6 Exponentials and logarithms — printable coverage pack

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

1. For the graph $y=3^x$ , state the $y$ -intercept and the horizontal asymptote.[2 marks]
Answer
- $y$ -intercept $(0,1)$
- Horizontal asymptote $y=0$
Method: 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 $y=0$ .

Tier 2 · Standard

1. For $y=2e^x-5$ , state the domain, range, horizontal asymptote and $y$ -intercept.[4 marks]
Answer
- Domain $x\in\mathbb{R}$
- Range $y>-5$
- Horizontal asymptote $y=-5$
- $y$ -intercept $(0,-3)$
Method: The function $e^x$ is defined and positive for every real $x$ . Multiplying by $2$ preserves positivity, then subtracting $5$ gives $y>-5$ and moves the asymptote to $y=-5$ . At $x=0$ , $y=2e^0-5=-3$ .

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 marks]
Answer
- Minimum value $6$ , attained when $x=1$
Method: 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 $(1,6)$ .

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. Differentiate $y=e^{4x}$ with respect to $x$ .[1 mark]
Answer
- $\dfrac{dy}{dx}=4e^{4x}$
Method: For $e^{kx}$ , differentiation multiplies the function by $k$ . Here $k=4$ , so $dy/dx=4e^{4x}$ .

Tier 2 · Standard

1. Find the equation of the tangent to $y=e^{2x}$ at $x=1/2$ .[4 marks]
Answer
- $y=2ex$
Method: At $x=1/2$ , the point is $(1/2,e)$ . Since $dy/dx=2e^{2x}$ , the gradient there is $2e$ . Thus $y-e=2e(x-1/2)$ , which simplifies to $y=2ex$ .

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 marks]
Answer
- $90$ units per hour
- The growth rate is proportional to the current population.
Method: 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.

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

1. Solve $\ln x=2$ exactly.[2 marks]
Answer
- $x=e^2$
Method: Apply $e$ to both sides: $e^{\ln x}=e^2$ , hence $x=e^2$ .

Tier 2 · Standard

1. The function $f(x)=e^{x-3}+2$ . Find $f^{-1}(x)$ and state the domain of the inverse.[4 marks]
Answer
- $f^{-1}(x)=\ln(x-2)+3$
- Domain $x>2$
Method: Write $y=e^{x-3}+2$ , so $y-2=e^{x-3}$ . Taking natural logarithms gives $\ln(y-2)=x-3$ , hence $x=\ln(y-2)+3$ . Interchanging labels gives $f^{-1}(x)=\ln(x-2)+3$ . The range of $f$ is $y>2$ , so the inverse domain is $x>2$ .

Tier 3 · Hard

1. For $g(x)=\ln(4-x^2)$ , determine the domain and range, and identify the maximum point.[5 marks]
Answer
- Domain $-2<x<2$
- Range $-\infty<g(x)\leq\ln4$
- Maximum point $(0,\ln4)$
Method: 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 $(-\infty,\ln4]$ .

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

1. Write $\ln12+\ln3-\ln2$ as a single logarithm.[2 marks]
Answer
- $\ln18$
Method: Use product and quotient laws: $\ln12+\ln3-\ln2=\ln(12\times3/2)=\ln18$ .

Tier 2 · Standard

1. Expand $\ln\!\left(\dfrac{x^3\sqrt{y}}{z^2}\right)$ , where $x$ , $y$ and $z$ are positive.[3 marks]
Answer
- $3\ln x+\dfrac12\ln y-2\ln z$
Method: The quotient gives subtraction, the product gives addition, and powers become coefficients: $\ln(x^3)+\ln(y^{1/2})-\ln(z^2)=3\ln x+\tfrac12\ln y-2\ln z$ .

Tier 3 · Hard

1. Solve $\log_3(x-1)+\log_3(x+3)=2$ , checking the domain of the original equation.[5 marks]
Answer
- $x=-1+\sqrt{13}$
Method: 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.

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

1. Solve $5^x=17$ , giving $x$ to $3$ decimal places.[2 marks]
Answer
- $x=1.760$
Method: Take natural logarithms: $x\ln5=\ln17$ , so $x=\ln17/\ln5=1.7603\ldots$ , which gives $1.760$ .

Tier 2 · Standard

1. Find the solution of $3^{2x-1}=20$ , giving $x$ to $3$ decimal places.[3 marks]
Answer
- $x=1.863$
Method: Taking logarithms gives $(2x-1)\ln3=\ln20$ . Therefore $2x-1=\ln20/\ln3$ and $x=\tfrac12(1+\ln20/\ln3)=1.8634\ldots$ , so $x=1.863$ .

Tier 3 · Hard

1. Determine all real solutions of $2^x+2^{-x}=5/2$ .[5 marks]
Answer
- $x=-1$ or $x=1$
Method: 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 $x=1$ .

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

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 marks]
Answer
- $y=2.51x^{1.7}$
Method: 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 $y=2.51x^{1.7}$ .

Tier 2 · Standard

1. 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 $x=5$ .[5 marks]
Answer
- $n=1.5$
- $a\approx2.24$
- $y\approx25.0$
Method: The gradient is $n=(1.55-0.65)/(0.8-0.2)=1.5$ . Using $(0.2,0.65)$ , the intercept is $\log_{10}a=0.65-1.5(0.2)=0.35$ , so $a=10^{0.35}=2.2387\ldots$ . Thus at $x=5$ , $y=2.2387\ldots(5^{1.5})=25.030\ldots$ , giving about $25.0$ .

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 marks]
Answer
- $k=5.47$
- $b=1.49$
- $y\approx18.2$
Method: 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 $y=e^{2.9}=18.174\ldots$ .

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

1. A savings balance of $\pounds1200$ earns $3.5\%$ compound interest each year. Find the balance after $4$ years, to the nearest penny.[2 marks]
Answer
- $\pounds1377.03$
Method: The annual multiplier is $1.035$ , so the balance is $1200(1.035)^4=1377.027\ldots$ . Rounded to the nearest penny, this is $\pounds1377.03$ .

Tier 2 · Standard

1. 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.[4 marks]
Answer
- Half-life $=3.01$ hours
- $C(5)=25.3\,\text{mg L}^{-1}$
Method: At half-life, $e^{-0.23t}=1/2$ , so $t=\ln2/0.23=3.0136\ldots$ hours. After $5$ hours, $C=80e^{-0.23(5)}=25.3309\ldots\,\text{mg L}^{-1}$ . The requested values are $3.01$ hours and $25.3\,\text{mg L}^{-1}$ .

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 marks]
Answer
- $k=\dfrac{\ln(1.5)}{4}\approx0.101$
- $t\approx11.9$ days
- For example, limited resources invalidate unlimited growth; a logistic model could include a carrying capacity.
Method: 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.