6 Exponentials and logarithms — 9MA0 spec coverage

6.1

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

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.

Tier 1 · Easy

2 marks

ORIGINAL

For the graph $y=3^x$ , state the $y$ -intercept and the horizontal asymptote.

Tier 2 · Standard

4 marks

ORIGINAL

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

Tier 3 · Hard

5 marks

ORIGINAL

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 $x$ .

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.

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 $y-y_0=m(x-x_0)$ .
Exponential modelling assumes a rate proportional to the current amount and a constant proportionality factor; changing conditions can make that assumption unsuitable.

Tier 1 · Easy

1 mark

ORIGINAL

Differentiate $y=e^{4x}$ with respect to $x$ .

Tier 2 · Standard

4 marks

ORIGINAL

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

Tier 3 · Hard

4 marks

ORIGINAL

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.

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.

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.

Tier 1 · Easy

2 marks

ORIGINAL

Solve $\ln x=2$ exactly.

Tier 2 · Standard

4 marks

ORIGINAL

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

Tier 3 · Hard

5 marks

ORIGINAL

For $g(x)=\ln(4-x^2)$ , determine the domain and range, and identify the maximum point.

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).

For positive arguments, $\log_a x+\log_a y=\log_a(xy)$ , $\log_a x-\log_a y=\log_a(x/y)$ and $k\log_a x=\log_a(x^k)$ .
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.

Tier 1 · Easy

2 marks

ORIGINAL

Write $\ln12+\ln3-\ln2$ as a single logarithm.

Tier 2 · Standard

3 marks

ORIGINAL

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

Tier 3 · Hard

5 marks

ORIGINAL

Solve $\log_3(x-1)+\log_3(x+3)=2$ , checking the domain of the original equation.

6.5

Solve equations of the form aˣ = b.

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.

Tier 1 · Easy

2 marks

ORIGINAL

Solve $5^x=17$ , giving $x$ to $3$ decimal places.

Tier 2 · Standard

3 marks

ORIGINAL

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

Tier 3 · Hard

5 marks

ORIGINAL

Determine all real solutions of $2^x+2^{-x}=5/2$ .

6.6

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

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.

Tier 1 · Easy

3 marks

ORIGINAL

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.

Tier 2 · Standard

5 marks

ORIGINAL

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 $x=5$ .

Tier 3 · Hard

6 marks

ORIGINAL

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 $x=3$ .

6.7

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.

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.

Tier 1 · Easy

2 marks

ORIGINAL

A savings balance of $\pounds1200$ earns $3.5\%$ compound interest each year. Find the balance after $4$ years, to the nearest penny.

Tier 2 · Standard

4 marks

ORIGINAL

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.

Tier 3 · Hard

6 marks

ORIGINAL

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.