Exponentials and Logarithms

------- | ---------- | ------------------------------------------------- | | AQA | Paper 1, 2 | Exponentials in P1;” date: 2025-06-02T16:25:28.480Z tags:

• Maths
• ALevel categories:
• Maths

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Intuition

Exponentials describe growth that accelerates, logarithms describe growth that decelerates: If something grows at a rate proportional to its current size (bacteria, compound interest, radioactive decay), it follows an exponential curve. Logarithms are the inverse, they answer “how long does it take to reach this size?” They’re the mathematical tools for understanding anything that grows or decays.

Why it matters: Exponential growth explains population dynamics, compound interest, and viral spread. Logarithmic scales (pH, Richter, decibels) compress huge ranges into manageable numbers. Together, they’re essential for modelling natural and financial phenomena.

The key insight: The number $e$ is special because $e^x$ is its own derivative, $\frac{d}{dx}e^x = e^x$. This makes it the natural choice for modelling continuous growth. When you see “continuous compounding” or “natural growth,” think $e$. The natural logarithm $\ln x$ is the inverse of $e^x$, and its derivative $\frac{1}{x}$ is the simplest interesting derivative.

Board Coverage

Board	Paper	Notes
AQA	Paper 1, 2	Exponentials in P1; modelling, natural logs in P2
Edexcel	P1, P2	Similar split
OCR (A)	Paper 1, 2	Includes $e^x$ and $\ln x$ graphs
CIE (9709)	P1, P2, P3	Exponentials/logarithms in P1; further in P3

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1. The Number $e$

1.1 Definition of $e$

Definition. The number $e$ is defined by the limit

$$e = \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$$

To see that this limit converges, consider a sequence of values:

$n$	$\left(1+\frac{1}{n}\right)^n$
1	2
2	2.25
10	2.5937…
100	2.7048…
1000	2.7169…
10000	2.7181…
$10^6$	2.71828…

The sequence is increasing and bounded above (by 3, as can be shown via the binomial theorem), so by the Monotone Convergence Theorem it converges. Its value is

$$e \approx 2.718281828459045...$$

1.2 Alternative characterisations

The number $e$ can equivalently be characterised as the unique positive real number such that

$$\frac{d}{dx}e^x \bigg|_{x=0} = 1$$

That is, the exponential function with base $e$ is its own derivative, the only exponential Function with this property. We prove this rigorously in the next section.

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2. The Derivative of $e^x$

2.1 Proof from the limit definition

Theorem. $\dfrac{d}{dx}e^x = e^x$.

Proof. By the limit definition of the derivative:

$$\frac{d}{dx}e^x = \lim_{h\to 0}\frac{e^{x+h}-e^x}{h} = \lim_{h\to 0}\frac{e^x(e^h - 1)}{h} = e^x \cdot \lim_{h\to 0}\frac{e^h - 1}{h}$$

So we need to show that $\displaystyle\lim_{h\to 0}\frac{e^h - 1}{h} = 1$.

Let $e^h - 1 = t$ So that $h = \ln(1+t)$. As $h \to 0$ We have $t \to 0$. Then:

$$\lim_{h\to 0}\frac{e^h - 1}{h} = \lim_{t\to 0}\frac{t}{\ln(1+t)} = \lim_{t\to 0}\frac{1}{\frac{\ln(1+t)}{t}}$$

We use the fundamental limit $\displaystyle\lim_{t\to 0}\frac{\ln(1+t)}{t} = 1$ (which follows from The definition of $\ln x$ as the area under $1/u$ from $1$ to $1+t$ Plus the squeeze theorem on $1/(1+t) \leq \ln(1+t)/t \leq 1$). Therefore:

$$\lim_{h\to 0}\frac{e^h - 1}{h} = \frac{1}{1} = 1$$

Hence $\dfrac{d}{dx}e^x = e^x \cdot 1 = e^x$. $\blacksquare$

2.2 Derivative of $a^x$

For a general base $a \gt 0$:

$$\frac{d}{dx}a^x = \frac{d}{dx}e^{x\ln a} = (\ln a)\, e^{x\ln a} = a^x \ln a$$

This follows immediately from the chain rule applied to $e^{x\ln a}$.

Intuition. The function $e^x$ is the unique function whose rate of change at any point equals Its value at that point. If a population of bacteria doubles every hour, its growth rate is Proportional to its current size, this is precisely the behaviour of $e^{kt}$. This is why $e$ Appears everywhere in nature: compound interest, radioactive decay, population dynamics, and cooling are all governed by exponential functions.

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3. Logarithms as Inverses of Exponentials

3.1 Definition

Definition. For $a \gt 0$, $a \neq 1$ The logarithm $\log_a x$ is the inverse function of $a^x$:

$$y = \log_a x \iff a^y = x$$

The natural logarithm is the logarithm with base $e$ Written $\ln x = \log_e x$.

3.2 Proof that $a^{\log_a x} = x$ and $\log_a(a^x) = x$

Theorem. For all $a \gt 0$, $a \neq 1$ And all $x \gt 0$:

$$(\mathrm i)\quad a^{\log_a x} = x \qquad (\mathrm{ii})\quad \log_a(a^x) = x$$

Proof of (i). Let $y = \log_a x$. By definition, $a^y = x$. Substituting $y = \log_a x$: $a^{\log_a x} = x$. $\blacksquare$

Proof of (ii). Let $y = a^x$. Then $\log_a y = \log_a(a^x)$. By definition of the logarithm as Inverse, $\log_a(a^x) = x$. $\blacksquare$

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4. Laws of Logarithms

4.1 Product law

Theorem. $\log_a(xy) = \log_a x + \log_a y$

Proof. Let $p = \log_a x$ and $q = \log_a y$ So $a^p = x$ and $a^q = y$.

$$xy = a^p \cdot a^q = a^{p+q}$$

Therefore $\log_a(xy) = p + q = \log_a x + \log_a y$. $\blacksquare$

4.2 Quotient law

Theorem. $\log_a\!\left(\dfrac{x}{y}\right) = \log_a x - \log_a y$

Proof. Similarly, with $a^p = x$ and $a^q = y$:

$$\frac{x}{y} = \frac{a^p}{a^q} = a^{p-q}$$

Therefore $\log_a(x/y) = p - q = \log_a x - \log_a y$. $\blacksquare$

4.3 Power law

Theorem. $\log_a(x^n) = n\log_a x$

Proof. Let $p = \log_a x$ So $a^p = x$. Then:

$$x^n = (a^p)^n = a^{pn}$$

Therefore $\log_a(x^n) = pn = n\log_a x$. $\blacksquare$

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5. Change of Base Formula

Theorem. For all $a, b \gt 0$ with $a, b \neq 1$:

$$\log_a b = \frac{\ln b}{\ln a} = \frac{\log_{10} b}{\log_{10} a}$$

Proof. Let $y = \log_a b$. Then $a^y = b$. Taking natural logarithms of both sides:

$\ln(a^y) = \ln b$ $y \ln a = \ln b$ $y = \frac{\ln b}{\ln a}$

Since $y = \log_a b$ We have $\log_a b = \dfrac{\ln b}{\ln a}$. $\blacksquare$

Intuition. The change of base formula tells us that logarithms in different bases are just Constant multiples of each other. If you think of $\log_a x$ as “how many times do I need to use Base $a$ to reach $x$?”, then $\log_a b / \log_a c$ tells you the ratio of the “number of steps” in Base $a$ to reach $b$ versus $c$. The formula shows this ratio is independent of $a$.

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6. The Natural Logarithm as an Area

6.1 Integral definition

Definition. The natural logarithm function is defined by:

$$\ln x = \int_1^x \frac{1}{t}\, dt, \quad x > 0$$

This means $\ln x$ is the signed area under the curve $y = 1/t$ from $t = 1$ to $t = x$.

6.2 Properties from the definition

From this definition, several properties follow immediately:

• $\ln 1 = 0$ (the integral from 1 to 1 of any function is zero)
• $\ln x \lt 0$ for $0 \lt x \lt 1$ (negative area when integrating backwards)
• $\ln x \gt 0$ for $x \gt 1$ (positive area)
• $\ln x$ is strictly increasing (the integrand $1/t \gt 0$)

6.3 Derivative of $\ln x$

$$\frac{d}{dx}\ln x = \frac{1}{x}$$

This follows directly from the Fundamental Theorem of Calculus applied to the integral definition.

Intuition. Think of $\ln x$ as measuring “how much area does $1/t$ sweep out from 1 to $x$?”. Since $1/t$ decreases as $t$ grows, each additional unit of $x$ contributes less area. This is why $\ln x$ grows so slowly, it takes $e^{10} \approx 22026$ to reach $\ln x = 10$.

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7. Solving Exponential Equations

7.1 Equations of the form $a^x = b$

Method. Take logarithms of both sides:

$$a^x = b \implies x \ln a = \ln b \implies x = \frac{\ln b}{\ln a}$$

7.2 Equations of the form $a^{f(x)} = b^{g(x)}$

Take logarithms of both sides:

$$a^{f(x)} = b^{g(x)} \implies f(x)\ln a = g(x)\ln b$$

This gives an equation in $x$ that can often be solved algebraically.

Example. Solve $3^{2x+1} = 5^{x-2}$.

$$\begin{aligned} (2x+1)\ln 3 &= (x-2)\ln 5 \\ 2x\ln 3 + \ln 3 &= x\ln 5 - 2\ln 5 \\ 2x\ln 3 - x\ln 5 &= -2\ln 5 - \ln 3 \\ X(2\ln 3 - \ln 5) &= -(2\ln 5 + \ln 3) \\ X &= \frac{2\ln 5 + \ln 3}{\ln 5 - 2\ln 3} \end{aligned}$$

7.3 Equations reducible to quadratic form

Sometimes we can use a substitution. For example, $e^{2x} + 3e^x - 4 = 0$.

Let $u = e^x$ (note $u \gt 0$). Then $u^2 + 3u - 4 = 0$ Giving $(u+4)(u-1) = 0$ So $u = 1$ (rejecting $u = -4$). Hence $e^x = 1$ Giving $x = 0$.

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8. Exponential Modelling

8.1 Exponential growth

A quantity $N$ grows exponentially when

$$N = N_0 e^{kt}, \quad k > 0$$

Where $N_0$ is the initial quantity and $k$ is the growth constant.

The doubling time $T_d$ satisfies $N_0 e^{kT_d} = 2N_0$ So:

$$e^{kT_d} = 2 \implies T_d = \frac{\ln 2}{k}$$

8.2 Exponential decay

$$N = N_0 e^{-kt}, \quad k > 0$$

The half-life $t_{1/2}$ satisfies $N_0 e^{-kt_{1/2}} = \frac{N_0}{2}$ So:

$$e^{-kt_{1/2}} = \frac{1}{2} \implies t_{1/2} = \frac{\ln 2}{k}$$

8.3 General exponential model

In many applications, the model $N = A \cdot b^{kt}$ or $N = A \cdot e^{kt}$ appears. Given two data Points, we can determine the parameters.

Example. A population is 500 at $t=0$ and 2000 at $t=6$ (hours). Assuming exponential growth $N = N_0 e^{kt}$:

\begin\{aligned\} N_0 &= 500 \\ 2000 &= 500 e^\{6k\} \\ 4 &= e^\{6k\} \\ 6k &= \ln 4 = 2\ln 2 \\ K &= \frac\{\ln 2\}\{3\} \end\{aligned\}

Doubling time: $T_d = \dfrac{\ln 2}{k} = \dfrac{\ln 2}{\ln 2 / 3} = 3$ hours.

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9. Graphs of Exponential and Logarithmic Functions

9.1 Key features of $y = e^x$

• Domain: $(-\infty, \infty)$ Range: $(0, \infty)$
• $y$-intercept: $(0, 1)$
• Horizontal asymptote: $y = 0$ (as $x \to -\infty$)
• Strictly increasing
• Passes through $(1, e)$, $(\ln 2, 2)$

9.2 Key features of $y = \ln x$

• Domain: $(0, \infty)$ Range: $(-\infty, \infty)$
• $x$-intercept: $(1, 0)$
• Vertical asymptote: $x = 0$
• Strictly increasing
• Passes through $(e, 1)$, $(2, \ln 2)$

9.3 Transformations

The graphs of $y = e^{-x}$ (reflection in $y$-axis), $y = e^{x} + c$ (vertical translation), and $y = \ln(x-a)$ (horizontal translation) follow from standard transformation rules.

Intuition. The graphs of $y = e^x$ and $y = \ln x$ are reflections of each other in the line $y = x$ Since they are inverse functions.

Explore the relationship between $y = e^x$ and $y = \ln x$ as inverse functions Reflected in $y = x$. Add transformations such as $y = e^{x+c}$ and $y = \ln(x - a)$ to see how they Shift the curves.

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10. Logarithmic Differentiation

For functions of the form $y = f(x)^{g(x)}$ Take logarithms first:

$$\ln y = g(x) \ln f(x)$$

Then differentiate implicitly:

$$\frac{1}{y}\frac{dy}{dx} = g'(x)\ln f(x) + \frac{g(x) f'(x)}{f(x)}$$ $$\frac{dy}{dx} = f(x)^{g(x)}\left[g'(x)\ln f(x) + \frac{g(x) f'(x)}{f(x)}\right]$$

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Problem Set

Problem 1

Evaluate $\displaystyle\lim_{n\to\infty}\left(1+\frac{3}{n}\right)^n$.

Solution 1

Let $m = n/3$ So $n = 3m$. As $n \to \infty$, $m \to \infty$.

$$\left(1+\frac{3}{n}\right)^n = \left(1+\frac{1}{m}\right)^{3m} = \left[\left(1+\frac{1}{m}\right)^m\right]^3 \to e^3$$

If you get this wrong, revise: The Number $e$, Section 1.1.

Problem 2

Prove that $\dfrac{d}{dx}(e^{3x^2}) = 6x\, e^{3x^2}$ using the chain rule.

Solution 2

Let $u = 3x^2$ So $y = e^u$ and $\dfrac{dy}{du} = e^u$, $\dfrac{du}{dx} = 6x$.

By the chain rule: $\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx} = e^u \cdot 6x = 6x\, e^{3x^2}$. $\blacksquare$

If you get this wrong, revise: The Derivative of $e^x$, Section 2.

Problem 3

Solve $\log_2(x+3) + \log_2(x-1) = 4$.

Solution 3

$$\log_2[(x+3)(x-1)] = 4 \implies (x+3)(x-1) = 16$$$$x^2 + 2x - 3 = 16 \implies x^2 + 2x - 19 = 0$$$$x = \frac{-2 \pm \sqrt{4+76}}{2} = \frac{-2 \pm \sqrt{80}}{2} = -1 \pm 2\sqrt{5}$$

Domain: $x \gt 1$ So $x = -1 + 2\sqrt{5}$.

If you get this wrong, revise: Laws of Logarithms, Section 4.

Problem 4

Given $\log_a 2 = 0.301$ and $\log_a 5 = 0.699$ Find $\log_a 200$.

Solution 4

$$\log_a 200 = \log_a(2 \times 100) = \log_a 2 + \log_a 100 = \log_a 2 + \log_a(4 \times 25)$$$$= \log_a 2 + 2\log_a 2 + 2\log_a 5 = 3(0.301) + 2(0.699) = 0.903 + 1.398 = 2.301$$

If you get this wrong, revise: Laws of Logarithms, Section 4.

Problem 5

Solve $5^{2x} - 6 \cdot 5^x + 5 = 0$.

Solution 5

Let $u = 5^x$ ($u \gt 0$). Then $u^2 - 6u + 5 = 0$ So $(u-1)(u-5) = 0$.

$u = 1$: $5^x = 1 \implies x = 0$. $u = 5$: $5^x = 5 \implies x = 1$.

If you get this wrong, revise: Solving Exponential Equations

• Section 7.3.

Problem 6

A radioactive substance decays with half-life 8 days. If initially there are 200g, find the mass after 25 days.

Solution 6

Model: $N = N_0 e^{-kt}$ where $N_0 = 200$.

Half-life: $t_{1/2} = \dfrac{\ln 2}{k} = 8 \implies k = \dfrac{\ln 2}{8}$.

$$N = 200 \cdot e^{-25\ln 2/8} = 200 \cdot 2^{-25/8} \approx 200 \cdot 0.1146 \approx 22.9 \mathrm{ g}$$

If you get this wrong, revise: Exponential Modelling, Section 8.2.

Problem 7

Differentiate $y = x^x$ using logarithmic differentiation.

Solution 7

$\ln y = x \ln x$. Differentiating: $\dfrac{1}{y}\dfrac{dy}{dx} = \ln x + x \cdot \dfrac{1}{x} = \ln x + 1$.

$$\frac{dy}{dx} = x^x(\ln x + 1)$$

If you get this wrong, revise: Logarithmic Differentiation, Section 10.

Problem 8

Solve $\ln(3x-1) = \ln(x+2) + \ln 4$.

Solution 8

$$\ln(3x-1) = \ln[4(x+2)] \implies 3x - 1 = 4x + 8 \implies x = -9$$

Check domain: $3(-9)-1 = -28 \lt 0$ and $-9+2 = -7 \lt 0$. Both logarithms undefined. No Solution.

If you get this wrong, revise: Solving Exponential Equations

• Section 7.1, and always check the domain.

Problem 9

Prove that $\log_a b \cdot \log_b a = 1$ for all $a, b \gt 0$, $a, b \neq 1$.

Solution 9

By the change of base formula:

$$\log_a b = \frac{\ln b}{\ln a}, \qquad \log_b a = \frac{\ln a}{\ln b}$$$$\log_a b \cdot \log_b a = \frac{\ln b}{\ln a} \cdot \frac{\ln a}{\ln b} = 1 \quad \blacksquare$$

If you get this wrong, revise: Change of Base Formula, Section 5.

Problem 10

The temperature $T$ of a cooling object follows $T = T_s + (T_0 - T_s)e^{-kt}$. A cup of tea at $90^\circ\mathrm C$ is placed in a room at $20^\circ\mathrm C$. After 10 minutes it is $60^\circ\mathrm C$. Find $k$ and determine when the tea reaches $35^\circ\mathrm C$.

Solution 10

$T_s = 20$, $T_0 = 90$. Model: $T = 20 + 70e^{-kt}$.

At $t = 10$: $60 = 20 + 70e^{-10k} \implies 40 = 70e^{-10k} \implies e^{-10k} = 4/7$.

$$-10k = \ln(4/7) \implies k = \frac{\ln(7/4)}{10} \approx 0.0560$$

For $T = 35$: $35 = 20 + 70e^{-kt} \implies 15 = 70e^{-kt} \implies e^{-kt} = 3/14$.

$$-kt = \ln(3/14) \implies t = \frac{\ln(14/3)}{k} = \frac{10\ln(14/3)}{\ln(7/4)} \approx 27.5 \mathrm{ minutes}$$

If you get this wrong, revise: Exponential Modelling, Section 8.

Problem 11

Express $y = e^{2x} + 3e^{-2x}$ in the form $y = 2\cosh(2x) + \cosh(2x)$ is not correct. Instead: Find the minimum value of $y = e^{2x} + 3e^{-2x}$.

Solution 11

$\dfrac{dy}{dx} = 2e^{2x} - 6e^{-2x} = 0 \implies 2e^{2x} = 6e^{-2x} \implies e^{4x} = 3 \implies x = \dfrac{\ln 3}{4}$.

At $x = \dfrac{\ln 3}{4}$: $e^{2x} = e^{\ln 3 / 2} = \sqrt{3}$, $e^{-2x} = 1/\sqrt{3}$.

$$y_{\min} = \sqrt{3} + \frac{3}{\sqrt{3}} = \sqrt{3} + \sqrt{3} = 2\sqrt{3}$$

If you get this wrong, revise: Solving Exponential Equations And Section 2.2.

Problem 12

Given $\ln(x^2 + 1) = 2\ln x + \ln 5$ Solve for $x$.

Solution 12

$$\ln(x^2+1) = \ln(x^2) + \ln 5 = \ln(5x^2)$$$$x^2 + 1 = 5x^2 \implies 4x^2 = 1 \implies x^2 = \frac{1}{4} \implies x = \frac{1}{2}$$

Reject $x = -1/2$ since $\ln x$ requires $x \gt 0$.

If you get this wrong, revise: Laws of Logarithms, Section 4 and Domain restrictions.

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Cross-References

• Differentiation — The derivative of $e^x$ is $e^x$ itself, and logarithmic differentiation handles functions of the form $f(x)^{g(x)}$.
• Integration — Integrating $1/x$ yields $\ln|x|$, and exponential functions integrate to themselves.
• Functions — Exponential and logarithmic functions are inverses of each other, a core concept in understanding function behaviour.
• Proof — The irrationality of $\log_2 3$ and similar results use proof by contradiction from this topic’s framework.