Exponentials & logarithms – A-level Maths exam questions with mark scheme

1CoreCalculator4 marks

(a) Write $2\log_{a}6-\log_{a}4$ as a single logarithm, simplifying your answer. [2]

(b) Solve $3^{x}=20$, giving your answer to 3 significant figures. [2]

(a) for using the power law: $2\log_a6=\log_a36$M1
(a) $\log_a9$A1
(b) for taking logs: $x=\dfrac{\log20}{\log3}$ or $x=\log_{3}20$M1
(b) $x=2.73$A1
Answer: (a) $\log_{a}9$ (b) $2.73$
Examiner tip: Apply the power law before combining: $\log_a 36-\log_a4=\log_a\frac{36}{4}$, not $\log_a 32$.

2StandardCalculator4 marks

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

Solve $$\log_{3}(x+4)-\log_{3}(x-2)=2.$$ [4]

for using the subtraction law: $\log_{3}\dfrac{x+4}{x-2}=2$M1
for removing logs correctly: $\dfrac{x+4}{x-2}=3^{2}$M1
for solving the resulting linear equation: $x+4=9x-18$M1
$x=\tfrac{11}{4}$A1
Answer: $x=\tfrac{11}{4}$
Examiner tip: $\log_{3}A=2$ means $A=3^{2}=9$, not $A=2^{3}$ or $A=6$.

3StandardCalculator5 marks

Solve the equation $$3^{2x}-7\left(3^{x}\right)+10=0,$$ giving your answers to 3 significant figures. [5]

for recognising a quadratic in $3^{x}$, e.g. $u^{2}-7u+10=0$ with $u=3^{x}$M1
$3^{x}=2$ and $3^{x}=5$A1
for using logs to solve $3^{x}=k$ with $k>0$, e.g. $x=\dfrac{\ln 2}{\ln3}$M1
$x=0.631$A1
$x=1.46$A1
Answer: $x=0.631,\ x=1.46$
Examiner tip: Spot $3^{2x}=(3^{x})^{2}$; don't stop at $3^{x}=2,5$ – the question wants $x$.

4HardCalculator7 marks

The number of bacteria, $P$, in a culture $t$ hours after it is first observed is modelled by $$P=Ae^{kt},$$ where $A$ and $k$ are constants. Initially there are 200 bacteria, and after 5 hours there are 350 bacteria.

(a) Find the value of $A$ and the value of $k$, giving $k$ to 3 significant figures. [3]

(b) Find the time taken, according to the model, for the number of bacteria to reach 1000. [2]

(a) $A=200$B1
(a) for $350=200e^{5k}$ and taking ln: $5k=\ln1.75$M1
(a) $k=0.112$ (awrt)A1
(b) for $1000=200e^{kt}\Rightarrow t=\dfrac{\ln5}{k}$ with their $k$M1
(b) $14.4$ hours (awrt)A1
(c) for $\dfrac{\mathrm{d}P}{\mathrm{d}t}=kAe^{kt}$ evaluated at $t=10$M1
(c) $68.6$ bacteria per hour (awrt; accept $68.5$–$68.7$, e.g. $68.7$ from $k=0.112$)A1
Answer: (a) $A=200,\ k=0.112$ (b) $14.4$ hours (c) $68.6$ bacteria per hour
Examiner tip: Keep $k$ unrounded ($\tfrac15\ln1.75$) in later parts – using $0.112$ can shift your final answer (e.g. $68.7$ instead of $68.6$ in (c)).

5HardCalculator5 marks

Two variables $x$ and $y$ are related by $y=ab^{x}$, where $a$ and $b$ are constants.

The graph of $\log_{10}y$ against $x$ is a straight line passing through the points $(0,\,1.2)$ and $(4,\,2.0)$, as shown.

Find the value of $a$ and the value of $b$, giving each answer to 3 significant figures. [5]

for taking logs: $\log_{10}y=\log_{10}a+x\log_{10}b$M1
for identifying $\log_{10}a$ with the intercept, $\log_{10}a=1.2$M1
$a=15.8$ (awrt)A1
for gradient $=\dfrac{2.0-1.2}{4}=0.2=\log_{10}b$M1
$b=1.58$ (awrt)A1
Answer: $a=15.8,\ b=1.58$
Examiner tip: The gradient is $\log_{10}b$, not $b$ – remember to undo the log: $b=10^{0.2}$.

6HardCalculator7 marks

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

(a) Show that the solution of $3^{x+1}=5^{x}$ can be written as $$x=\frac{\ln3}{\ln5-\ln3}.$$ [3]

(b) Solve $$2\log_{3}x-\log_{3}(x-2)=2.$$ [4]

(a) for taking logs of both sides and using the power law: $(x+1)\ln3=x\ln5$M1
(a) for collecting the $x$ terms: $x\ln5-x\ln3=\ln3$M1
(a) $x=\dfrac{\ln3}{\ln5-\ln3}$ with no errorsA1*
(b) for using the power and subtraction laws: $\log_{3}\dfrac{x^{2}}{x-2}=2$M1
(b) for removing logs: $x^{2}=9(x-2)$M1
(b) for solving the 3-term quadratic $x^{2}-9x+18=0$M1
(b) $x=3$ and $x=6$ (both valid as $x>2$)A1
Answer: (a) shown (b) $x=3,\ x=6$
Examiner tip: In (b) check each answer in the original equation – here both work, but say so, because examiners look for it.

7HardCalculator7 marks

A cup of tea is left to cool. Its temperature, $T\,^\circ\text{C}$, $t$ minutes after it was made is modelled by $$T=20+70e^{-kt},$$ where $k$ is a positive constant. After 10 minutes the temperature is $55\,^\circ\text{C}$.

(a) Show that $k=\dfrac{\ln2}{10}$. [3]

(b) Find the rate at which the temperature is decreasing at the instant when $T=41$. Give your answer to 3 significant figures. [3]

(a) for substituting: $55=20+70e^{-10k}$M1
(a) for rearranging to $e^{-10k}=\tfrac12$ and taking lnM1
(a) $-10k=\ln\tfrac12\Rightarrow k=\dfrac{\ln2}{10}$ with no errorsA1*
(b) for differentiating to the form $Ce^{-kt}$, e.g. $\dfrac{\mathrm{d}T}{\mathrm{d}t}=-70ke^{-kt}$M1
(b) for using $T=41$ to obtain $70e^{-kt}=21$ (or finding $t$ and substituting)M1
(b) $1.46\,^\circ\text{C}$ per minute (awrt; accept $\dfrac{\mathrm{d}T}{\mathrm{d}t}=-1.46$)A1
(c) $70e^{-kt}>0$ for all $t$, so $T>20$ (allow: as $t\to\infty$, $e^{-kt}\to0$ so $T\to20$ but $e^{-kt}$ is never zero)B1
Answer: (b) $1.46\,^\circ\text{C}$ per minute (c) $e^{-kt}>0$ so $T>20$
Examiner tip: For (b), write $70e^{-kt}=T-20=21$ – it saves finding $t$ and avoids rounding errors.

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