Exponentials and logarithms

Mathematics 2210 · CCEA AS Level · 8 min read · Updated: 31 Aug 2026

Welcome to Exponentials and Logarithms

Welcome to one of the most powerful and exciting topics in AS 1: Pure Mathematics (CCEA 2210)! If you have ever wondered how scientists model viral outbreaks, how archaeologists determine the age of ancient fossils using carbon dating, or how earthquakes are measured on the Richter scale, you are looking at exponentials and logarithms in action.

Don't worry if this seems tricky at first. At its heart, a logarithm is simply the reverse operation of a power or index. In these notes, we will break down the rules step-by-step, master the essential formal proofs required by CCEA, learn how to linearise curved graphs, and explore real-world exponential growth and decay.

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1. The Foundations: What is a Logarithm?

You already know how indices (powers) work: \(2^3 = 8\). A logarithm simply asks the question in reverse: "To what power must I raise 2 to get 8?" The answer is \(3\).

The Golden Definition

For any base \(a > 0\) (where \(a \neq 1\)) and any positive number \(y > 0\):

\(y = a^x \iff x = \log_a y\)

Here is how to translate between forms:

• Index / Exponential form: \(a^x = y\) (Base \(a\), power \(x\), result \(y\))
• Logarithmic form: \(\log_a y = x\) (Base \(a\), argument \(y\), result \(x\))

Special Base Values

Because of how basic indices work, three special results hold true for any base \(a\):

• \(\log_a 1 = 0\) (because \(a^0 = 1\))
• \(\log_a a = 1\) (because \(a^1 = a\))
• \(a^{\log_a x} = x\) (the exponential function and logarithm undo each other)

The Natural Number \(e\) and Natural Logarithms (\(\ln\))

In advanced mathematics and real-world modelling, one particular base is special: Euler's number, denoted by \(e \approx 2.71828\dots\). Logarithms with base \(e\) are called natural logarithms and are written as \(\ln x\) rather than \(\log_e x\).

• \(\ln x \equiv \log_e x\) (defined strictly for \(x > 0\))
• \(\ln(e^x) = x\)
• \(e^{\ln x} = x\) (for \(x > 0\))

Key Takeaway: A logarithm is just an index in disguise. Always remember: \(\text{Base}^{\text{Answer}} = \text{Argument}\).

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2. The Three Laws of Logarithms (Examinable Formal Proofs)

In the CCEA AS 1 exam, you are required not only to use the three laws of logarithms, but also to be able to write out their formal proofs rigorously. Candidates often lose marks by skipping the initial definitions, so ensure you write out every single line shown below.

Law 1: The Multiplication Law

\(\log_a(xy) = \log_a x + \log_a y\)

Proof:
1. Let \(x = a^p \implies p = \log_a x\)
2. Let \(y = a^q \implies q = \log_a y\)
3. Multiply the two expressions: \(xy = a^p \cdot a^q = a^{p+q}\)
4. Convert back to logarithmic form: \(\log_a(xy) = p + q\)
5. Substitute \(p\) and \(q\) back in: \(\log_a(xy) = \log_a x + \log_a y\) (as required).

Law 2: The Division Law

\(\log_a\left(\frac{x}{y}\right) = \log_a x - \log_a y\)

Proof:
1. Let \(x = a^p \implies p = \log_a x\)
2. Let \(y = a^q \implies q = \log_a y\)
3. Divide the two expressions: \(\frac{x}{y} = \frac{a^p}{a^q} = a^{p-q}\)
4. Convert back to logarithmic form: \(\log_a\left(\frac{x}{y}\right) = p - q\)
5. Substitute \(p\) and \(q\) back in: \(\log_a\left(\frac{x}{y}\right) = \log_a x - \log_a y\) (as required).

Law 3: The Power Law

\(\log_a(x^k) = k \log_a x\)

Proof:
1. Let \(x = a^p \implies p = \log_a x\)
2. Raise both sides to the power \(k\): \(x^k = (a^p)^k = a^{kp}\)
3. Convert back to logarithmic form: \(\log_a(x^k) = kp\)
4. Substitute \(p\) back in: \(\log_a(x^k) = k\log_a x\) (as required).

Common Pitfall to Avoid: Watch out for illegal algebra! Never write \(\log(A + B) = \log A + \log B\) or \(\frac{\log A}{\log B} = \log(A - B)\). These are invalid shortcuts.

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3. Graphs of Exponential and Logarithmic Functions

The Exponential Graph: \(y = a^x\) (where \(a > 1\))

• \(y\)-intercept: Always passes through \((0, 1)\) because \(a^0 = 1\).
• Asymptote: Has a horizontal asymptote along the \(x\)-axis (\(y = 0\)) as \(x \to -\infty\).
• Domain & Range: Domain is all real numbers (\(x \in \mathbb{R}\)); Range is strictly positive (\(y > 0\)).

The Natural Logarithmic Graph: \(y = \ln x\)

• \(x\)-intercept: Always passes through \((1, 0)\) because \(\ln(1) = 0\).
• Asymptote: Has a vertical asymptote along the \(y\)-axis (\(x = 0\)) as \(x \to 0^+\).
• Domain & Range: Domain is strictly positive (\(x > 0\)); Range is all real numbers (\(y \in \mathbb{R}\)).
• Reflection: The curve \(y = \ln x\) is the exact reflection of \(y = e^x\) across the line \(y = x\).

Key Takeaway: Notice how the domain and range swap between \(y = e^x\) and \(y = \ln x\). This confirms that they are inverse functions of each other.

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4. Solving Equations Involving Exponentials and Logarithms

Type 1: Unknown in the Exponent (\(a^{f(x)} = b\))

Take the natural logarithm (\(\ln\)) of both sides and apply the power law to bring the variable down.

Example: Solve \(3^{2x-1} = 7\).
\(\ln(3^{2x-1}) = \ln 7\)
\((2x - 1)\ln 3 = \ln 7\)
\(2x - 1 = \frac{\ln 7}{\ln 3}\)
\(2x = \frac{\ln 7}{\ln 3} + 1\)
\(x = \frac{1}{2}\left(\frac{\ln 7}{\ln 3} + 1\right) \approx 1.39\) (to 3 s.f.)

Type 2: Equations Reducible to Quadratics (Hidden Quadratics)

Equations containing terms like \(e^{2x}\) and \(e^x\), or \(a^{2x}\) and \(a^x\), can be transformed into standard quadratic equations using substitution.

Example: Solve \(e^{2x} - 2e^x - 15 = 0\).
1. Let \(u = e^x\) (where \(u > 0\)).
2. Substitute into the equation: \(u^2 - 2u - 15 = 0\).
3. Factorise: \((u - 5)(u + 3) = 0 \implies u = 5\) or \(u = -3\).
4. Substitute back: \(e^x = 5 \implies x = \ln 5 \approx 1.61\).
5. Critical Step: Reject \(e^x = -3\) because \(e^x > 0\) for all real \(x\) (and \(\ln(-3)\) is undefined).
Thus, the only valid solution is \(x = \ln 5\).

Type 3: Combining Multiple Logarithms

Use the laws of logarithms to combine separate terms into a single log before converting to index form.

Example: Solve \(\log_2(x + 6) - \log_2(x) = 3\).
\(\log_2\left(\frac{x + 6}{x}\right) = 3\)
Convert to index form: \(\frac{x + 6}{x} = 2^3 = 8\)
\(x + 6 = 8x \implies 7x = 6 \implies x = \frac{6}{7}\).
Always check that your answer makes the original log terms positive: \(\frac{6}{7} > 0\) and \(\frac{6}{7} + 6 > 0\), so \(x = \frac{6}{7}\) is valid.

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5. Reduction to Linear Form (Linearisation / Log Graphs)

When scientists collect non-linear experimental data, plotting the raw variables makes it difficult to determine unknown constants. By taking logarithms of both sides, we can transform curved relationships into a straight line of the form \(Y = mX + c\).

Case A: Power Relationship \(y = a x^n\)

1. Take natural logarithms of both sides: \(\ln y = \ln(a x^n)\)
2. Apply the multiplication and power laws: \(\ln y = \ln a + \ln(x^n) \implies \ln y = n\ln x + \ln a\)
3. Compare with \(Y = mX + c\):
• Vertical axis (\(Y\)): \(\ln y\)
• Horizontal axis (\(X\)): \(\ln x\)
• Gradient (\(m\)): \(n\)
• Vertical Intercept (\(c\)): \(\ln a \implies a = e^c\)

Case B: Exponential Relationship \(y = k b^x\) (or \(y = A e^{kx}\))

1. Take natural logarithms of both sides: \(\ln y = \ln(k b^x)\)
2. Apply log laws: \(\ln y = \ln k + \ln(b^x) \implies \ln y = (\ln b)x + \ln k\)
3. Compare with \(Y = mX + c\):
• Vertical axis (\(Y\)): \(\ln y\)
• Horizontal axis (\(X\)): \(x\)
• Gradient (\(m\)): \(\ln b \implies b = e^m\)
• Vertical Intercept (\(c\)): \(\ln k \implies k = e^c\)

Quick Summary Table:

• Model \(y = ax^n\) \(\implies\) Plot \(\ln y\) against \(\ln x\) \(\implies\) Gradient \(= n\), Intercept \(= \ln a\)
• Model \(y = kb^x\) \(\implies\) Plot \(\ln y\) against \(x\) \(\implies\) Gradient \(= \ln b\), Intercept \(= \ln k\)

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6. Exponential Growth and Decay Modelling

Exponential functions naturally model quantities whose rate of increase or decrease is proportional to the amount currently present.

The standard model is: \(N(t) = N_0 e^{kt}\) or \(P(t) = P_0 a^t\)

• \(N_0\) or \(P_0\): The initial quantity at time \(t = 0\).
• \(k > 0\): Exponential growth rate.
• \(k < 0\): Exponential decay rate.

Worked Example: Half-Life Calculation

A radioactive substance decays according to the formula \(N(t) = N_0 e^{kt}\), where \(t\) is measured in years. The half-life is the time taken for the quantity to reduce to half its original amount (\(\frac{1}{2}N_0\)).

Set \(N(t) = \frac{1}{2}N_0\):
\(\frac{1}{2}N_0 = N_0 e^{kt}\)
Divide both sides by \(N_0\): \(\frac{1}{2} = e^{kt}\)
Take natural logs: \(\ln(0.5) = kt \implies \ln\left(2^{-1}\right) = kt \implies -\ln 2 = kt\)
Therefore, the half-life is: \(t = -\frac{\ln 2}{k}\).

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7. Essential Exam Tips & Pitfalls Checklist

• Check for Extraneous Roots: Whenever you solve a hidden quadratic in \(e^x\) or \(a^x\), remember that \(e^x > 0\). If you get a negative value for \(u\), you must explicitly reject it.
• Domain Restrictions: You cannot evaluate the logarithm of a negative number or zero. Always check that your final solutions keep every original logarithm argument positive (\(> 0\)).
• Full Proof Steps: In formal proofs, always begin with "Let \(x = a^p\)" and "Let \(y = a^q\)". Do not skip steps.
• Linearisation Axes: Look carefully at the horizontal axis. If it is \(\ln x\), the original model was a power model (\(y = ax^n\)). If it is \(x\), the original model was an exponential model (\(y = kb^x\)).
• Accuracy & Rounding: Keep exact values in your calculator during intermediate steps to avoid premature rounding errors. State final answers to 3 significant figures unless an exact expression is requested.

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What is the fundamental definition linking an exponential statement to a logarithmic statement?

For any base \(a > 0\) (where \(a \neq 1\)) and \(y > 0\), \(y = a^x \iff x = \log_a y\).

What are natural logarithms and Euler's number \(e\)?

Natural logarithms are logarithms with base \(e \approx 2.71828\dots\), written as \(\ln x \equiv \log_e x\) for \(x > 0\).

What are the three laws of logarithms?

\(\log_a(xy) = \log_a x + \log_a y\), \(\log_a\left(\frac{x}{y}\right) = \log_a x - \log_a y\), and \(\log_a(x^k) = k \log_a x\).

How do you formally prove the multiplication law \(\log_a(xy) = \log_a x + \log_a y\)?

Let \(x = a^p\) (so \(p = \log_a x\)) and \(y = a^q\) (so \(q = \log_a y\)). Then \(xy = a^p a^q = a^{p+q}\). Converting to logarithmic form gives \(\log_a(xy) = p + q = \log_a x + \log_a y\).

What are the key graphical features of \(y = a^x\) (for \(a > 1\)) and \(y = \ln x\)?

\(y = a^x\) has a \(y\)-intercept at \((0, 1)\) and a horizontal asymptote \(y = 0\). \(y = \ln x\) has an \(x\)-intercept at \((1, 0)\) and a vertical asymptote \(x = 0\).

How do you solve an exponential equation with the unknown in the index, such as \(a^{f(x)} = b\)?

Take the natural logarithm (\(\ln\)) of both sides and apply the power law to bring down the exponent: \(f(x)\ln a = \ln b\), then solve for \(x\).

Why must negative solutions for the substitution variable \(u = e^x\) be rejected when solving hidden quadratics?

Because \(e^x > 0\) for all real \(x\); an exponential term cannot equal a negative value, and \(\ln(\text{negative})\) is undefined.

How is a power relationship \(y = ax^n\) reduced to linear form \(Y = mX + c\)?

Taking natural logs gives \(\ln y = n\ln x + \ln a\). Plotting \(\ln y\) against \(\ln x\) produces a line with gradient \(m = n\) and vertical intercept \(c = \ln a\) (so \(a = e^c\)).

How is an exponential relationship \(y = kb^x\) reduced to linear form \(Y = mX + c\)?

Taking natural logs gives \(\ln y = (\ln b)x + \ln k\). Plotting \(\ln y\) against \(x\) produces a line with gradient \(m = \ln b\) (so \(b = e^m\)) and intercept \(c = \ln k\) (so \(k = e^c\)).

In the exponential decay model \(N(t) = N_0 e^{kt}\), how is the half-life calculated?

Set \(N(t) = \frac{1}{2}N_0\), simplify to \(e^{kt} = 0.5\), take natural logarithms to get \(kt = -\ln 2\), giving \(t = -\frac{\ln 2}{k}\).

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