Question: Variables $x$ and $y$ are related by $y = ab^x$. Show how to linearise this relationship, and explain how $a$ and $b$ are obtained from the plot.

Step 1. Take $\lg$ of both sides:

$$\lg y = \lg(ab^x) = \lg a + x \lg b$$

Step 2. This is in the form $Y = mX + c$ with:

• $Y = \lg y$ (plot on vertical axis)
• $X = x$ (plot on horizontal axis)
• Gradient $m = \lg b$
• $Y$-intercept $c = \lg a$

Step 3. Recover constants:

• $a = 10^c$ (from $y$-intercept)
• $b = 10^m$ (from gradient)

Common variants:

• $y = ae^{kt}$ → plot $\ln y$ vs $t$; gradient $= k$, intercept $= \ln a$
• $y = ax^n$ → plot $\lg y$ vs $\lg x$; gradient $= n$, intercept $= \lg a$
• $T = T_0 + ae^{-kt}$ → plot $\ln(T - T_0)$ vs $t$

Question: Express $y = ax + \dfrac{b}{x}$ in a form suitable for drawing a straight-line graph. State what to plot and how to find $a$ and $b$.

Step 1. Multiply both sides by $x$:

$$xy = ax^2 + b$$

Step 2. This is in the form $Y = mX + c$ with:

• $Y = xy$
• $X = x^2$
• Gradient $m = a$
• $Y$-intercept $c = b$

To find constants: read off the gradient from the line for $a$, and the $Y$-intercept for $b$.

Other algebraic linearisations:

• $y = \dfrac{a}{x + b}$ → $\dfrac{1}{y} = \dfrac{1}{a}x + \dfrac{b}{a}$, plot $\dfrac{1}{y}$ vs $x$
• $y = a\sqrt{x} + \dfrac{b}{\sqrt{x}}$ → $y\sqrt{x} = ax + b$, plot $y\sqrt{x}$ vs $x$

Tip: Choose new variables so that ONE side has only $x$- or $y$-dependent terms — the constants $a$, $b$ should end up as gradient and intercept.

Question: Variables $x$ and $y$ are related by $y = ab^x$. A graph of $\lg y$ against $x$ is a straight line passing through $(1, 0.7)$ and $(3, 1.5)$. Find $a$ and $b$.

Step 1. From the linearisation $\lg y = \lg a + x\lg b$, the line has gradient $\lg b$ and $y$-intercept $\lg a$.

Step 2. Gradient from two points:

$$\lg b = \dfrac{1.5 - 0.7}{3 - 1} = \dfrac{0.8}{2} = 0.4$$ $$b = 10^{0.4} \approx 2.51$$

Step 3. $y$-intercept from point-slope: $\lg y = 0.4x + c$. Sub $(1, 0.7)$:

$$0.7 = 0.4(1) + c \Rightarrow c = 0.3$$ $$a = 10^{0.3} \approx 2.00$$

Answer: $a \approx 2.00$, $b \approx 2.51$.

Tip: Two points → compute gradient first, then sub one point back for the intercept. Always identify what the gradient/intercept REPRESENTS in your linearisation.

Question: Values of $\lg y$ are tabulated against $x$:

$x$	1	2	3	4	5
$\lg y$	0.5	0.8	1.6	1.4	1.7

One reading is incorrect. Identify which, and estimate the correct value.

Step 1. Plot the points. Expected linear trend gives $\lg y \approx 0.3x + 0.2$ from the other points.

Step 2. Check each $x$ against the predicted line:

• $x=1$: predicted $0.5$, actual $0.5$ ✓
• $x=2$: predicted $0.8$, actual $0.8$ ✓
• $x=3$: predicted $1.1$, actual $1.6$ ✗ (outlier)
• $x=4$: predicted $1.4$, actual $1.4$ ✓
• $x=5$: predicted $1.7$, actual $1.7$ ✓

Step 3. The $x=3$ reading is the outlier. Correct $\lg y$ value from the line: $\lg y = 0.3(3) + 0.2 = 1.1$.

Tip: When you see a kink in an expected-linear plot, the rogue point sits noticeably above or below the line. Read its corrected value off the line.

Question: A graph of $\lg y$ against $x$ has been drawn as a straight line through $(0, 1)$ with gradient $0.4$, modelling $y = ab^x$. By drawing a suitable straight line on the same axes, find $x$ such that $y = 100 \cdot 2^x$.

Step 1. From the existing line: $\lg y = 0.4x + 1$ (so $a = 10$, $b = 10^{0.4}$).

Step 2. The target equation $y = 100 \cdot 2^x$ gives:

$$\lg y = \lg 100 + x\lg 2 = 2 + 0.301x$$

Step 3. Draw the line $\lg y = 0.301x + 2$ on the same axes. Where it intersects the existing line, the $x$-value satisfies both:

$$0.4x + 1 = 0.301x + 2 \Rightarrow 0.099x = 1 \Rightarrow x \approx 10.1$$

Tip: "Hence find" via drawing a second line means rearranging the target equation into the SAME transformed variables, then equating the two linear expressions.

Question: The population $P$ of a town (in thousands) over years $t$ is modelled by $P = ab^t$. From a $\lg P$ vs $t$ graph, gradient $= 0.04$ and $y$-intercept $= 1.0$. Estimate the year (from $t = 0$) when $P$ first exceeds 13 thousand.

Step 1. From the linearisation: $\lg P = 0.04 t + 1.0$.

Step 2. Set $P = 13$:

$$\lg 13 = 0.04 t + 1.0$$ $$1.114 = 0.04 t + 1.0$$ $$0.04 t = 0.114$$ $$t = 2.85$$

Step 3. $P$ first exceeds 13 when $t > 2.85$, so during year $t = 3$.

Tip: Convert your target value into the LINEAR variable ($\lg P$) before solving — much cleaner than working in original $P$ then taking logs of the answer.

Question: Explain how a straight-line graph can be drawn to represent $y = ax(x + b)$, and state how $a$ and $b$ can be obtained from the line.

Step 1. Expand: $y = ax^2 + abx$.

Step 2. Divide both sides by $x$ (assuming $x \neq 0$):

$$\dfrac{y}{x} = ax + ab$$

Step 3. Plot $\dfrac{y}{x}$ (vertical) against $x$ (horizontal). This is a straight line with:

• Gradient $= a$ → read off $a$ directly.
• $y$-intercept $= ab$ → once $a$ is known, $b = \dfrac{\text{intercept}}{a}$.

Tip: "Explain how" questions usually want THREE things stated: (1) what's plotted on each axis, (2) the linear form, (3) how each constant comes from the gradient and intercept. Skip the actual graph — just the strategy.