The Mission
Eight rapid questions — one for each skill from Lessons 1–5: log laws, natural logarithms, exponential equations, linearising data and growth models. No calculator: use the laws and the structure, not decimals. Target pace: about 7 seconds per question.
$3^4 = 81$, so $\log_3 81 = 4$. The notation asks: which power of $3$ gives $81$?
Product law: $\log_{10}(2 \times 3) = \log_{10}2 + \log_{10}3 = \log_{10}6$. The trap: $\log_{10}(2+3)$ is not $\log_{10}2 + \log_{10}3$ — no log law splits a sum inside the argument.
$\ln$ is the logarithm with base $e$, so $\ln e^7 = 7$ — in general $\log_a a^n = n$.
Power law: $\ln(x^2) = 2\ln x$. The trap: $(\ln x)^2$ is the square of the logarithm — a different object that cannot be simplified.
$\dfrac{1}{27} = 27^{-1} = 3^{-3}$, so $3^x = 3^{-3}$ gives $x = -3$. A negative exponent means a reciprocal.
Take logs of both sides: $x\ln 2 = \ln 10$, so $x = \dfrac{\ln 10}{\ln 2}$. Flipping the fraction or subtracting the logs are the two classic slips.
$y = ab^t \Rightarrow \ln y = \ln a + t\ln b$. So the gradient is $\ln b$ and the intercept is $\ln a$. (For the power model $y = ax^b$, plot $\ln y$ against $\ln x$: gradient $b$, intercept $\ln a$.)
Growth of 8% multiplies by $1.08$ each year: $P = 2000(1.08)^t$. Adding $0.08t$ is linear growth, and $0.92$ would model 8% decay.
The Bridge — Where This Review Goes
You have just retrieved every tool from this chapter: log laws, natural logarithms, exponential equations, linearising and growth models. In this lesson we chain them together the way exam questions do — alongside the W1 algebra toolkit (modulus, factor theorem, partial fractions, binomial expansion). Any question that felt slow is your personal revision target for today.
For $a, b, x, y > 0$ with $a, b \neq 1$:
Remember: $\log_a a = 1$, $\log_a 1 = 0$. These laws apply to products and quotients of arguments, never to sums inside the argument.
Adjust x and y to verify the product, quotient, and power laws.
To solve $a^x = b$, take logarithms of both sides:
For equations like $a^{2x} + pa^x + q = 0$, use the substitution $u = a^x$ to convert to a quadratic in $u$, solve, then back-substitute (rejecting $u \leq 0$ since $a^x > 0$).
The natural logarithm $\ln x = \log_e x$ and the exponential $e^x$ are inverses:
To solve $\ln f(x) = c$, exponentiate both sides: $f(x) = e^c$. Always check the domain: the argument of $\ln$ must be positive.
eˣ = | ln(x) =
Green: y = eˣ. Amber: y = ln(x). The curves are reflections of each other in y = x.
Non-linear relationships can be tested by plotting logarithms:
Power model $y = ax^b$: $\log y = \log a + b\log x$ → plot $\log y$ vs $\log x$ (straight line, gradient $b$, intercept $\log a$).
If the log-log or log-linear plot is a straight line, the model fits. Use the gradient and intercept to recover $a$ and $b$.
Many exam questions combine log/exp skills with the W1 algebra toolkit:
Factor theorem: $(x - k)$ is a factor of $f(x)$ $\iff f(k) = 0$. Used to factorise cubics before solving $a^{3y} + \ldots = 0$ via substitution.
Partial fractions: $\dfrac{px + q}{(x-a)(x-b)} = \dfrac{A}{x-a} + \dfrac{B}{x-b}$. Often needed before integrating or expanding.
Binomial expansion: $(1+x)^n = 1 + nx + \dfrac{n(n-1)}{2!}x^2 + \cdots$ for $|x| < 1$.
Key connection: Substitution $u = a^x$ turns an exponential equation into a polynomial equation — then the factor theorem and polynomial skills from W1 become essential.
Before splitting into expert groups, predict together: which topic area do you feel MOST confident about, and which feels LEAST secure? Note your prediction - you will revisit it after combining the jigsaw.
How it works: Form 4 expert groups. Each group studies one topic area, then all groups combine their knowledge to solve a master challenge. Click your group's panel to reveal the key concepts and your expert challenge.
- Product Law: $\log_a(mn) = \log_a m + \log_a n$
- Quotient Law: $\log_a\!\left(\dfrac{m}{n}\right) = \log_a m - \log_a n$
- Power Law: $\log_a(m^n) = n \log_a m$
$2\log 4 = \log 16$ and $\dfrac{1}{2}\log 144 = \log 12$. So $\log 16 + \log 9 - \log 12 = \log\!\left(\dfrac{16 \times 9}{12}\right) = \log 12$.
- $e^x$ and $\ln x$ are inverse functions: $e^{\ln x} = x$ and $\ln(e^x) = x$
- Domain of $e^x$: all real $x$; Range: $y > 0$
- Domain of $\ln x$: $x > 0$; Range: all real $y$
- $y = e^{kx}$: growth when $k > 0$, decay when $k < 0$
$\ln(3 - x) - \ln x = 4 \Rightarrow \ln\!\left(\dfrac{3 - x}{x}\right) = 4 \Rightarrow \dfrac{3 - x}{x} = e^4 \Rightarrow 3 - x = x e^4 \Rightarrow x = \dfrac{3}{e^4 + 1} \approx 0.0525$.
- To solve $a^x = b$: take logs $\Rightarrow x = \dfrac{\log b}{\log a}$
- To solve $a^{f(x)} = b^{g(x)}$: take logs $\Rightarrow f(x) \log a = g(x) \log b$
- Substitution for $e^{2x} + e^x = c$: let $u = e^x$
Let $u = e^x$ (so $u > 0$): $u^2 + u - 12 = 0 \Rightarrow (u + 4)(u - 3) = 0$. Since $u = e^x > 0$, take $u = 3 \Rightarrow x = \ln 3 \approx 1.099$.
- $y = kx^n \Rightarrow \ln y = \ln k + n \ln x$ (plot $\ln y$ vs $\ln x$)
- $y = ka^x \Rightarrow \ln y = \ln k + x \ln a$ (plot $\ln y$ vs $x$)
- Gradient $= n$ or $\ln a$; Intercept $= \ln k$; $k = e^{\text{intercept}}$
Exponentiate both sides: $y = e^{1.5 + 0.4x} = e^{1.5} \times e^{0.4x}$.
$\ln P = \ln P_0 + kt$. At $t = 0$: $\ln P_0 = 2.3 \Rightarrow P_0 = e^{2.3} \approx 9.97$. At $t = 10$: $4.1 = 2.3 + 10k \Rightarrow k = 0.18$.
Doubling means $P = 2P_0$: $2P_0 = P_0 e^{0.18t} \Rightarrow 2 = e^{0.18t} \Rightarrow t = \dfrac{\ln 2}{0.18} \approx 3.85$ years.
$\ln 1000 = 2.3 + 0.18t \Rightarrow t = \dfrac{\ln 1000 - 2.3}{0.18} = \dfrac{6.908 - 2.3}{0.18} \approx 25.6$ years.
Before you begin, gauge your current level across the chapter: where are you strongest, and where might you stumble? The quiz will adapt to your answers, so answer honestly - there is no penalty for an easier path.
How it works: Test your mastery of the entire Log/Exp chapter! This quiz adapts to your level - answer correctly to get harder questions, or incorrectly to get easier ones. 5 questions stand between you and mastery!
Solve the equation $2\lvert 3^x - 1 \rvert = 3^x$, giving your answers correct to 3 significant figures.
2|3ˣ − 1| = | 3ˣ =
Green: y = 2|3ˣ − 1|. Amber: y = 3ˣ. Move the slider to find where the curves meet.
Both roots exist: one positive (where $3^x > 1$) and one negative (where $3^x < 1$). The negative root is $-\!0.369$, not $+\!0.369$, because $3^x = \tfrac{2}{3} < 1$ gives a negative $x$.
Let $u = 3^x$ (so $u > 0$). The equation becomes $2|u - 1| = u$, i.e. $|u - 1| = \dfrac{u}{2}$. Split into two cases based on whether $u \geq 1$ or $u < 1$.
$$2|u - 1| = u \implies |u - 1| = \dfrac{u}{2}$$
$u - 1 = \dfrac{u}{2} \implies u - \dfrac{u}{2} = 1 \implies \dfrac{u}{2} = 1 \implies u = 2$
Check: $u = 2 \geq 1$ ✓ So $3^x = 2 \implies x = \log_3 2 = \dfrac{\ln 2}{\ln 3} = \dfrac{0.6931}{1.0986} \approx 0.6309$
$1 - u = \dfrac{u}{2} \implies 1 = \dfrac{3u}{2} \implies u = \dfrac{2}{3}$
Check: $u = \dfrac{2}{3} < 1$ ✓ So $3^x = \dfrac{2}{3} \implies x = \log_3\!\left(\dfrac{2}{3}\right) = \dfrac{\ln(2/3)}{\ln 3} = \dfrac{-0.4055}{1.0986} \approx -0.3691$
$$x \approx 0.631 \quad \text{or} \quad x \approx -0.369 \quad \text{(3 s.f.)}$$
The polynomial $\mathrm{f}(x)$ is defined by $\mathrm{f}(x) = 12x^3 + 25x^2 - 4x - 12$.
(i) Show that $\mathrm{f}(-2) = 0$ and factorise $\mathrm{f}(x)$ completely.
(ii) Given that $12 \times 27^y + 25 \times 9^y - 4 \times 3^y - 12 = 0$, state the value of $3^y$ and hence find $y$ correct to 3 significant figures.
Cambridge International AS & A Level Mathematics 9709 Paper 31 Q4 June 2011
f(u) = 12u³ + 25u² − 4u − 12 =
Move the slider to find the roots of f(u). The substitution is u = 3y (so u > 0).
Only $u = \tfrac{2}{3}$ is positive (the other roots $u = -2$ and $u = -\tfrac{3}{4}$ are rejected since $u = 3^y > 0$). So $3^y = \tfrac{2}{3}$, giving $y = \log_3(2/3) = \dfrac{\ln(2/3)}{\ln 3} \approx -0.369$.
For (i), substitute $x = -2$ into f(x) and verify it equals 0, then divide f(x) by $(x + 2)$ to get a quadratic — factorise that quadratic. For (ii), let $u = 3^y$ so that $27^y = u^3$ and $9^y = u^2$. The equation becomes $\mathrm{f}(u) = 0$. Use your factorisation from (i) and remember $u = 3^y > 0$.
$$\mathrm{f}(-2) = 12(-2)^3 + 25(-2)^2 - 4(-2) - 12 = -96 + 100 + 8 - 12 = 0 \quad \checkmark$$
So $(x + 2)$ is a factor. Divide $\mathrm{f}(x)$ by $(x + 2)$:
$$\mathrm{f}(x) = (x + 2)(12x^2 + x - 6)$$
Solve $12x^2 + x - 6 = 0$:
$$x = \dfrac{-1 \pm \sqrt{1 + 288}}{24} = \dfrac{-1 \pm 17}{24}$$
$x = \dfrac{16}{24} = \dfrac{2}{3}$ or $x = \dfrac{-18}{24} = -\dfrac{3}{4}$
So $12x^2 + x - 6 = (3x - 2)(4x + 3)$, giving:
$$\mathrm{f}(x) = (x + 2)(3x - 2)(4x + 3)$$
Since $27^y = (3^y)^3 = u^3$ and $9^y = (3^y)^2 = u^2$, the equation becomes:
$$12u^3 + 25u^2 - 4u - 12 = 0 \quad \text{which is } \mathrm{f}(u) = 0$$
Using the factorisation from (i):
$$(u + 2)(3u - 2)(4u + 3) = 0$$
$u = -2$, $u = \dfrac{2}{3}$, or $u = -\dfrac{3}{4}$
Since $u = 3^y > 0$, reject $u = -2$ and $u = -\dfrac{3}{4}$. The only valid value is:
$$3^y = \dfrac{2}{3}$$
$$y = \log_3\!\left(\dfrac{2}{3}\right) = \dfrac{\ln(2/3)}{\ln 3} = \dfrac{\ln 2 - \ln 3}{\ln 3} = \dfrac{0.6931 - 1.0986}{1.0986} = \dfrac{-0.4055}{1.0986} \approx -0.3691$$
(i) $\mathrm{f}(x) = (x + 2)(3x - 2)(4x + 3)$
(ii) $3^y = \dfrac{2}{3}$, $y \approx -0.369$ (3 s.f.)
Test your understanding across the W1–W2 syllabus. Each question targets one objective — try each before moving on!
$|2x - 1| = 3$ gives $2x - 1 = 3$ or $2x - 1 = -3$. So $2x = 4 \Rightarrow x = 2$, or $2x = -2 \Rightarrow x = -1$. Both solutions are valid.
By the factor theorem, $(x - k)$ is a factor of $\mathrm{f}(x)$ if and only if $\mathrm{f}(k) = 0$. Here $(x + 1) = (x - (-1))$, so $k = -1$ and we need $\mathrm{f}(-1) = 0$.
Product law: $\log_2 8 + \log_2 4 = \log_2(8 \times 4) = \log_2 32 = 5$ (since $2^5 = 32$). Alternatively, $\log_2 8 = 3$ and $\log_2 4 = 2$, so $3 + 2 = 5$.
Substitute $n = -1$: $(1+x)^{-1} = 1 + (-1)x + \dfrac{(-1)(-2)}{2!}x^2 + \cdots = 1 - x + x^2 - \cdots$ Valid for $|x| < 1$. This is the geometric series $\dfrac{1}{1-x} = 1 + x + x^2 + \cdots$ with $x$ replaced by $-x$.
Since $32 = 2^5$, we have $2^x = 2^5$, so $x = 5$. In general, $2^x = b \Rightarrow x = \log_2 b = \dfrac{\ln b}{\ln 2}$.
Solve the inequality $2 \leqslant 5^x < 8$.
Since $\log_5$ is an increasing function, the inequality signs are preserved:
$$\log_5 2 \leqslant x < \log_5 8$$
$$\log_5 2 = \dfrac{\ln 2}{\ln 5} = \dfrac{0.6931}{1.6094} \approx 0.4307$$
$$\log_5 8 = \dfrac{\ln 8}{\ln 5} = \dfrac{2.0794}{1.6094} \approx 1.2920$$
$$0.431 \leqslant x < 1.29 \quad \text{(3 s.f.)}$$
Solve the equation $6^{2x} - 5(6^x) + 6 = 0$.
Since $6^{2x} = (6^x)^2 = u^2$, the equation becomes:
$$u^2 - 5u + 6 = 0$$
$$(u - 2)(u - 3) = 0 \implies u = 2 \text{ or } u = 3$$
Both values are positive, so both are valid.
$6^x = 2 \implies x = \log_6 2 = \dfrac{\ln 2}{\ln 6} = \dfrac{0.6931}{1.7918} \approx 0.3869$
$6^x = 3 \implies x = \log_6 3 = \dfrac{\ln 3}{\ln 6} = \dfrac{1.0986}{1.7918} \approx 0.6131$
$$x \approx 0.387 \quad \text{or} \quad x \approx 0.613 \quad \text{(3 s.f.)}$$
Using the substitution $u = 3^x$, or otherwise, solve, correct to 3 significant figures, the equation $3^x = 2 + 3^{-x}$.
Let $u = 3^x$ (so $3^{-x} = \dfrac{1}{u}$). Multiply through by $u$ to clear the fraction, giving a quadratic in $u$. Remember $u > 0$.
Since $3^{-x} = \dfrac{1}{u}$, the equation becomes:
$$u = 2 + \dfrac{1}{u}$$
$$u^2 = 2u + 1 \implies u^2 - 2u - 1 = 0$$
$$u = \dfrac{2 \pm \sqrt{4 + 4}}{2} = \dfrac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}$$
Since $u = 3^x > 0$, reject $u = 1 - \sqrt{2} \approx -0.414$ (negative).
So $u = 1 + \sqrt{2} \approx 2.414$.
$$3^x = 1 + \sqrt{2} \implies x = \log_3(1 + \sqrt{2}) = \dfrac{\ln(1 + \sqrt{2})}{\ln 3} = \dfrac{\ln(2.4142)}{\ln 3} = \dfrac{0.88137}{1.0986} \approx 0.80222$$
$$x \approx 0.802 \quad \text{(3 s.f.)}$$
(i) After the introduction of a vaccine, the numbers of new cases of a virus infection in successive weeks are given in the table below.
$\begin{array}{|l|c|c|c|c|c|} \hline \text{Week number, } t & 1 & 2 & 3 & 4 & 5 \\ \hline \text{No. of new cases, } y & 240 & 150 & 95 & 58 & 38 \\ \hline \end{array}$
Plot the points $(t, y)$ on graph paper and join them with a smooth curve.
(ii) Decide on a suitable model to relate $t$ and $y$, and plot an appropriate graph to find this model. (You may wish to use a spreadsheet or graphing software.)
(iii) Find the value of $y$ when $t = 15$, and explain what you find.
Check whether the ratios $\dfrac{y_{t+1}}{y_t}$ are roughly constant — if so, an exponential model $y = ab^t$ is appropriate. Take logs: $\log y = \log a + t\log b$, then plot $\log y$ against $t$.
The points $(1, 240)$, $(2, 150)$, $(3, 95)$, $(4, 58)$, $(5, 38)$ form a smooth decreasing curve — consistent with exponential decay.
Compute $\log_{10} y$ for each data point:
$\begin{array}{|c|c|c|c|c|c|} \hline t & 1 & 2 & 3 & 4 & 5 \\ \hline \log_{10} y & 2.380 & 2.176 & 1.978 & 1.763 & 1.580 \\ \hline \end{array}$
The values of $\log_{10} y$ decrease roughly linearly, confirming the exponential model $y = ab^t$ where $\log_{10} y = \log_{10} a + t \log_{10} b$.
Using the endpoints $(1, 2.380)$ and $(5, 1.580)$:
$$\text{Gradient} = \log_{10} b = \dfrac{1.580 - 2.380}{5 - 1} = \dfrac{-0.800}{4} = -0.200$$
$$b = 10^{-0.200} \approx 0.631$$
$$\log_{10} a = 2.380 - 1 \times (-0.200) = 2.580 \implies a = 10^{2.580} \approx 380$$
So the model is: $y \approx 380 \times 0.631^t$.
$$y = 380 \times 0.631^{15}$$
Compute $0.631^{15}$: $\ln(0.631^{15}) = 15 \times \ln(0.631) = 15 \times (-0.4605) = -6.908$
$$0.631^{15} = e^{-6.908} \approx 0.00100$$
$$y \approx 380 \times 0.00100 \approx 0.38$$
When $t = 15$, $y \approx 0.38$, which is less than 1. This means the model predicts fewer than one new case per week — the number of new cases has effectively dropped to zero. The vaccine has successfully brought the infection under control.
Caution: the model is based on only 5 data points. Beyond the data range, the prediction assumes the exponential decay continues at the same rate — in reality, the decay may slow as residual cases take longer to resolve.
(ii) $y \approx 380 \times 0.631^t$ (exponential decay model).
(iii) $y(15) \approx 0.38$ — fewer than 1 new case per week, indicating the vaccine is working.
Exit Ticket — W1–W2 Review
Check the learning objectives you feel confident about: