6 Exponentials and logarithms — revision question pack

7 specification points · notes, questions, answers and worked methods

Checked against Edexcel 9MA0 section 6. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Mathematics (9MA0) specification; registry verification recorded 11 July 2026.

How this checking works

6.1 · Know and use the function aˣ and its graph, where a is positive; know and use the function eˣ and its graph.

Explanation

For $a>0$ with $a\neq1$ , the exponential function $y=a^x$ has domain $\mathbb{R}$ , range $y>0$ , horizontal asymptote $y=0$ and intercept $(0,1)$ .
If $a>1$ the graph is increasing, while if $0<a<1$ it is decreasing; $a=1$ gives the constant function $y=1$ .
The natural exponential $e^x$ follows the same graph facts with base $e$ , and transformations such as $Ae^{kx}+c$ scale, reflect and translate the basic graph.
An exponential graph never reaches its horizontal asymptote; treating $a^0$ as $0$ instead of $1$ is a common error.

The basic exponential graphs approach the

x

-axis, their horizontal asymptote

y=0

; the base determines whether the curve rises or falls.

Worked example

For $y=2e^x-5$ , state the domain, range, horizontal asymptote and $y$ -intercept.

1.The function $e^x$ is defined and positive for every real $x$ .
2.Multiplying by $2$ preserves positivity, then subtracting $5$ gives $y>-5$ and moves the asymptote to $y=-5$ .
3.At $x=0$ , $y=2e^0-5=-3$ .

Answer: Domain $x\in\mathbb{R}$ Range $y>-5$ Horizontal asymptote $y=-5$ $y$ -intercept $(0,-3)$

Common mistakes

Don't write $a^0=0$ instead of $1$ , so the $y$ -intercept is wrong.
Don't draw an exponential curve crossing its horizontal asymptote.

Exam tip

State the asymptote, intercept, domain and range before sketching the transformed exponential graph.

Tier 1 · Easy

1.
For the graph $y=3^x$ , state the $y$ -intercept and the horizontal asymptote.
(2)
(Total for Question 1 is 2 marks)
2.
For the graph $y=\left(\dfrac13\right)^x$ , state whether the function is increasing or decreasing, and state its range.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
Describe the geometrical relationship between the graphs $y=2^x$ and $y=2^{-x}$ , and find their point of intersection.
(3)
(Total for Question 1 is 3 marks)
2.
The graph of $y=2^x$ is transformed to the graph of $y=3-2^{x-1}$ . Describe the transformations, and state the range and horizontal asymptote of the transformed graph.
(4)
(Total for Question 2 is 4 marks)
3.
The graph of $y=a^x$ , where $a>0$ , passes through $\left(3,\dfrac1{64}\right)$ . Find $a$ . State whether the function is increasing or decreasing, and state its $y$ -intercept and horizontal asymptote.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
The curve $y=3^x+3^{2-x}$ has a minimum point. Prove that its minimum value is $6$ and find the corresponding value of $x$ .
(5)
(Total for Question 1 is 5 marks)
2.
Find the exact coordinates of both intersections of the curves $y=4^x$ and $y=5-4^{-x}$ .
(5)
(Total for Question 2 is 5 marks)
3.
The curve $y=Aq^x+c$ , where $A>0$ and $0<q<1$ , passes through $(0,11)$ , $(1,7)$ and $(2,5)$ . Determine $A$ , $q$ and $c$ . Hence state the range and horizontal asymptote of the curve.
(6)
(Total for Question 3 is 6 marks)
4.
Starting with $y=e^x$ , consider the two transformations $A$ : a stretch parallel to the $y$ -axis with scale factor $3$ , and $B$ : a translation by the vector $\begin{pmatrix}\ln2\\-4\end{pmatrix}$ . Find the equations of the images obtained by applying $A$ then $B$ , and by applying $B$ then $A$ . Show that the two images do not intersect. Hence find the exact values of $x$ for which the images lie on opposite sides of the $x$ -axis.
(6)
(Total for Question 4 is 6 marks)
5.
Starting with the graph of $y=e^x$ , apply, in order, a stretch parallel to the $x$ -axis with scale factor $2$ , a reflection in the $x$ -axis, and a translation by the vector $\begin{pmatrix}-1\\4\end{pmatrix}$ . Find the equation, range, horizontal asymptote and exact $x$ -intercept of the resulting graph. Find also the image of the point $(0,1)$ under these transformations.
(7)
(Total for Question 5 is 7 marks)

6.2 · Know that the gradient of e^(kx) is equal to k·e^(kx) and hence understand why the exponential model is suitable in many applications.

Explanation

Differentiating gives $\dfrac{d}{dx}e^{kx}=ke^{kx}$ , so the gradient of an exponential is a constant multiple of the function itself.
For $Q=Q_0e^{kt}$ , the rate satisfies $dQ/dt=kQ$ : positive $k$ models growth and negative $k$ models decay.
A tangent at $x=x_0$ uses the point $(x_0,e^{kx_0})$ and gradient $ke^{kx_0}$ in $y-y_0=m(x-x_0)$ .
This proportional-rate property is the mathematical reason for choosing an exponential model.
Changing conditions can make its constant proportionality assumption unsuitable, so compare the model's rate with observed changes before accepting it as a valid model.

Worked example

Find the equation of the tangent to $y=e^{2x}$ at $x=1/2$ .

1.At $x=1/2$ , the point is $(1/2,e)$ .
2.Since $dy/dx=2e^{2x}$ , the gradient there is $2e$ .
3.Thus $y-e=2e(x-1/2)$ , which simplifies to $y=2ex$ .

Answer: $y=2ex$

Common mistakes

Don't differentiate $e^{kx}$ as $e^{kx}$ and omit the chain-rule factor $k$ .
Don't use an exponential model when the rate of change is not proportional to the current amount.

Exam tip

Show both $\mathrm dy/\mathrm dx=ky$ and the constant of proportionality when justifying an exponential model.

Tier 1 · Easy

1.
Differentiate $y=e^{4x}$ with respect to $x$ .
(1)
(Total for Question 1 is 1 mark)
2.
Given that $y=7e^{-0.4x}$ , show that $\dfrac{\mathrm dy}{\mathrm dx}=-0.4y$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
For the curve $y=e^{2x}$ , find the exact coordinates of the point where the gradient is $10$ .
(3)
(Total for Question 1 is 3 marks)
2.
The curve $y=Ae^{3x}$ passes through $(\ln2,40)$ . Determine, in exact form, its tangent at this point.
(4)
(Total for Question 2 is 4 marks)
3.
The point $P$ on the curve $y=5e^{2x}$ has $x$ -coordinate $p$ . Show that the tangent at $P$ meets the $x$ -axis at a point whose $x$ -coordinate is $p-\dfrac12$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
A culture is modelled by $P=320e^{0.18t}$ , where $t$ is in hours. Find its instantaneous growth rate when $P=500$ , and explain the feature of the model that makes this calculation direct.
(4)
(Total for Question 1 is 4 marks)
2.
A quantity is modelled by $Q=Ce^{-kt}$ , where $C>0$ and $k>0$ . When $Q=180$ , it is decreasing at $27$ units per hour. Given also that $Q=600$ when $t=0$ , find the time when $Q=100$ , giving your answer to $3$ significant figures.
(6)
(Total for Question 2 is 6 marks)
3.
The curve $y=Ae^{kx}$ passes through $(2,12)$ and has gradient $9$ there. Find $A$ and $k$ in exact form. The tangent at this point meets the coordinate axes at $R$ and $S$ . Find the exact area of triangle $ORS$ , where $O$ is the origin.
(6)
(Total for Question 3 is 6 marks)
4.
The curve $y=Ae^{kx}$ , where $A>0$ and $k>0$ , passes through $P(1,8)$ . The normal to the curve at $P$ passes through $Q(5,6)$ . Find $A$ and $k$ in exact form. Hence find the exact coordinates of the point on the curve where the gradient is $5$ .
(6)
(Total for Question 4 is 6 marks)
5.
A curve has equation $y=Ae^{kx}+c$ , where $A>0$ and $k>0$ . At $x=0$ its value is $7$ and its gradient is $6$ . At $x=\ln3$ its value is $19$ and its gradient is $18$ . Determine $A$ , $k$ and $c$ . Hence find the exact value of $x$ for which $y=37$ , and find the tangent at $x=0$ .
(7)
(Total for Question 5 is 7 marks)

6.3 · Know and use the definition of logₐx as the inverse of aˣ, where a is positive and x ≥ 0; know and use the function ln x and its graph; know and use ln x as the inverse function of eˣ.

Explanation

For $a>0$ with $a\neq1$ , $\log_a x=y$ means exactly that $a^y=x$ ; a real logarithm requires $x>0$ .
The graph of $y=\log_a x$ is the reflection of $y=a^x$ in $y=x$ , so its domain is $x>0$ , range is $\mathbb{R}$ and vertical asymptote is $x=0$ .
Natural logarithms use base $e$ , giving the inverse relations $\ln(e^x)=x$ for every real $x$ and $e^{\ln x}=x$ for $x>0$ .
The specification's inverse relationship does not make $\ln0$ defined: $e^x$ approaches zero but never equals it, so logarithm arguments must be strictly positive.

Worked example

The function $f(x)=e^{x-3}+2$ . Find $f^{-1}(x)$ and state the domain of the inverse.

1.Write $y=e^{x-3}+2$ , so $y-2=e^{x-3}$ .
2.Taking natural logarithms gives $\ln(y-2)=x-3$ , hence $x=\ln(y-2)+3$ .
3.Interchanging labels gives $f^{-1}(x)=\ln(x-2)+3$ .
4.The range of $f$ is $y>2$ , so the inverse domain is $x>2$ .

Answer: $f^{-1}(x)=\ln(x-2)+3$ Domain $x>2$

Common mistakes

Don't include zero in the domain of $\ln x$ , even though $e^x$ approaches but never reaches zero.
Don't find an inverse formula and fail to interchange the exponential function's range with the logarithm's domain.

Exam tip

Before taking a logarithm, require its argument to be strictly positive and state the resulting domain.

Tier 1 · Easy

1.
Solve $\ln x=2$ exactly.
(2)
(Total for Question 1 is 2 marks)
2.
Find the exact value of $\log_5\!\left(\dfrac1{125}\right)$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
For $y=\ln(x-2)+1$ , state the domain, range and vertical asymptote, and find the exact $x$ -intercept.
(4)
(Total for Question 1 is 4 marks)
2.
Let $f(x)=3-e^{-x}$ for $x\in\mathbb{R}$ . Find $f^{-1}(x)$ and state the domain of $f^{-1}$ .
(4)
(Total for Question 2 is 4 marks)
3.
Given that $\log_a8=\dfrac32$ , find $a$ . Hence solve $\log_a(x-1)=-\dfrac12$ , giving the exact value of $x$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
For $g(x)=\ln(4-x^2)$ , determine the domain and range, and identify the maximum point.
(5)
(Total for Question 1 is 5 marks)
2.
The function $f(x)=\ln\!\left(\dfrac{x+2}{4-x}\right)$ is defined on its largest possible interval. Find its domain, range and vertical asymptotes. Hence find $f^{-1}(x)$ and state its domain.
(6)
(Total for Question 2 is 6 marks)
3.
Sketch the graph of $y=|\ln x|$ for $x>0$ , marking its $x$ -intercept, vertical asymptote and minimum point. State its range. Hence solve $|\ln x|=\ln7$ exactly.
(6)
(Total for Question 3 is 6 marks)
4.
A function $f$ has inverse $f^{-1}(x)=3^{2x-1}+4$ for $x\in\mathbb{R}$ . Find $f(x)$ . State the domain, range and vertical asymptote of $f$ , and find its exact $x$ -intercept.
(6)
(Total for Question 4 is 6 marks)
5.
A point $P$ lies on the curve $y=3^x$ , and $Q$ is the reflection of $P$ in the line $y=x$ . Prove that $Q$ lies on $y=\log_3x$ . Given that the midpoint of $PQ$ is $M\left(\dfrac{11}{2},\dfrac{11}{2}\right)$ , find the coordinates of $P$ and $Q$ without using numerical methods. Hence find the equation of $PQ$ and its exact length.
(6)
(Total for Question 5 is 6 marks)

6.4 · Understand and use the laws of logarithms: logₐx + logₐy = logₐ(xy); logₐx − logₐy = logₐ(x/y); k logₐx = logₐxᵏ (including, for example, k = −1 and k = −½).

Explanation

For positive arguments, $\log_a x+\log_a y=\log_a(xy)$ , $\log_a x-\log_a y=\log_a(x/y)$ and $k\log_a x=\log_a(x^k)$ ; also $\log_a a=1$ and $\log_a1=0$ .
To expand a logarithm, turn products into sums, quotients into differences and powers into coefficients; to condense, apply those steps in reverse.
A negative coefficient represents a reciprocal power, for example $-\tfrac12\log_a x=\log_a(x^{-1/2})=\log_a(1/\sqrt{x})$ .
The laws combine logarithms, not their arguments: $\log(x+y)$ cannot be split into $\log x+\log y$ , and every original logarithm argument must remain positive.
Check the final form by expanding it back again.

Worked example

Expand $\ln\!\left(\dfrac{x^3\sqrt{y}}{z^2}\right)$ , where $x$ , $y$ and $z$ are positive.

1.The quotient gives subtraction, the product gives addition, and powers become coefficients: $\ln(x^3)+\ln(y^{1/2})-\ln(z^2)=3\ln x+\tfrac12\ln y-2\ln z$ .

Answer: $3\ln x+\dfrac12\ln y-2\ln z$

Common mistakes

Don't split $\log(x+y)$ into $\log x+\log y$ , although there is no logarithm law for a sum.
Don't combine logarithms without retaining the condition that every original argument is positive.

Exam tip

For “single logarithm”, apply powers first, then combine products and quotients while keeping domain restrictions.

Tier 1 · Easy

1.
Write $\ln12+\ln3-\ln2$ as a single logarithm.
(2)
(Total for Question 1 is 2 marks)
2.
Given that $p>0$ and $q>0$ , express $3\log_a p-\dfrac12\log_a q$ as a single logarithm.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
Solve $2\ln(x+1)-\ln x=\ln8$ , giving all solutions allowed by the original equation.
(5)
(Total for Question 1 is 5 marks)
2.
Given that $\log_a2=p$ and $\log_a3=q$ , express $\log_{\sqrt a}72$ in terms of $p$ and $q$ .
(4)
(Total for Question 2 is 4 marks)
3.
Solve $\ln(x+6)=2\ln x$ , giving all solutions permitted by the original logarithms.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Solve $\log_3(x-1)+\log_3(x+3)=2$ , checking the domain of the original equation.
(5)
(Total for Question 1 is 5 marks)
2.
Solve $\log_x16=\log_2x$ , giving all real values of $x$ allowed by the logarithms.
(5)
(Total for Question 2 is 5 marks)
3.
Positive numbers $p$ and $q$ satisfy $\log_2p+\log_4q=5$ and $\log_8p-\log_2q=-1$ . Find $p$ and $q$ exactly.
(5)
(Total for Question 3 is 5 marks)
4.
Solve the inequality $\ln\!\left(\dfrac{x-1}{x+2}\right)>\ln\!\left(\dfrac12\right)$ , giving the complete set of real values of $x$ permitted by the original logarithm.
(5)
(Total for Question 4 is 5 marks)
5.
The equation $\ln(x-1)+\ln(5-x)=k$ is considered for real $x$ and real $k$ . Determine the values of $k$ for which the equation has two distinct real solutions, exactly one real solution, or no real solutions. Find the solutions when $k=\ln3$ .
(6)
(Total for Question 5 is 6 marks)

6.5 · Solve equations of the form aˣ = b.

Explanation

For $a>0$ , $a\neq1$ and $b>0$ , taking logarithms gives $a^x=b\Rightarrow x=\log_a b=\ln b/\ln a$ .
If the exponent is linear, isolate the exponential or take logarithms first, then solve the resulting linear equation without rounding intermediate values.
When both $a^x$ and $a^{-x}$ occur, substitute $u=a^x>0$ , so $a^{-x}=1/u$ , and solve the resulting algebraic equation before converting back.
A positive-base exponential cannot equal zero or a negative number; after a substitution, reject non-positive values because $a^x$ is always positive.
Substitute the final value into the original equation to verify it.

Worked example

Find the solution of $3^{2x-1}=20$ , giving $x$ to $3$ decimal places.

1.Taking logarithms gives $(2x-1)\ln3=\ln20$ .
2.Therefore $2x-1=\ln20/\ln3$ and $x=\tfrac12(1+\ln20/\ln3)=1.8634\ldots$ , so $x=1.863$ .

Answer: $x=1.863$

Common mistakes

Don't take logarithms but forget to multiply the exponent by $\log a$ .
Don't accept a non-positive substituted value for $a^x$ , even though a positive-base exponential is always positive.

Exam tip

Use one logarithm base consistently and keep full calculator precision until the requested final rounding.

Tier 1 · Easy

1.
Solve $5^x=17$ , giving $x$ to $3$ decimal places.
(2)
(Total for Question 1 is 2 marks)
2.
Solve $7^{x+1}=19$ , giving your answer to $3$ decimal places.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
Solve $5^{2x-1}=40$ , giving your answer exactly in logarithmic form and to $3$ decimal places.
(3)
(Total for Question 1 is 3 marks)
2.
Solve $2^{x+1}=5^{2-x}$ . Give your answer exactly in logarithmic form and to $3$ decimal places.
(4)
(Total for Question 2 is 4 marks)
3.
The equation $5^{2x+k}=13$ has the solution $x=1.2$ . Find the value of $k$ , giving your answer to three decimal places.
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1.
Determine all real solutions of $2^x+2^{-x}=5/2$ .
(5)
(Total for Question 1 is 5 marks)
2.
Find all real solutions of $9^x-9\cdot3^x+14=0$ , giving exact answers.
(5)
(Total for Question 2 is 5 marks)
3.
The equation $4^x-k\,2^x+9=0$ , where $k$ is a real constant, has exactly one real solution. Find $k$ and the exact solution for $x$ .
(6)
(Total for Question 3 is 6 marks)
4.
Solve $5^{2x-1}>7^{x+2}$ , giving the exact boundary for $x$ . Hence find the smallest integer value of $x$ satisfying the inequality.
(5)
(Total for Question 4 is 5 marks)
5.
Solve simultaneously for the real numbers $x$ and $y$ , given that $2^x3^y=12$ and $\dfrac{4^x}{3^y}=\dfrac{16}{3}$ . Give the exact values of $x$ and $y$ .
(4)
(Total for Question 5 is 4 marks)

6.6 · Use logarithmic graphs to estimate parameters in relationships of the form y = axⁿ and y = kbˣ, given data for x and y.

Explanation

For $y=ax^n$ , taking logarithms gives $\log y=\log a+n\log x$ , so a plot of $\log y$ against $\log x$ has gradient $n$ and intercept $\log a$ .
For $y=kb^x$ , $\log y=\log k+x\log b$ , so a plot of $\log y$ against $x$ has gradient $\log b$ and intercept $\log k$ .
Read two well-separated points from the fitted line, calculate its gradient and intercept, then undo the chosen logarithm base to recover $a$ , $k$ or $b$ .
The transformed coordinates and the logarithm base must be identified: confusing a log-log graph with a semi-log graph, or forgetting to exponentiate the intercept, gives incorrect parameters.

For

y=ax^n

, plotting

\log y

against

\log x

gives a straight line with gradient

n

and intercept

\log a

Worked example

For a model $y=ax^n$ , a fitted graph of $\log_{10}y$ against $\log_{10}x$ passes through $(0.2,0.65)$ and $(0.8,1.55)$ . Estimate $a$ and $n$ , then predict $y$ when $x=5$ .

1.The gradient is $n=(1.55-0.65)/(0.8-0.2)=1.5$ .
2.Using $(0.2,0.65)$ , the intercept is $\log_{10}a=0.65-1.5(0.2)=0.35$ , so $a=10^{0.35}=2.2387\ldots$ .
3.Thus at $x=5$ , $y=2.2387\ldots(5^{1.5})=25.030\ldots$ , giving about $25.0$ .

Answer: $n=1.5$ $a\approx2.24$ $y\approx25.0$

Common mistakes

Don't use the intercept itself as $a$ or $k$ instead of undoing the logarithm.
Don't plot $(\log x,\log y)$ for an exponential model $y=kb^x$ , which requires $(x,\log y)$ .

Exam tip

Name the transformed axes, then identify exactly which model parameter is the gradient and which is obtained from the intercept.

Tier 1 · Easy

1.
A straight-line fit on base- $10$ logarithmic axes has equation $\log_{10}y=0.4+1.7\log_{10}x$ . State the corresponding model $y=ax^n$ , giving $a$ to $3$ significant figures.
(3)
(Total for Question 1 is 3 marks)
2.
A relationship is believed to have the form $y=kb^x$ . State the quantities that should be plotted to obtain a straight line, and state what its gradient and vertical-axis intercept represent when natural logarithms are used.
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
For a model $y=kb^x$ , a fitted graph of $\ln y$ against $x$ passes through $(2,1.3)$ and $(6,2.5)$ . Estimate $k$ and $b$ to $3$ significant figures, then predict $y$ when $x=4$ .
(5)
(Total for Question 1 is 5 marks)
2.
For a model $y=kb^x$ , the fitted straight line has equation $\log_{10}y=0.52+0.28x$ . Estimate $k$ and $b$ to $3$ significant figures. Hence find the value of $x$ for which $y=50$ , giving $x$ to $3$ significant figures.
(5)
(Total for Question 2 is 5 marks)
3.
For a model $y=ax^n$ , a fitted graph of $\log_{10}y$ against $\log_{10}x$ passes through $(-1,1.2)$ and $(2,3.0)$ . Find $n$ and $a$ , giving $a$ to $3$ significant figures. Find the value of $\log_{10}y$ when $\log_{10}x=0.5$ .
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
An exponential relationship $y=kb^x$ is analysed by plotting $\ln y$ against $x$ . The fitted line goes through $(1,2.1)$ and $(5,3.7)$ . Find $k$ and $b$ to $3$ significant figures, and estimate $y$ at $x=3$ .
(6)
(Total for Question 1 is 6 marks)
2.
A relationship is modelled by $y=ax^n$ . Because of a rescaled vertical axis, the fitted line is $\log_{10}(2y)=1.4-0.8\log_{10}x$ . Find $a$ and $n$ , giving $a$ to $3$ significant figures. Hence estimate the value of $x$ when $y=4$ , giving $x$ to $3$ significant figures. Use unrounded values in your working.
(6)
(Total for Question 2 is 6 marks)
3.
Three data pairs are $(x,y)=(4,16)$ , $(9,54)$ and $(16,128)$ . Use logarithms to decide whether $y=ax^n$ or $y=kb^x$ fits all three pairs exactly. Determine the parameters of the chosen model and hence predict $y$ when $x=25$ .
(6)
(Total for Question 3 is 6 marks)
4.
A relationship has the form $y=ax^n$ , where $x>0$ . Doubling $x$ multiplies $y$ by $8$ , and $y=20$ when $x=2$ . Determine $a$ and $n$ . Write the corresponding straight-line equation for a graph of $\log_{10}y$ against $\log_{10}x$ , and find $x$ when $y=540$ .
(6)
(Total for Question 4 is 6 marks)
5.
For positive $x$ and $y$ , a fitted graph of $\log_{10}x$ against $\log_{10}y$ has equation $\log_{10}x=0.4\log_{10}y-0.2$ . Express the relationship in the form $y=ax^n$ , giving $a$ and $n$ exactly. Hence predict the exact value of $y$ when $x=40$ .
(5)
(Total for Question 5 is 5 marks)

6.7 · Understand and use exponential growth and decay; use in modelling (e.g. compound interest, radioactive decay, drug concentration decay, population growth); consideration of limitations and refinements of exponential models.

Explanation

Repeated percentage change over whole periods uses $Q_n=Q_0(1+r)^n$ , while continuous change at a rate proportional to the amount uses $Q=Q_0e^{kt}$ .
For decay $k<0$ ; the half-life in $Q=Q_0e^{-\lambda t}$ is $t_{1/2}=\ln2/\lambda$ , independent of the starting amount.
Use two observations to determine an unknown multiplier or exponent, preserve full precision, and interpret a calculated time or amount in the units and practical setting of the model.
Unlimited exponential growth and a constant decay parameter may fail when resources, competition, dosage cycles or environmental conditions change; a carrying-capacity or piecewise model can be a refinement.

Worked example

A medicine concentration is modelled by $C=80e^{-0.23t}$ , where $C$ is in $\text{mg L}^{-1}$ and $t$ is in hours. Calculate the half-life and the concentration after $5$ hours, each to $3$ significant figures.

1.At half-life, $e^{-0.23t}=1/2$ , so $t=\ln2/0.23=3.0136\ldots$ hours.
2.After $5$ hours, $C=80e^{-0.23(5)}=25.3309\ldots\,\text{mg L}^{-1}$ .
3.The requested values are $3.01$ hours and $25.3\,\text{mg L}^{-1}$ .

Answer: Half-life $=3.01$ hours $C(5)=25.3\,\text{mg L}^{-1}$

Common mistakes

Don't use a positive exponent for a decay model, so the predicted quantity increases.
Don't assume unlimited growth or a constant decay rate when resources, dosage cycles or conditions change.

Exam tip

Interpret the initial value at $t=0$ , then link a stated model limitation to a concrete carrying-capacity or piecewise refinement.

Tier 1 · Easy

1.
A savings balance of $\pounds1200$ earns $3.5\%$ compound interest each year. Find the balance after $4$ years, to the nearest penny.
(2)
(Total for Question 1 is 2 marks)
2.
A machine is worth $\pounds4500$ and loses $12\%$ of its value each year. Write down a model for its value $V$ pounds after $n$ years.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
An investment of $\pounds5000$ grows by $4.2\%$ each year. Find the smallest whole number of years after which its value exceeds $\pounds8000$ , and state one limitation of this exponential model.
(4)
(Total for Question 1 is 4 marks)
2.
The value of equipment is modelled by $V=15000q^t$ , where $t$ is measured in years. After $4$ years its value is $\pounds9800$ . Find the annual percentage depreciation to $3$ significant figures, and predict the value after $7$ years to the nearest pound. Use unrounded values in your working.
(4)
(Total for Question 2 is 4 marks)
3.
A substance is modelled by $M=250e^{-kt}$ , where $t$ is measured in hours and $k>0$ . After $5$ hours, $82\%$ of the initial mass remains. Find $k$ and the half-life, each to $3$ significant figures. Use unrounded values in your working.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
A colony is modelled by $P=600e^{kt}$ . Its measured population at $t=4$ days is $900$ . Find $k$ , predict when the model first reaches $2000$ , and state one limitation with a suitable refinement.
(6)
(Total for Question 1 is 6 marks)
2.
A chemical mass is modelled by $M=M_0e^{-kt}$ , where $t$ is in days. Measurements give $M=240$ mg when $t=3$ and $M=85$ mg when $t=9$ . Find $k$ and $M_0$ , then find when the model first predicts $M<30$ mg. Give numerical answers to $3$ significant figures and state one limitation of the model.
(6)
(Total for Question 2 is 6 marks)
3.
Two populations are modelled by $A=900(1.08)^t$ and $B=1500(0.96)^t$ , where $t$ is measured in years. Find when the models predict equal populations, giving $t$ to $3$ significant figures. Hence find the first whole number of years for which $A>B$ . State one limitation of this comparison and a suitable refinement. Use unrounded values in your working.
(6)
(Total for Question 3 is 6 marks)
4.
A town population is modelled by $P=2000e^{kt}$ , where $t$ is in years. The measured population at $t=3$ is $2500$ . Find $k$ exactly and use the model to predict $P$ at $t=6$ . The actual population at $t=6$ is $2900$ . Calculate the percentage by which the model overestimates this value, giving your answer to $3$ significant figures, and assess the constant-rate assumption. Use unrounded values in your working.
(6)
(Total for Question 4 is 6 marks)
5.
Before a treatment begins, a population is modelled by $P=1200e^{0.18t}$ for $0\leq t\leq4$ , where $t$ is in days. From $t=4$ onwards, the treatment makes the population decay continuously at a rate of $12\%$ of its current value per day. Construct a continuous model for $P$ when $t\geq4$ . Find the maximum modelled population and the time when the model first returns to $1200$ .
(6)
(Total for Question 5 is 6 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

6.1 · Know and use the function aˣ and its graph, where a is positive; know and use the function eˣ and its graph.

Tier 1 · Easy

Mark scheme for 6.1 Tier 1 · Easy
Question	Scheme	Marks
1	$y$ -intercept $(0,1)$ Horizontal asymptote $y=0$	2
(2 marks)		2
Notes
At $x=0$ , $y=3^0=1$ , giving the intercept $(0,1)$ . As $x\to-\infty$ , $3^x\to0$ without reaching zero, so the horizontal asymptote is $y=0$ .
2	The function is decreasing Range $y>0$	2
(2 marks)		2
Notes
The base $1/3$ lies between $0$ and $1$ , so the exponential function decreases as $x$ increases. A positive-base exponential is always positive, so its range is $y>0$ .

Tier 2 · Standard

Mark scheme for 6.1 Tier 2 · Standard
Question	Scheme	Marks
1	The graphs are reflections of each other in the $y$ -axis. Intersection $(0,1)$	3
(3 marks)		3
Notes
Replacing $x$ by $-x$ reflects a graph in the $y$ -axis, so $y=2^{-x}$ is the reflection of $y=2^x$ . At an intersection, $2^x=2^{-x}$ , hence $x=-x$ and $x=0$ . Both functions then have value $1$ , so the intersection is $(0,1)$ .
2	Translate $1$ unit to the right, reflect in the $x$ -axis, then translate $3$ units upwards Range $y<3$ Horizontal asymptote $y=3$	4
(4 marks)		4
Notes
Replacing $x$ by $x-1$ translates the graph $1$ unit right. Multiplication by $-1$ reflects it in the $x$ -axis, and adding $3$ translates it $3$ units upwards. Since $2^{x-1}>0$ , the transformed values satisfy $y<3$ and approach the horizontal asymptote $y=3$ .
3	$a=\dfrac14$ The function is decreasing $y$ -intercept $(0,1)$ Horizontal asymptote $y=0$	4
(4 marks)		4
Notes
The point gives $a^3=1/64=(1/4)^3$ . Since $a>0$ , $a=1/4$ . As $0<a<1$ , the function is decreasing. Also $a^0=1$ , giving the $y$ -intercept $(0,1)$ , and $a^x\to0$ as $x\to\infty$ , so the horizontal asymptote is $y=0$ .

Tier 3 · Hard

Mark scheme for 6.1 Tier 3 · Hard
Question	Scheme	Marks
1	Minimum value $6$ , attained when $x=1$	5
(5 marks)		5
Notes
Let $u=3^x$ , so $u>0$ and $3^{2-x}=9/u$ . Then $y=u+9/u$ . Since $y-6=(u^2-6u+9)/u=(u-3)^2/u\geq0$ , it follows that $y\geq6$ . Equality occurs when $u=3$ , so $3^x=3$ and $x=1$ . Hence the minimum point is $(1,6)$ .
2	$\left(\log_4\!\left(\dfrac{5-\sqrt{21}}2\right),\dfrac{5-\sqrt{21}}2\right)$ and $\left(\log_4\!\left(\dfrac{5+\sqrt{21}}2\right),\dfrac{5+\sqrt{21}}2\right)$	5
(5 marks)		5
Notes
Let $u=4^x$ , so $u>0$ and $4^{-x}=1/u$ . At an intersection, $u=5-1/u$ , hence $u^2-5u+1=0$ . Thus $u=(5\pm\sqrt{21})/2$ ; both roots are positive. Since $u=4^x=y$ , the two corresponding coordinates are the stated points.
3	$A=8$ , $q=\dfrac12$ and $c=3$ Range $y>3$ Horizontal asymptote $y=3$	6
(6 marks)		6
Notes
The three points give $A+c=11$ , $Aq+c=7$ and $Aq^2+c=5$ . The quantities $11-c$ , $7-c$ and $5-c$ are consecutive terms of a geometric progression, and each is nonzero since $A>0$ and $q>0$ , so $\dfrac{7-c}{11-c}=\dfrac{5-c}{7-c}$ . Hence $(7-c)^2=(11-c)(5-c)$ , which gives $c=3$ . It follows that $A=8$ and $q=(7-3)/8=1/2$ . Since $8(1/2)^x>0$ for all real $x$ and tends to zero as $x\to\infty$ , the range is $y>3$ and the horizontal asymptote is $y=3$ .
4	$A$ then $B$ : $y=\dfrac32e^x-4$ $B$ then $A$ : $y=\dfrac32e^x-12$ The images do not intersect $\ln\!\left(\dfrac83\right)<x<\ln8$	6
(6 marks)		6
Notes
Applying $A$ first gives $y=3e^x$ ; translating right by $\ln2$ and down by $4$ then gives $y=3e^{x-\ln2}-4=\tfrac32e^x-4$ . Applying $B$ first gives $y=e^{x-\ln2}-4=\tfrac12e^x-4$ ; the vertical stretch then gives $y=\tfrac32e^x-12$ . The first image is exactly $8$ units above the second for every $x$ , so they cannot intersect. They are on opposite sides of the $x$ -axis exactly when the first is positive and the second is negative: $\tfrac32e^x>4$ and $\tfrac32e^x<12$ . Thus $8/3<e^x<8$ , and since $\ln$ is increasing, $\ln(8/3)<x<\ln8$ .
5	$y=4-e^{(x+1)/2}$ Range $y<4$ and horizontal asymptote $y=4$ $x$ -intercept $\left(2\ln4-1,0\right)$ The image of $(0,1)$ is $(-1,3)$	7
(7 marks)		7
Notes
The horizontal stretch gives $y=e^{x/2}$ , reflection gives $y=-e^{x/2}$ , and translating one unit left and four units up gives $y=4-e^{(x+1)/2}$ . The exponential term is positive, so $y<4$ and approaches the horizontal asymptote $y=4$ . Setting $y=0$ gives $e^{(x+1)/2}=4$ , hence $x=2\ln4-1$ . The point $(0,1)$ maps successively to $(0,1)$ , $(0,-1)$ and $(-1,3)$ .

6.2 · Know that the gradient of e^(kx) is equal to k·e^(kx) and hence understand why the exponential model is suitable in many applications.

Tier 1 · Easy

Mark scheme for 6.2 Tier 1 · Easy
Question	Scheme	Marks
1	$\dfrac{dy}{dx}=4e^{4x}$	1
(1 mark)		1
Notes
For $e^{kx}$ , differentiation multiplies the function by $k$ . Here $k=4$ , so $dy/dx=4e^{4x}$ .
2	$\dfrac{\mathrm dy}{\mathrm dx}=-2.8e^{-0.4x}=-0.4y$	2
(2 marks)		2
Notes
Differentiating gives $\dfrac{\mathrm dy}{\mathrm dx}=7(-0.4)e^{-0.4x}=-2.8e^{-0.4x}$ . Since $y=7e^{-0.4x}$ , this is $-0.4y$ .

Tier 2 · Standard

Mark scheme for 6.2 Tier 2 · Standard
Question	Scheme	Marks
1	$\left(\dfrac12\ln5,5\right)$	3
(3 marks)		3
Notes
Differentiate to obtain $dy/dx=2e^{2x}$ . Setting the gradient equal to $10$ gives $2e^{2x}=10$ , so $e^{2x}=5$ and $x=\tfrac12\ln5$ . At this value, $y=e^{2x}=5$ , giving the point $\left(\tfrac12\ln5,5\right)$ .
2	$y-40=120(x-\ln2)$ (or $y=120x+40-120\ln2$ )	4
(4 marks)		4
Notes
Using the point, $40=Ae^{3\ln2}=8A$ , so $A=5$ . Also $\dfrac{\mathrm dy}{\mathrm dx}=3Ae^{3x}=3y$ , so the gradient at the point is $3(40)=120$ . The tangent is therefore $y-40=120(x-\ln2)$ .
3	The tangent meets the $x$ -axis at $\left(p-\dfrac12,0\right)$	4
(4 marks)		4
Notes
At $P$ , $y=5e^{2p}$ and $\dfrac{\mathrm dy}{\mathrm dx}=10e^{2x}$ , so the tangent gradient is $10e^{2p}$ . Its equation is $y-5e^{2p}=10e^{2p}(x-p)$ . At the $x$ -axis, $y=0$ , so $-5e^{2p}=10e^{2p}(x-p)$ . Dividing by the non-zero factor $5e^{2p}$ gives $-1=2(x-p)$ , hence $x=p-1/2$ .

Tier 3 · Hard

Mark scheme for 6.2 Tier 3 · Hard
Question	Scheme	Marks
1	$90$ units per hour The growth rate is proportional to the current population.	4
(4 marks)		4
Notes
Differentiate: $dP/dt=0.18(320e^{0.18t})=0.18P$ . When $P=500$ , $dP/dt=0.18(500)=90$ units per hour. The exponential model makes the derivative a constant multiple of the current value.
2	$k=0.15$ $t=11.9$ hours	6
(6 marks)		6
Notes
Differentiating gives $\dfrac{\mathrm dQ}{\mathrm dt}=-kQ$ . Hence $-27=-180k$ , so $k=0.15$ . Since $Q(0)=C=600$ , setting $Q=100$ gives $100=600e^{-0.15t}$ . Thus $e^{-0.15t}=1/6$ and $t=\ln6/0.15=11.945\ldots$ , so $t=11.9$ hours to $3$ significant figures.
3	$k=\dfrac34$ and $A=12e^{-3/2}$ Area of triangle $ORS=2$ square units	6
(6 marks)		6
Notes
Since $\dfrac{\mathrm dy}{\mathrm dx}=kAe^{kx}=ky$ , the data at $(2,12)$ give $9=12k$ , so $k=3/4$ . Then $12=Ae^{(3/4)(2)}$ , giving $A=12e^{-3/2}$ . The tangent is $y-12=9(x-2)$ , or $y=9x-6$ . Its intercepts are $R=(2/3,0)$ and $S=(0,-6)$ , so the area of triangle $ORS$ is $\tfrac12(2/3)(6)=2$ square units.
4	$k=\dfrac14$ and $A=8e^{-1/4}$ $\left(1+4\ln\!\left(\dfrac52\right),20\right)$	6
(6 marks)		6
Notes
The normal gradient is $(6-8)/(5-1)=-1/2$ , so the tangent gradient at $P$ is $2$ . Since $\mathrm dy/\mathrm dx=ky$ , $2=8k$ and $k=1/4$ . Using $P$ gives $8=Ae^{1/4}$ , so $A=8e^{-1/4}$ . Where the gradient is $5$ , $ky=5$ , hence $y=20$ . Writing the curve as $y=8e^{(x-1)/4}$ gives $20=8e^{(x-1)/4}$ , so $x=1+4\ln(5/2)$ .
5	$A=6$ , $k=1$ and $c=1$ $x=\ln6$ Tangent $y=6x+7$	7
(7 marks)		7
Notes
For $y=Ae^{kx}+c$ , $\mathrm dy/\mathrm dx=kAe^{kx}=k(y-c)$ . The two point-gradient pairs give $6=k(7-c)$ and $18=k(19-c)$ . Subtracting gives $12=12k$ , so $k=1$ ; then $c=1$ . At $x=0$ , $7=A+1$ , hence $A=6$ . Thus $y=6e^x+1$ . Setting $y=37$ gives $e^x=6$ and $x=\ln6$ . The tangent uses point $(0,7)$ and gradient $6$ , so it is $y=6x+7$ .

6.3 · Know and use the definition of logₐx as the inverse of aˣ, where a is positive and x ≥ 0; know and use the function ln x and its graph; know and use ln x as the inverse function of eˣ.

Tier 1 · Easy

Mark scheme for 6.3 Tier 1 · Easy
Question	Scheme	Marks
1	$x=e^2$	2
(2 marks)		2
Notes
Apply $e$ to both sides: $e^{\ln x}=e^2$ , hence $x=e^2$ .
2	$-3$	2
(2 marks)		2
Notes
Since $125=5^3$ , $1/125=5^{-3}$ . By the definition of a logarithm, $\log_5(5^{-3})=-3$ .

Tier 2 · Standard

Mark scheme for 6.3 Tier 2 · Standard
Question	Scheme	Marks
1	Domain $x>2$ Range $y\in\mathbb{R}$ Vertical asymptote $x=2$ $x$ -intercept $\left(2+e^{-1},0\right)$	4
(4 marks)		4
Notes
The logarithm requires $x-2>0$ , so the domain is $x>2$ and the vertical asymptote is $x=2$ . A translation does not restrict the range of the logarithm, so the range is all real numbers. At the $x$ -intercept, $\ln(x-2)+1=0$ , so $x-2=e^{-1}$ and $x=2+e^{-1}$ .
2	$f^{-1}(x)=-\ln(3-x)$ Domain $x<3$	4
(4 marks)		4
Notes
Write $y=3-e^{-x}$ . Then $e^{-x}=3-y$ , so $-x=\ln(3-y)$ and $x=-\ln(3-y)$ . Interchanging $x$ and $y$ gives $f^{-1}(x)=-\ln(3-x)$ . Since $e^{-x}>0$ , the range of $f$ is $y<3$ , which is the inverse domain.
3	$a=4$ $x=\dfrac32$	4
(4 marks)		4
Notes
The definition of a logarithm gives $a^{3/2}=8$ , so $(\sqrt a)^3=8$ . Since a logarithm base is positive, $\sqrt a=2$ and $a=4$ . Then $\log_4(x-1)=-1/2$ gives $x-1=4^{-1/2}=1/2$ , so $x=3/2$ , which satisfies $x-1>0$ .

Tier 3 · Hard

Mark scheme for 6.3 Tier 3 · Hard
Question	Scheme	Marks
1	Domain $-2<x<2$ Range $-\infty<g(x)\leq\ln4$ Maximum point $(0,\ln4)$	5
(5 marks)		5
Notes
The logarithm requires $4-x^2>0$ , so $x^2<4$ and $-2<x<2$ . The argument $4-x^2$ has maximum $4$ at $x=0$ , giving the maximum value $\ln4$ . As $x\to2^-$ or $x\to-2^+$ , the argument tends to $0^+$ and $g(x)\to-\infty$ . Therefore the range is $(-\infty,\ln4]$ .
2	Domain $-2<x<4$ Range $\mathbb{R}$ Vertical asymptotes $x=-2$ and $x=4$ $f^{-1}(x)=\dfrac{4e^x-2}{e^x+1}$ , with domain $x\in\mathbb{R}$	6
(6 marks)		6
Notes
The quotient is positive on $-2<x<4$ , which is the required interval. It tends to $0^+$ as $x\to-2^+$ and to $+\infty$ as $x\to4^-$ , so $f$ has the two vertical asymptotes and range $\mathbb{R}$ . If $y=\ln((x+2)/(4-x))$ , then $e^y(4-x)=x+2$ . Hence $x=(4e^y-2)/(e^y+1)$ , giving the stated inverse after interchanging variables. Its domain is the range of $f$ , namely $\mathbb{R}$ .
3	$y=-\ln x$ for $0<x\leq1$ and $y=\ln x$ for $x\geq1$ , with $x$ -intercept and minimum point $(1,0)$ and vertical asymptote $x=0$ Range $y\geq0$ $x=\dfrac17$ or $x=7$	6
(6 marks)		6
Notes
For $0<x<1$ , $\ln x<0$ , so $\|\ln x\|=-\ln x$ ; for $x\geq1$ , $\|\ln x\|=\ln x$ . The two branches meet at $(1,0)$ , the minimum point and only $x$ -intercept. As $x\to0^+$ , $\|\ln x\|\to\infty$ , so $x=0$ is a vertical asymptote; as $x\to\infty$ , $\|\ln x\|\to\infty$ . Hence the range is $y\geq0$ . Finally, $\|\ln x\|=\ln7$ gives $\ln x=\pm\ln7$ , so $x=7$ or $x=1/7$ .
4	$f(x)=\dfrac{1+\log_3(x-4)}2$ Domain $x>4$ , range $\mathbb{R}$ and vertical asymptote $x=4$ $x$ -intercept $\left(\dfrac{13}{3},0\right)$	6
(6 marks)		6
Notes
Write $y=3^{2x-1}+4$ . Then $y-4=3^{2x-1}$ , so $\log_3(y-4)=2x-1$ and $x=(1+\log_3(y-4))/2$ . Interchanging variables gives the stated $f$ . Its logarithm requires $x>4$ , its range is all real numbers, and its vertical asymptote is $x=4$ . For the $x$ -intercept, $1+\log_3(x-4)=0$ , so $x-4=1/3$ and $x=13/3$ .
5	$P=(2,9)$ and $Q=(9,2)$ $PQ$ : $y=11-x$ $\|PQ\|=7\sqrt2$	6
(6 marks)		6
Notes
Write $P=(p,3^p)$ . Reflection in $y=x$ interchanges the coordinates, so $Q=(3^p,p)$ . Since $\log_3(3^p)=p$ , the point $Q$ lies on $y=\log_3x$ . The midpoint condition gives $p+3^p=11$ . The value $p=2$ satisfies this equation, and it is the unique solution because both $p$ and $3^p$ are strictly increasing functions of $p$ . Hence $P=(2,9)$ and $Q=(9,2)$ . The line through them has gradient $-1$ , so its equation is $y=11-x$ . Finally, $\|PQ\|=\sqrt{(9-2)^2+(2-9)^2}=7\sqrt2$ .

6.4 · Understand and use the laws of logarithms: logₐx + logₐy = logₐ(xy); logₐx − logₐy = logₐ(x/y); k logₐx = logₐxᵏ (including, for example, k = −1 and k = −½).

Tier 1 · Easy

Mark scheme for 6.4 Tier 1 · Easy
Question	Scheme	Marks
1	$\ln18$	2
(2 marks)		2
Notes
Use product and quotient laws: $\ln12+\ln3-\ln2=\ln(12\times3/2)=\ln18$ .
2	$\log_a\!\left(\dfrac{p^3}{\sqrt q}\right)$	2
(2 marks)		2
Notes
Use the power law first: $3\log_a p=\log_a(p^3)$ and $\tfrac12\log_a q=\log_a(\sqrt q)$ . The subtraction law then gives $\log_a(p^3/\sqrt q)$ .

Tier 2 · Standard

Mark scheme for 6.4 Tier 2 · Standard
Question	Scheme	Marks
1	$x=3-2\sqrt2$ or $x=3+2\sqrt2$	5
(5 marks)		5
Notes
The logarithms require $x>0$ . Combine them to get $\ln\!\left((x+1)^2/x\right)=\ln8$ , so $(x+1)^2=8x$ . Hence $x^2-6x+1=0$ , giving $x=3\pm2\sqrt2$ . Both values are positive, so both satisfy the domain and are valid.
2	$6p+4q$	4
(4 marks)		4
Notes
Since $72=2^3\cdot3^2$ , $\log_a72=3p+2q$ . Also $\log_{\sqrt a}72=\dfrac{\log_a72}{\log_a\sqrt a}=\dfrac{3p+2q}{1/2}=6p+4q$ .
3	$x=3$	4
(4 marks)		4
Notes
The original logarithms require $x>0$ . By the power law, $2\ln x=\ln(x^2)$ , so $x+6=x^2$ . Hence $(x-3)(x+2)=0$ , giving the candidates $x=3$ and $x=-2$ . The value $x=-2$ is outside the original domain, so the only solution is $x=3$ .

Tier 3 · Hard

Mark scheme for 6.4 Tier 3 · Hard
Question	Scheme	Marks
1	$x=-1+\sqrt{13}$	5
(5 marks)		5
Notes
The original arguments require $x>1$ . Combine the logarithms: $\log_3((x-1)(x+3))=2$ , so $(x-1)(x+3)=9$ . Hence $x^2+2x-12=0$ , giving $x=-1\pm\sqrt{13}$ . Only $-1+\sqrt{13}>1$ , so the other algebraic root is rejected.
2	$x=\dfrac14$ or $x=4$	5
(5 marks)		5
Notes
The base requires $x>0$ and $x\neq1$ . Let $p=\log_2x$ , so $p\neq0$ . By change of base, $\log_x16=\dfrac{\log_216}{\log_2x}=4/p$ . Hence $4/p=p$ , so $p^2=4$ and $p=\pm2$ . Therefore $x=2^p$ gives $x=1/4$ or $x=4$ , both valid.
3	$p=2^{27/7}$ and $q=2^{16/7}$	5
(5 marks)		5
Notes
Let $P=\log_2p$ and $Q=\log_2q$ . Change of base gives $\log_4q=Q/2$ and $\log_8p=P/3$ . The equations become $P+Q/2=5$ and $P/3-Q=-1$ . The second gives $Q=1+P/3$ . Substitution into the first gives $P+(1+P/3)/2=5$ , so $P=27/7$ and then $Q=16/7$ . Therefore $p=2^P=2^{27/7}$ and $q=2^Q=2^{16/7}$ .
4	$x<-2$ or $x>4$	5
(5 marks)		5
Notes
The logarithm requires $(x-1)/(x+2)>0$ , giving $x<-2$ or $x>1$ . Since $\ln x$ is increasing, the inequality is equivalent on this domain to $(x-1)/(x+2)>1/2$ . This rearranges without losing the denominator sign by writing $(x-1)/(x+2)-1/2=(x-4)/(2(x+2))>0$ . A sign analysis gives $x<-2$ or $x>4$ , both of which satisfy the original domain.
5	Two distinct solutions when $k<\ln4$ Exactly one solution when $k=\ln4$ No solutions when $k>\ln4$ When $k=\ln3$ , $x=2$ or $x=4$	6
(6 marks)		6
Notes
The original logarithms require $1<x<5$ . Combining them gives $\ln((x-1)(5-x))=k$ . Now $(x-1)(5-x)=4-(x-3)^2$ , which is positive on the domain and has maximum $4$ at $x=3$ . Therefore its logarithm has range $(-\infty,\ln4]$ . Values below the maximum occur twice, the maximum once, and larger values never. If $k=\ln3$ , then $4-(x-3)^2=3$ , so $(x-3)^2=1$ and $x=2$ or $x=4$ .

6.5 · Solve equations of the form aˣ = b.

Tier 1 · Easy

Mark scheme for 6.5 Tier 1 · Easy
Question	Scheme	Marks
1	$x=1.760$	2
(2 marks)		2
Notes
Take natural logarithms: $x\ln5=\ln17$ , so $x=\ln17/\ln5=1.7603\ldots$ , which gives $1.760$ .
2	$x=0.513$	2
(2 marks)		2
Notes
Taking logarithms gives $(x+1)\ln7=\ln19$ . Hence $x=\ln19/\ln7-1=0.513142\ldots$ , so $x=0.513$ to $3$ decimal places.

Tier 2 · Standard

Mark scheme for 6.5 Tier 2 · Standard
Question	Scheme	Marks
1	$x=\dfrac12\left(1+\dfrac{\ln40}{\ln5}\right)$ , or any equivalent exact form: $\tfrac12\left(1+\log_5 40\right)$ , $\tfrac12\log_5 200$ and $\log_5\sqrt{200}$ are all accepted, in any base. $x=1.646$ to $3$ decimal places	3
(3 marks)		3
Notes
Taking natural logarithms gives $(2x-1)\ln5=\ln40$ . Hence $2x-1=\ln40/\ln5$ , so $x=\frac12(1+\ln40/\ln5)=1.646014\ldots$ .
2	$x=\dfrac{2\ln5-\ln2}{\ln10}$ $x=1.097$ to $3$ decimal places	4
(4 marks)		4
Notes
Taking logarithms gives $(x+1)\ln2=(2-x)\ln5$ . Hence $x(\ln2+\ln5)=2\ln5-\ln2$ , so $x=(2\ln5-\ln2)/\ln10=1.0969\ldots$ .
3	$k=-0.806$	3
(3 marks)		3
Notes
Substituting $x=1.2$ gives $5^{2.4+k}=13$ . Taking logarithms gives $(2.4+k)\ln5=\ln13$ , so $k=\ln13/\ln5-2.4=-0.806307\ldots$ . Therefore $k=-0.806$ to three decimal places.

Tier 3 · Hard

Mark scheme for 6.5 Tier 3 · Hard
Question	Scheme	Marks
1	$x=-1$ or $x=1$	5
(5 marks)		5
Notes
Let $u=2^x$ , so $u>0$ and $2^{-x}=1/u$ . Then $u+1/u=5/2$ . Multiplying by $2u$ gives $2u^2-5u+2=0=(2u-1)(u-2)$ . Thus $u=1/2$ or $u=2$ , so $2^x=2^{-1}$ or $2^x=2^1$ , giving $x=-1$ or $x=1$ .
2	$x=\log_3 2$ or $x=\log_3 7$	5
(5 marks)		5
Notes
Let $u=3^x$ , so $u>0$ and $9^x=u^2$ . The equation becomes $u^2-9u+14=0=(u-2)(u-7)$ . Thus $3^x=2$ or $3^x=7$ , giving $x=\log_3 2$ or $x=\log_3 7$ .
3	$k=6$ $x=\log_2 3$	6
(6 marks)		6
Notes
Let $u=2^x$ , so $u>0$ and $4^x=u^2$ . The equation becomes $u^2-ku+9=0$ . If its discriminant is negative, there is no real value of $u$ and hence no real solution for $x$ . If its discriminant is positive, the two roots have product $9>0$ and therefore have the same sign: for $k>0$ both are positive and give two values of $x$ , while for $k<0$ both are negative and give none. Exactly one positive root must therefore be repeated, so $k^2-36=0$ and $k=\pm6$ . If $k=-6$ , the repeated root is $u=-3$ , which is impossible. Hence $k=6$ , giving $(u-3)^2=0$ . Thus $2^x=3$ and $x=\log_2 3$ .
4	$x>\dfrac{\ln5+2\ln7}{2\ln5-\ln7}$ Smallest integer $x=5$	5
(5 marks)		5
Notes
Taking logarithms gives $(2x-1)\ln5>(x+2)\ln7$ . Hence $x(2\ln5-\ln7)>\ln5+2\ln7$ . The coefficient is positive because $25>7$ , so division preserves the inequality and gives the stated boundary. This boundary is $4.32\ldots$ , so the smallest integer solution is $5$ .
5	$x=2$ and $y=1$	4
(4 marks)		4
Notes
Multiplying the equations eliminates $3^y$ and gives $8^x=64=8^2$ , so $x=2$ . Substitution into $2^x3^y=12$ gives $4(3^y)=12$ , hence $3^y=3$ and $y=1$ . These values satisfy both original equations.

6.6 · Use logarithmic graphs to estimate parameters in relationships of the form y = axⁿ and y = kbˣ, given data for x and y.

Tier 1 · Easy

Mark scheme for 6.6 Tier 1 · Easy
Question	Scheme	Marks
1	$y=2.51x^{1.7}$	3
(3 marks)		3
Notes
Comparing with $\log_{10}y=\log_{10}a+n\log_{10}x$ gives $n=1.7$ and $\log_{10}a=0.4$ . Hence $a=10^{0.4}=2.511\ldots$ , so $y=2.51x^{1.7}$ .
2	Plot $\ln y$ against $x$ Gradient $=\ln b$ and vertical-axis intercept $=\ln k$	3
(3 marks)		3
Notes
Taking logarithms gives $\ln y=\ln k+x\ln b$ . This is linear in $x$ , so plotting $\ln y$ vertically against $x$ horizontally gives gradient $\ln b$ and intercept $\ln k$ .

Tier 2 · Standard

Mark scheme for 6.6 Tier 2 · Standard
Question	Scheme	Marks
1	$k\approx2.01$ $b\approx1.35$ $y\approx6.69$ when unrounded parameter values are carried; $y\approx6.68$ is also accepted if the stated $3$ s.f. values $k=2.01$ and $b=1.35$ are used	5
(5 marks)		5
Notes
The straight-line form is $\ln y=\ln k+x\ln b$ . Its gradient is $(2.5-1.3)/(6-2)=0.3$ , so $b=e^{0.3}=1.3498\ldots$ . The intercept is $1.3-0.3(2)=0.7$ , so $k=e^{0.7}=2.0137\ldots$ . Prefer carrying these unrounded values: at $x=4$ , $\ln y=0.7+0.3(4)=1.9$ , so $y=e^{1.9}=6.6858\ldots\approx6.69$ . However, following the instruction to use the reported $3$ s.f. parameters gives $y=2.01(1.35)^4=6.676\ldots\approx6.68$ , which is also valid.
2	$k=3.31$ $b=1.91$ $x=4.21$	5
(5 marks)		5
Notes
Comparing with $\log_{10}y=\log_{10}k+x\log_{10}b$ gives $k=10^{0.52}=3.31131\ldots$ and $b=10^{0.28}=1.90546\ldots$ . When $y=50$ , $\log_{10}50=0.52+0.28x$ , so $x=(\log_{10}50-0.52)/0.28=4.21061\ldots$ .
3	$n=0.6$ $a=63.1$ $\log_{10}y=2.1$	5
(5 marks)		5
Notes
The straight-line form is $\log_{10}y=\log_{10}a+n\log_{10}x$ . Its gradient is $n=(3.0-1.2)/(2-(-1))=0.6$ . Using $(-1,1.2)$ gives $\log_{10}a=1.2-0.6(-1)=1.8$ , so $a=10^{1.8}=63.0957\ldots=63.1$ to $3$ significant figures. When $\log_{10}x=0.5$ , $\log_{10}y=1.8+0.6(0.5)=2.1$ .

Tier 3 · Hard

Mark scheme for 6.6 Tier 3 · Hard
Question	Scheme	Marks
1	$k=5.47$ $b=1.49$ $y\approx18.2$	6
(6 marks)		6
Notes
The straight-line form is $\ln y=\ln k+x\ln b$ . Its gradient is $(3.7-2.1)/(5-1)=0.4$ , so $\ln b=0.4$ and $b=e^{0.4}=1.4918\ldots$ . The intercept is $2.1-0.4(1)=1.7$ , so $k=e^{1.7}=5.4739\ldots$ . At $x=3$ , $\ln y=1.7+0.4(3)=2.9$ , giving $y=e^{2.9}=18.174\ldots$ .
2	$a=12.6$ $n=-0.8$ $x=4.18$	6
(6 marks)		6
Notes
Since $\log_{10}(2y)=\log_{10}2+\log_{10}a+n\log_{10}x$ , the gradient gives $n=-0.8$ and the intercept gives $\log_{10}(2a)=1.4$ . Hence $a=10^{1.4}/2=12.5594\ldots$ . When $y=4$ , $\log_{10}8=1.4-0.8\log_{10}x$ , so $x=(10^{1.4}/8)^{1/0.8}=4.17963\ldots$ .
3	The power model fits $a=2$ and $n=\dfrac32$ $y=250$ when $x=25$	6
(6 marks)		6
Notes
For a power model, the gradient between the first two log-log points is $\ln(54/16)/\ln(9/4)=\ln(27/8)/\ln(9/4)=3/2$ . Between the next two it is $\ln(128/54)/\ln(16/9)=\ln(64/27)/\ln(16/9)=3/2$ again, so the three log-log points are collinear. In contrast, the semi-log gradients $\ln(54/16)/(9-4)$ and $\ln(128/54)/(16-9)$ are unequal, so one exponential model does not fit all three pairs. Using $(4,16)$ in $y=ax^{3/2}$ gives $16=8a$ , so $a=2$ . Hence at $x=25$ , $y=2(25^{3/2})=2(125)=250$ .
4	$a=\dfrac52$ and $n=3$ $\log_{10}y=\log_{10}\!\left(\dfrac52\right)+3\log_{10}x$ $x=6$ when $y=540$	6
(6 marks)		6
Notes
For a power model, $y(2x)/y(x)=2^n$ . The multiplier $8$ gives $2^n=8$ , so $n=3$ . Using $(x,y)=(2,20)$ gives $20=8a$ , hence $a=5/2$ . Taking base- $10$ logarithms gives the stated line. Finally, $540=(5/2)x^3$ , so $x^3=216$ and, since $x>0$ , $x=6$ .
5	$y=\sqrt{10}\,x^{5/2}$ , so $a=\sqrt{10}$ and $n=\dfrac52$ $y=32000$ when $x=40$	5
(5 marks)		5
Notes
Rearranging the fitted line gives $\log_{10}y=2.5\log_{10}x+0.5$ . Therefore $y=10^{0.5}x^{2.5}=\sqrt{10}\,x^{5/2}$ . At $x=40$ , $y=\sqrt{10}(40^2\sqrt{40})=\sqrt{10}(1600)(2\sqrt{10})=32000$ .

6.7 · Understand and use exponential growth and decay; use in modelling (e.g. compound interest, radioactive decay, drug concentration decay, population growth); consideration of limitations and refinements of exponential models.

Tier 1 · Easy

Mark scheme for 6.7 Tier 1 · Easy
Question	Scheme	Marks
1	$\pounds1377.03$	2
(2 marks)		2
Notes
The annual multiplier is $1.035$ , so the balance is $1200(1.035)^4=1377.027\ldots$ . Rounded to the nearest penny, this is $\pounds1377.03$ .
2	$V=4500(0.88)^n$	2
(2 marks)		2
Notes
Losing $12\%$ leaves $88\%=0.88$ of the value after each year. Repeated multiplication therefore gives $V=4500(0.88)^n$ .

Tier 2 · Standard

Mark scheme for 6.7 Tier 2 · Standard
Question	Scheme	Marks
1	$12$ years Any valid contextual limitation, for example: the annual interest rate may change; charges, deposits or withdrawals are ignored; or interest may be compounded on a different schedule	4
(4 marks)		4
Notes
The value after $n$ years is $5000(1.042)^n$ . Solving $5000(1.042)^n>8000$ gives $n>\ln(1.6)/\ln(1.042)=11.423\ldots$ , so the smallest whole number is $12$ . Indeed, the model gives about $\pounds7861.67$ after $11$ years and $\pounds8191.86$ after $12$ . A limitation is that it assumes the interest rate remains fixed (and ignores charges or withdrawals).
2	Annual depreciation $=10.1\%$ Value after $7$ years $=\pounds7122$	4
(4 marks)		4
Notes
From $9800=15000q^4$ , $q=(9800/15000)^{1/4}=0.899050\ldots$ . The annual percentage loss is $100(1-q)=10.0950\ldots\%$ , giving $10.1\%$ . After $7$ years, $V=15000q^7=7121.59\ldots$ , so the predicted value is $\pounds7122$ .
3	$k=0.0397\,\text{h}^{-1}$ Half-life $=17.5$ hours	4
(4 marks)		4
Notes
The observation gives $e^{-5k}=0.82$ , so $k=-\ln(0.82)/5=0.0396901\ldots\,\text{h}^{-1}$ . If $T$ is the half-life, then $e^{-kT}=1/2$ , so $T=\ln2/k=17.4639\ldots$ hours. Therefore $k=0.0397\,\text{h}^{-1}$ and the half-life is $17.5$ hours, each to $3$ significant figures.

Tier 3 · Hard

Mark scheme for 6.7 Tier 3 · Hard
Question	Scheme	Marks
1	$k=\dfrac{\ln(1.5)}{4}\approx0.101$ $t\approx11.9$ days For example, limited resources invalidate unlimited growth; a logistic model could include a carrying capacity.	6
(6 marks)		6
Notes
From $900=600e^{4k}$ , $e^{4k}=1.5$ , so $k=\ln(1.5)/4=0.10137\ldots$ . For $P=2000$ , $e^{kt}=2000/600=10/3$ , hence $t=\ln(10/3)/k=11.877\ldots$ days. The model assumes a constant proportional growth rate and no resource limit; a logistic model with a carrying capacity would refine this.
2	$k=0.173\text{ day}^{-1}$ $M_0=403$ mg $M<30$ mg when $t>15.0$ days For example, the model assumes a constant proportional decay rate under unchanged conditions	6
(6 marks)		6
Notes
Dividing the observations gives $85/240=e^{-6k}=17/48$ , so $k=\ln(48/17)/6=0.172998\ldots$ . Then $M_0=240e^{3k}=403.281\ldots$ mg. The inequality $M_0e^{-kt}<30$ gives $t>\ln(M_0/30)/k=15.0200\ldots$ days. A limitation is that the proportional decay constant is assumed not to change with environmental conditions or composition.
3	$t=4.34$ years The first whole number of years for which $A>B$ is $5$ For example, both percentage changes are assumed constant; refine the models using updated or piecewise rates	6
(6 marks)		6
Notes
Equality gives $900(1.08)^t=1500(0.96)^t$ , so $(1.08/0.96)^t=5/3$ . Hence $t=\ln(5/3)/\ln(1.125)=4.33700\ldots$ , giving $4.34$ years. At $t=4$ , $A/B=(3/5)(1.125)^4=0.961\ldots<1$ , while at $t=5$ the ratio is $1.081\ldots>1$ , so the first whole number is $5$ . The comparison assumes both percentage changes remain constant; a piecewise model using rates updated from later observations would be a suitable refinement.
4	$k=\dfrac{\ln(5/4)}3$ Model prediction $P(6)=3125$ Percentage overestimate $=7.76\%$ The constant proportional growth assumption is not supported: the successive three-year growth factors are $1.25$ and $1.16$	6
(6 marks)		6
Notes
From $2500=2000e^{3k}$ , $e^{3k}=5/4$ and $k=\ln(5/4)/3$ . Hence $P(6)=2000(e^{3k})^2=2000(5/4)^2=3125$ . Relative to the actual value, the overestimate is $100(3125-2900)/2900=225/29=7.7586\ldots\%$ , giving $7.76\%$ . The observed three-year factors are $2500/2000=1.25$ and $2900/2500=1.16$ , so a constant proportional rate does not fit both intervals.
5	$P=1200e^{0.72}e^{-0.12(t-4)}$ for $t\geq4$ Maximum population $1200e^{0.72}$ at $t=4$ The model first returns to $1200$ at $t=10$ days	6
(6 marks)		6
Notes
At treatment time, $P(4)=1200e^{0.18(4)}=1200e^{0.72}$ . A continuous decay rate of $12\%$ gives the factor $e^{-0.12(t-4)}$ after treatment, producing the stated continuous model. The first branch increases and the second decreases, so the maximum occurs at $t=4$ . Setting the second model equal to $1200$ gives $0.72-0.12(t-4)=0$ , hence $t-4=6$ and $t=10$ .