6 Exponentials and logarithms — notes on one page

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.

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

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 $y-y_0=m(x-x_0)$.
• 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.

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

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.

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

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).

Definition. Specification 6.4: understand and use the laws of logarithms: $\log_a x+\log_a y=\log_a(xy)$; $\log_a x-\log_a y=\log_a\left(\dfrac xy\right)$; $k\log_a x=\log_a x^k$, including for example $k=-1$ and $k=-\frac12$.

• The laws convert between one logarithm and several: a product becomes a sum, a quotient a difference, and a power a coefficient, so $\ln\dfrac{x^3\sqrt y}{z^2}=3\ln x+\frac12\ln y-2\ln z$.
• Read backwards they collect an equation into a single logarithm, $\log_2(x+6)+\log_2(x-2)=3$ becoming $\log_2\big((x+6)(x-2)\big)=3$, which is then removed by writing the argument equal to $2^3$; “taking logs” of $y=ab^x$ gives $\log y=\log a+x\log b$, a straight line in $x$ whose intercept and gradient give $a$ and $b$.
• What the laws do not do: $\log(x+y)$ is not $\log x+\log y$, and $\log x\cdot\log y$ is not $\log(xy)$.
• Because a logarithm needs a positive argument, every solution is checked in the original equation and any that makes an argument zero or negative is rejected with that reason. $\log_a a=1$ and $\log_a1=0$.

What the examiner wants. The laws applied as written, and a check that the arguments stay positive. The June 2024 Paper 1 report says attempts at using logarithms were mostly unsuccessful because the laws were not adhered to, and that candidates whose model gave the logarithm of a negative value should have taken it as a sign of an earlier error rather than forcing a positive answer.

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

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.
• Substitute the final value into the original equation to verify it.

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

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.

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

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.

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