Exponentials and Logarithms (Edexcel IAL P3)

Edexcel International A-Level Mathematics· 2018 Specification Issue 3, P3 §3.1–3.3· 25 min read

1. The Natural Exponential Function $e^x$★★☆☆☆⏱ 7 min

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📘 Definition

Natural Exponential Function

$e^{x}$ or $exp (x)$

Exponential function with base $e \approx 2.718$ , where the gradient of the graph at any point equals the value of the function at that point. It is strictly increasing for all real $x$ .

Example:

$e^{0} = 1$ , $e^{1} \approx 2.718$

You will need to sketch and interpret transformed exponential functions of the form $y = e^{a x + b} + c$ , where $a, b, c$ are constants. Key features to identify include: the y-intercept (set $x = 0$ ), horizontal asymptote (as $x \to \pm \infty$ , $e^{a x + b}$ tends to 0, so $y \to c$ ), and monotonicity (increasing if $a > 0$ , decreasing if $a < 0$ ).

📐 Worked Example

Sketch the graph of $y = 2 e^{3 x - 1} - 4$ , labelling all intercepts and asymptotes.

1
1. Find the horizontal asymptote: as $x \to - \infty$ , $e^{3 x - 1} \to 0$ , so $y \to - 4$ . Asymptote: $y = - 4$
2
1. Find y-intercept (set $x = 0$ ):
$y=2e^{-1} -4 \approx 2(0.3679) -4 = -3.26 \text{ (3 sf)}$
3
1. Find x-intercept (set $y = 0$ ):
$2e^{3x-1} -4 = 0 \implies e^{3x-1}=2 \implies 3x-1 = \ln 2 \implies x = \frac{\ln 2 + 1}{3} \approx 0.564 \text{ (3 sf)}$
4
1. Since $a = 3 > 0$ , the graph is strictly increasing, starting close to $y = - 4$ for negative $x$ , rising through the intercepts, and increasing without bound as $x$ increases.

Exam tip:

Always label asymptotes with their full equation, not just a dashed line, and give intercept values to 3 significant figures unless told otherwise.

2. The Natural Logarithm Function $\ln x$★★☆☆☆⏱ 7 min

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📘 Definition

Natural Logarithm

$ln x$

Inverse function of the natural exponential $e^{x}$ , so $ln (e^{x}) = x$ for all real $x$ , and $e^{l n x} = x$ for $x > 0$ . It is only defined for positive $x$ , with a vertical asymptote at $x = 0$ .

Example:

$ln 1 = 0$ , $ln e = 1$

You are required to solve equations of the form $e^{a x + b} = p$ and $ln (a x + b) = q$ , using the inverse relationship between $e^{x}$ and $ln x$ . Always check your solutions are valid: for $ln (a x + b) = q$ , the argument $a x + b$ must be positive.

📐 Worked Example

Solve (a) $5 e^{2 x - 3} = 12$ , (b) $3 ln (4 x + 1) = 9$ , giving your answers to 3 significant figures.

1
Part (a): Isolate the exponential term:
$e^{2x-3} = \frac{12}{5} = 2.4$
2
Take natural log of both sides:
$2x - 3 = \ln (2.4)$
3
Rearrange for x:
$x = \frac{\ln (2.4) + 3}{2} \approx \frac{0.8755 + 3}{2} = 1.94 \text{ (3 sf)}$
4
Part (b): Isolate the log term:
$\ln (4x+1) = 3$
5
Exponentiate both sides using base e:
$4x + 1 = e^3$
6
Check validity: $e^{3} \approx 20.085 > 0$ , so solution is valid. Rearrange for x:
$x = \frac{e^3 - 1}{4} \approx \frac{20.085 -1}{4} = 4.77 \text{ (3 sf)}$

Exam tip:

Always substitute your solution back into the original log expression to confirm the argument is positive, as invalid solutions will lose marks.

3. Log-Linear Graphs for Parameter Estimation★★★★☆⏱ 9 min

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Many real-world non-linear relationships can be linearised using logarithms, allowing you to estimate unknown parameters using a straight line graph of form $Y = m X + c$ , where $m$ is the gradient and $c$ is the y-intercept. Two common models are tested in P3: $y = a x^{n}$ and $y = k b^{x}$ .

Model	Log Transformation	Y Axis	X Axis	Gradient ( $m$ )	Y-Intercept ( $c$ )
$y = a x^{n}$	$lo g_{10} y = lo g_{10} a + n lo g_{10} x$	$lo g_{10} y$	$lo g_{10} x$	$n$	$lo g_{10} a$
$y = k b^{x}$	$lo g_{10} y = lo g_{10} k + x lo g_{10} b$	$lo g_{10} y$	$x$	$lo g_{10} b$	$lo g_{10} k$

📐 Worked Example

The relationship between variables $x$ and $y$ is modelled by $y = k b^{x}$ . Experimental data gives: $x = 1, y = 12$ ; $x = 3, y = 43.2$ . Estimate $k$ and $b$ to 2 significant figures.

1
1. Calculate base-10 log y values for each data point:
$x=1: \log_{10} 12 \approx 1.079; x=3: \log_{10} 43.2 \approx 1.635$
2
1. Calculate gradient of the line between $(1, 1.079)$ and $(3, 1.635)$ :
$m = \frac{1.635 - 1.079}{3 - 1} = 0.278 = \log_{10} b$
3
1. Solve for $b$ :
$b = 10^{0.278} \approx 1.9 \text{ (2 sf)}$
4
1. Find intercept $c = lo g_{10} k$ using the point $(1, 1.079)$ :
$1.079 = 0.278(1) + c \implies c = 0.801 = \log_{10} k$
5
1. Solve for $k$ :
$k = 10^{0.801} \approx 6.3 \text{ (2 sf)}$
6
1. Verify: $y = 6.3 \times 1. 9^{x}$ gives $y \approx 12$ for $x = 1$ and $y \approx 43.2$ for $x = 3$ , matching the experimental data.

Exam tip:

You can use either base 10 or natural logarithm for log-linear plots, but you must use the same base consistently for all calculations in a question. Edexcel accepts both bases as valid.

4. Common Pitfalls

Wrong move:

Forgetting the horizontal asymptote of $y = e^{a x + b} + c$ is $y = c$ , not $y = 0$

Why:

Students assume all exponential functions asymptote to 0, but vertical shifts change the asymptote value

Correct move:

For transformed exponentials, take the limit as $x \to - \infty$ (if $a > 0$ ) to find the correct asymptote value $c$

Wrong move:

Giving a solution for $ln (a x + b) = q$ where $a x + b \leq 0$

Why:

Logarithms are only defined for positive arguments, so invalid solutions are not accepted

Correct move:

After solving, substitute back into the log argument to confirm it is positive, discard any invalid solutions

Wrong move:

Mixing up axes for log-linear plots for $y = a x^{n}$ and $y = k b^{x}$

Why:

Swapping axes leads to incorrect gradient and intercept values, and thus wrong parameter estimates

Correct move:

Check the model: if $x$ is in the exponent, X-axis is $x$ ; if $x$ is raised to a power, X-axis is $ln x$

Wrong move:

Forgetting to convert intercept/gradient from log form to original parameter values

Why:

The intercept of the log-linear plot is $ln a$ or $ln k$ , not the parameter itself

Correct move:

After finding intercept $c$ , calculate $a = 1 0^{c}$ (base 10) or $a = e^{c}$ (natural log) to get the actual parameter value

Wrong move:

Using different log bases for gradient and intercept calculations in the same question

Why:

Inconsistent bases lead to incorrect parameter estimates that do not match the original model

Correct move:

Choose either base 10 or natural log at the start of the question, and use it for all log calculations

5. Quick Reference Cheatsheet

Concept	Key Rule / Formula	Exam Check
$y = e^{a x + b} + c$	Asymptote at $y = c$ , y-intercept = $e^{b} + c$	Label asymptote equation, give intercepts to 3 sf
Solve $e^{a x + b} = p$	Take ln both sides: $a x + b = ln p$	No real solution if $p \leq 0$
Solve $ln (a x + b) = q$	Exponentiate both sides: $a x + b = e^{q}$	Check $a x + b > 0$ to confirm valid solution
$y = a x^{n}$ log-linear plot	Plot $ln y$ vs $ln x$ , gradient = $n$ , intercept = $ln a$	Convert intercept from log to get $a$
$y = k b^{x}$ log-linear plot	Plot $ln y$ vs $x$ , gradient = $ln b$ , intercept = $ln k$	Convert gradient and intercept from log to get $b$ and $k$

6. Frequently Asked

Do I need to memorize log laws for P3?

Yes, log laws from P2 are assumed knowledge for log-linear graph questions. Only the identity $e^{x l n a} = a^{x}$ is provided in the P3 formula booklet.

Can I use any base for log-linear plots?

Edexcel accepts either base 10 or natural logarithm for log-linear work, as long as you use the same base consistently across all calculations for a given question.

Going deeper

What's Next

Now that you have mastered P3 exponentials and logarithms, you can apply these skills to upcoming P3 topics including differentiation of exponential and log functions, and integration involving $e^{x}$ and $1/ x$ . Log-linear modelling is also frequently tested in Statistics 1 and 2 units for real-world data analysis. Make sure you practice past paper questions on this topic to familiarize yourself with exam phrasing and mark scheme expectations, as these questions are often worth 5+ marks each and are high-yield for scoring well in P3.