\frac{9}{16}}\) | M1 | AO 1.1a — uses exponential as inverse of logarithm; condone use of \(10^{\frac{9}{16}}\) for M1 |
| \(x = 1.755\) | A1F | AO 3.4 — AWFW 1.75 to 1.76. Or FT \(x = 3.65\) from using \(10^{\frac{9}{16}}\), AWFW 3.65 to 3.66 |
| \(= 21\) months | A1F | AO 3.2a — or 1 year 9 months. Or FT: 3 years 8 months / 44 months |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| As \(x\) approaches zero, \(\ln(x)\) becomes negative | E1 | AO 3.5b — states what happens to logarithm for values approaching zero, or gives appropriate example |
| As the dog age approaches zero the equivalent human age can become negative | E1 | AO 3.5a — states the consequence for the dog's equivalent human age |
## Question 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $16\ln 5 + 31 = 57$ years | B1 | AO 3.3 — obtains 57 (AWFW 56 to 57) |
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## Question 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\ln x = \frac{40-31}{16}$, so $x = e^{\frac{9}{16}}$ | M1 | AO 1.1a — uses exponential as inverse of logarithm; condone use of $10^{\frac{9}{16}}$ for M1 |
| $x = 1.755$ | A1F | AO 3.4 — AWFW 1.75 to 1.76. Or FT $x = 3.65$ from using $10^{\frac{9}{16}}$, AWFW 3.65 to 3.66 |
| $= 21$ months | A1F | AO 3.2a — or 1 year 9 months. Or FT: 3 years 8 months / 44 months |
---
## Question 6(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| As $x$ approaches zero, $\ln(x)$ becomes negative | E1 | AO 3.5b — states what happens to logarithm for values approaching zero, or gives appropriate example |
| As the dog age approaches zero the equivalent human age can become negative | E1 | AO 3.5a — states the consequence for the dog's equivalent human age |
---6 An on-line science website states:\\
'To find a dog's equivalent human age in years, multiply the natural logarithm of the dog's age in years by 16 then add 31.'
6
\begin{enumerate}[label=(\alph*)]
\item Calculate the equivalent age to the nearest human year of a dog aged 5 years.
6
\item A dog's equivalent age in human years is 40 years. Find the dog's actual age to the nearest month.\\
6
\item Explain why the behaviour of the natural logarithm for values close to zero means that the formula given on the website cannot be true for very young dogs.
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 2 2022 Q6 [6]}}