2023 Paper 1, Q19
Jaxon and Shari each own a shop and have recorded the number of customers entering their shop on two consecutive days.
The number of daily customers for each shop can be modelled by the equation $y=A\ln(Bx)$, where $x$ is the day and $y$ is the number of customers. The constants $A$ and $B$ are different for each shop.
Determine algebraically whether the total number of customers for Jaxon and Shari’s shops will be the same every day in the future.
Two shops trade customers, and the total never changes
Day by day, Jaxon loses customers at exactly the rate Shari gains them. Watch the split inside the column move while the stack stays pinned to the 50 line.
eight days, then a short hold · the stack always reaches the dashed line. Only the split moves
The two log curves are the same curve, shifted
$\ln(2x)$ and $\ln\!\left(\frac{x}{16}\right)$ differ by a constant $\ln 32$ at every $x$. Multiply that constant gap by $\frac{10}{\ln 2}$ and you get exactly 50.
black: Jaxon · orange: Shari
orange: $\ln(2x)$ · black: $\ln\!\left(\frac{x}{16}\right)$
QCAA marking guide · 7 marks
Two data points give two equations per shop. For Shari, $x=1$ and $x=2$:
$$10=A\ln B\ \ (1)\qquad 20=A\ln(2B)\ \ (2)$$
Expand (2) with the log law: $20=A\ln 2+A\ln B=A\ln 2+10$, so $A\ln 2=10$ and
$$A=\frac{10}{\ln 2}$$
Marker: correctly determines $A$ (or $B$) for Shari’s shop. The mark may be awarded if $B$ is correctly found before $A$; equivalent statements are accepted.
Substitute back into (1): $10=\frac{10}{\ln 2}\ln B$, so $\ln B=\ln 2$ and $B=2$.
$$y=\left(\frac{10}{\ln 2}\right)\ln(2x)\quad\text{Shari's shop}$$
Marker: determines the model for Shari’s shop.
Same two equations for Jaxon, with 40 and 30:
$$30=A\ln 2+40\ \Rightarrow\ A\ln 2=-10\ \Rightarrow\ A=\frac{-10}{\ln 2}$$
A negative $A$, the shop is losing trade.
Marker: correctly determines $A$ (or $B$) for Jaxon’s shop. Follow-through marks are allowed for errors in prior working.
$$40=\frac{-10}{\ln 2}\ln B\ \Rightarrow\ \ln B=-4\ln 2=\ln 2^{-4}\ \Rightarrow\ B=\frac{1}{16}$$
$$y=\left(\frac{-10}{\ln 2}\right)\ln\!\left(\frac{x}{16}\right)\quad\text{Jaxon's shop}$$
Marker: determines the model for Jaxon’s shop.
On any day $x$, add the two models:
$$\left(\frac{10}{\ln 2}\right)\ln(2x)+\left(\frac{-10}{\ln 2}\right)\ln\!\left(\frac{x}{16}\right)$$
Marker: determines an expression for the total number of daily customers in both shops.
$$=\frac{10}{\ln 2}\left[\ln(2x)-\ln\!\left(\frac{x}{16}\right)\right]=\frac{10}{\ln 2}\ln\!\left(\frac{2x}{\frac{x}{16}}\right)=\frac{10}{\ln 2}\ln 32$$
The $x$ cancels inside the logarithm, leaving no $x$ at all. Then $\ln 32=\ln 2^5=5\ln 2$:
$$=\frac{10\times 5\ln 2}{\ln 2}=50$$
Marker: justifies the sum obtained by explaining mathematical reasoning. Trial and error and substitution are not appropriate for this mark.
Yes, 50 customers, on any day
The sum of the models is independent of $x$, so the two shops together will always take 50 customers.
Marker: provides an appraisal by interpreting the result of the analysis of the sum of the two models performed. Alternative methods are accepted provided they use the models.
“Determine algebraically” is an instruction, not a decoration
Combining the two logarithms into $\ln\!\left(\frac{2x}{x/16}\right)$ and watching the $x$ cancel. The result holds for every day because no $x$ survives.
Testing days 3, 4 and 5, getting 50 each time, and stopping there. The marking guide rules out trial and error, because three examples are not a proof.
One thing worth noticing is that Jaxon’s model reaches zero at $x=16$ and goes negative after that, so the model stops making sense before the mathematics does. The question asks about the sum of the models, and the sum is 50 forever. A good appraisal can still point out that the model stops describing a real shop.
There are four unknowns and a table of four numbers, and the way through is the log law $\ln(2B)=\ln 2+\ln B$. It turns each pair of equations into one linear equation in $A\ln 2$, and nothing in the stem hints at it. The second hidden step is that the question wants a general statement, so you have to simplify the sum algebraically until the $x$ disappears. Substituting a few days looks like an answer, but it earns nothing. I see a lot of students test days 3, 4 and 5 and stop there. When the QCAA write “determine algebraically”, they mean it!
Question wording and marking-guide steps are from the 2023 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2023, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.