CAIE A-Level Mathematics 3.2 Logarithmic and Exponential Functions Question Bank

Practise solving logarithmic and exponential equations and transforming growth or decay relationships into straight-line form to calculate model constants and times.

Exam points

apply product, quotient and power laws before isolating the unknown in a log equation
take logarithms of exponential relations and solve the resulting linear equation accurately
match a transformed straight line to Y = mX + c and recover original model constants

Covered syllabus points

3.2.1—Log laws 3.2.2—Exponential and ln 3.2.3—Log equations 3.2.4—Linearising relationships

[Maximum number: 5]

(a)

Show that the equation $\log _{3}(2 x+1)=1+2 \log _{3}(x-1)$ can be written as a quadratic equation in x.

[ 3 ]

Use law of logarithm of a power

M1

$\log _{3}(2 x+1)=1+\log _{3}(x-1)^{2}$

Use $\log _{3} 3=1$

B1

$\begin{aligned} & \log _{3}(2 x+1)=\log _{3} 3+2 \log _{3}(x-1) & {\left[\log 3\left(\frac{2 x+1}{(x-1)^{2}}\right)=\log _{3} 3 \text { or }\left(\frac{2 x+1}{(x-1)^{2}}\right)=3\right]} \end{aligned}$

SC For candidates scoring M0 B0 due to combining logs before dealing with coefficient 2, and confusing coefficients, allow $\log _{3}(\ldots)=c$ leading to $(\ldots)=3^{c} \quad \mathbf{B1}$ .

Obtain $3 x^{2}-8 x+2=0$ or $1.5 x^{2}-4 x+1=0$

A1

OE 3 terms only and =0 required.

(b)

Hence solve the equation $\log _{3}(4 y+1)=1+2 \log _{3}(2 y-1)$ , giving your answer correct to 2 decimal places.

[ 2 ]

Solve 3-term quadratic equation from part 3(a) or restart to find y

M1

$\begin{aligned} & y=\frac{4 \pm \sqrt{10}}{6} \text { or } y=1.1937 \ldots \text { or } y=0.1396 \ldots & (x=2.3874 \text { or } x=0.2792) \end{aligned}$

May solve for x but must find $y=\frac{x}{2}$ to gain M1.

Obtain answer 1.19

A1

CAO. 2 dp required.

[Maximum number: 5]

The number of bacteria in a population, P, at time t hours is modelled by the equation $P=ae^{kt}$ , where a and k are constants. The graph of $\ln P$ against t, shown in the diagram, has gradient $\frac1{20}$ and intersects the vertical axis at (0,3).

(a)

State the value of k and find the value of a correct to 2 significant figures.

[ 3 ]

From the graph,
$\ln P=\ln a+kt=\frac1{20}t+3.$
B1

$k=\frac1{20}.$
B1

$\ln a=3,\quad a=e^3=20$
to 2 significant figures.
B1

(b)

Figure for Question (b) — CAIE A-Level Mathematics A2

Find the time taken for P to double. Give your answer correct to the nearest hour.

[ 2 ]

Use
$2a=ae^{kt},$
so
$2=e^{kt},\quad kt=\ln2.$
M1

With $k=\frac1{20}$ ,
$t=20\ln2=13.86\ldots\approx14\text{ hours}.$
A1

Guidance: allow 13.75 hours to 14 hours.

[Maximum number: 4]

Figure for Question 3.2 Logarithmic and exponential functions question 3 — CAIE A-Level Mathematics A2

The variables x and y are related by the equation $y=a b^{x}$ , where a and b are constants. The diagram shows the result of plotting $\ln y$ against x for two pairs of values of x and y. The coordinates of these points are ( 1,3.7 ) and ( 2.2,6.46 ).

Use this information to find the values of a and b.

State or imply that $\ln y=\ln a+x \ln b$

B1

Carry out a completely correct method for finding $\ln a$ or $\ln b$

M1

$3.7=\ln a+\ln b$ and $6.46=\ln a+2.2 \ln b$
leading to $\ln a=1.4, \ln b=2.3$ .

Obtain value a=4.06

A1

Obtain value b=9.97

A1

SC B1 for $a=\mathrm{e}^{1.4}$ and $b=\mathrm{e}^{2.3}$ .

Alternative Method for Question 3

$e^{3.7}=a b^{1}$ and $e^{6.46}=a b^{2.2}$

B1

Divide to obtain $\mathrm{e}^{2.76}=b^{1.2}$ and state or imply $2.76=1.2 \ln b$

M1

Obtain value a=4.06

A1

Obtain value b=9.97

A1

All question bank results loaded

Covered syllabus points

3.2 Logarithmic and exponential functions question 1

Question (a)

Question (b)

3.2 Logarithmic and exponential functions question 2

Question (a)

Question (b)

3.2 Logarithmic and exponential functions question 3