Change of Base

This lesson covers the change of base formula for logarithms, which allows you to convert between different logarithm bases. This is needed for the Edexcel 9MA0 specification and is useful for solving equations, evaluating logarithms on a calculator, and proving results.

The Change of Base Formula

The change of base formula states:

log_a(x) = log_b(x) / log_b(a)

This allows you to express a logarithm in any base b. The most common choices are b = 10 (common logarithm) and b = e (natural logarithm).

Proof

Let log_a(x) = n. Then aⁿ = x.

Take log_b of both sides: log_b(aⁿ) = log_b(x)

By the power law: n × log_b(a) = log_b(x)

Therefore: n = log_b(x) / log_b(a)

Since n = log_a(x), we have: log_a(x) = log_b(x) / log_b(a)

Using the Formula with a Calculator

Calculators typically have buttons for log₁₀ and ln, but not for other bases. The change of base formula lets you evaluate any logarithm.

Example 1: Find log₃(20)

log₃(20) = log₁₀(20) / log₁₀(3) = 1.3010 / 0.4771 ≈ 2.727

Or using natural logarithms:

log₃(20) = ln(20) / ln(3) = 2.9957 / 1.0986 ≈ 2.727

Both give the same answer.

Example 2: Find log₇(100)

log₇(100) = log₁₀(100) / log₁₀(7) = 2 / 0.8451 ≈ 2.366

The Reciprocal Relationship

A useful special case of the change of base formula is:

log_a(b) = 1 / log_b(a)

This follows from setting x = b in the formula:

log_a(b) = log_b(b) / log_b(a) = 1 / log_b(a)

Example: If log₂(5) = p, find log₅(2).

log₅(2) = 1 / log₂(5) = 1/p

This relationship is frequently tested in exams.

Solving Equations Using Change of Base

Example 1: Solve log₂(x) = 3 + log₈(x)

Convert log₈(x) to base 2:

log₈(x) = log₂(x) / log₂(8) = log₂(x) / 3

So the equation becomes:

log₂(x) = 3 + log₂(x)/3
log₂(x) - log₂(x)/3 = 3
(2/3) × log₂(x) = 3
log₂(x) = 9/2
x = 2^(9/2) = (√2)⁹ = 2⁴ × √2 = 16√2

Example 2: Solve log₃(x) + log₉(x) = 6

Convert log₉(x) to base 3:

log₉(x) = log₃(x) / log₃(9) = log₃(x) / 2

So:

log₃(x) + log₃(x)/2 = 6
(3/2) × log₃(x) = 6
log₃(x) = 4
x = 3⁴ = 81

Proving Logarithmic Identities

The change of base formula is also used to prove identities.

Example: Prove that log_a(b) × log_b(c) = log_a(c)

Using change of base (converting to base a):

log_a(b) × log_b(c) = log_a(b) × [log_a(c) / log_a(b)] = log_a(c)

The log_a(b) terms cancel, giving log_a(c). Proved.

Example: Prove that log_a(b) × log_b(c) × log_c(a) = 1

Using the result above:

log_a(b) × log_b(c) = log_a(c)
So log_a(b) × log_b(c) × log_c(a) = log_a(c) × log_c(a) = log_a(a) = 1

Converting Between ln and log₁₀

Using change of base:

ln(x) = log₁₀(x) / log₁₀(e) ≈ log₁₀(x) / 0.4343 ≈ 2.303 × log₁₀(x)
log₁₀(x) = ln(x) / ln(10) ≈ ln(x) / 2.303 ≈ 0.4343 × ln(x)

Exam Tip: You rarely need to convert between ln and log₁₀ explicitly, but knowing the relationship helps check your work. The key skill is converting unusual bases (like log₃ or log₅) to log₁₀ or ln for calculation.

Summary

The change of base formula: log_a(x) = log_b(x) / log_b(a).
Use it to evaluate logarithms on a calculator (convert to base 10 or base e).
The reciprocal relationship: log_a(b) = 1/log_b(a).
Use change of base to solve equations where different log bases appear.
The formula is essential for proving logarithmic identities.

A-Level Deep Dive: Change of Base

Spec mapping

Edexcel 9MA0-01 specification section 6 — Exponentials and logarithms, change-of-base sub-strand covers the laws of logarithms... including the change-of-base formula $\log_a x = \dfrac{\log_b x}{\log_b a}$ (refer to the official specification document for exact wording). Although nominally a single bullet point, change of base is the synoptic glue that makes section 6 work: it lets you collapse an equation with two different bases into one, evaluate a logarithm whose base is not on a calculator (calculators have only $\log_{10}$ and $\ln$ ), and translate between mathematical and computer-science conventions ( $\log_2$ in algorithm analysis, $\ln$ in calculus). It is examined directly in 9MA0-01 (in pure log equations) and indirectly in 9MA0-02 (in exponential models where rate constants must be re-expressed). The Edexcel formula booklet does list the change-of-base identity, so the M1 mark is awarded for selecting and applying it correctly — not for memorising it.

Synoptically, change of base connects to solving exponential equations (where mixed-base exponents force a base conversion), calculator computation (every base- $a$ logarithm on a non-base-10 calculator routes through $\ln$ ), computer science ( $\log_2 n$ governs binary-search and divide-and-conquer complexity), and information theory (entropy is naturally measured in bits via $\log_2$ but in nats via $\ln$ — translation is pure change of base).

Worked example with full mark scheme

Question (8 marks): Solve, giving $x$ as an exact value:

$\log_4 x + \log_2 x = 9$

Solution with mark scheme:

Step 1 — choose a common base and convert.

The two logarithms use bases 4 and 2. Since $4 = 2^2$ , base 2 is the natural common choice. Apply change of base to $\log_4 x$ :

$\log_4 x = \dfrac{\log_2 x}{\log_2 4} = \dfrac{\log_2 x}{2}$

M1 — selecting change-of-base formula and applying it with a sensible common base. Equally valid: convert $\log_2 x$ to base 4 via $\log_2 x = \log_4 x / \log_4 2 = 2\log_4 x$ . Either route earns M1.

A1 — correct evaluation $\log_2 4 = 2$ and clean substitution.

Step 2 — substitute and combine.

The equation becomes:

$\dfrac{\log_2 x}{2} + \log_2 x = 9$

Step 3 — clear the fraction.

Multiply through by 2:

$\log_2 x + 2\log_2 x = 18$

$3\log_2 x = 18$

M1 — combining the two terms in $\log_2 x$ with consistent algebra.

A1 — correct simplified equation $3\log_2 x = 18$ .

Step 4 — solve for $\log_2 x$ .

$\log_2 x = 6$

M1 — isolating the logarithmic term.

Step 5 — exponentiate.

$x = 2^6 = 64$

A1 — exact value $x = 64$ .

Step 6 — domain check.

For $\log_4 x$ and $\log_2 x$ to be defined we require $x > 0$ . Since $64 > 0$ , the solution is valid.

B1 — explicit domain check, awarded for stating the constraint and verifying the candidate solution satisfies it.

A1 — final answer $x = 64$ with full justification.

Total: 8 marks (M3 A4 B1).

Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Question (6 marks): Given that $\log_9 y = k$ , where $k$ is a constant:

(a) Express $\log_3 y$ in terms of $k$ . (2)

(b) Hence solve $\log_9 y + \log_3 y = 6$ , giving $y$ as an exact value. (4)

Mark scheme decomposition by AO:

(a)

M1 (AO1.1a) — applying change of base: $\log_3 y = \dfrac{\log_9 y}{\log_9 3} = \dfrac{k}{\log_9 3}$ , then recognising $\log_9 3 = \tfrac{1}{2}$ since $9^{1/2} = 3$ .
A1 (AO1.1b) — clean answer $\log_3 y = 2k$ .

(b)

M1 (AO1.1b) — substituting from (a): $k + 2k = 6$ .
A1 (AO1.1b) — solving $3k = 6$ to give $k = 2$ .
M1 (AO1.1b) — back-substituting: $\log_9 y = 2 \implies y = 9^2$ .
A1 (AO2.5) — exact value $y = 81$ , with brief domain justification ( $y > 0$ , satisfied).

Total: 6 marks split AO1 = 5, AO2 = 1. This is an AO1-dominated question — the AO2 mark rewards the candidate who notices that "exact value" forbids a decimal and that the back-substitution step deserves explicit working rather than a one-line numerical jump.

Synoptic links

Connects to:

Solving exponential equations (section 6): equations like $5^x = 3^{x+1}$ require taking a logarithm of some base on both sides. Choosing $\ln$ is conventional, but the resulting expression $x = \dfrac{\ln 3}{\ln 5 - \ln 3}$ is just $\log_{5/3} 3$ in disguise — change of base unifies these solution forms.
Calculator computation (section 6): to evaluate $\log_7 50$ on a calculator that only has $\ln$ and $\log_{10}$ , change of base gives $\log_7 50 = \dfrac{\ln 50}{\ln 7} \approx 2.011$ . Every "log to an unusual base" calculator question routes through this identity.
Computer science complexity ( $O(\log n)$ ): binary search runs in $\log_2 n$ steps; merge sort in $n \log_2 n$ . But in big-O notation, $O(\log_2 n) = O(\log_e n) = O(\log_{10} n)$ because the change-of-base formula shows they differ only by a constant factor $1/\log_b a$ . Change of base is why the base inside big-O is irrelevant.
Information theory (entropy): the Shannon entropy $H = -\sum p_i \log_2 p_i$ is measured in bits. Switching to natural log gives entropy in nats: $H_{\text{nats}} = H_{\text{bits}} \cdot \ln 2$ . The conversion factor is exactly the change-of-base reciprocal $\log_2 e = 1/\ln 2 \approx 1.4427$ .
Combinatorics and counting: the number of binary digits needed to represent $n$ is $\lceil \log_2(n+1) \rceil$ . Translating a counting argument expressed in decimal digits ( $\log_{10}$ ) to bits ( $\log_2$ ) — for example, "a 6-digit decimal number requires at most how many bits?" — is an immediate change-of-base application: $6 \log_2 10 \approx 19.93$ , so 20 bits.

Mark-scheme literacy

Change-of-base questions on 9MA0 split AO marks as follows:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	65–75%	Recalling the formula, selecting an appropriate common base, applying it correctly, simplifying $\log_b a$ where $a$ is a power of $b$
AO2 (reasoning / interpretation)	20–30%	Justifying the choice of base ("base 2 is natural since $4 = 2^2$ "), recognising when change of base is required (mixed-base equations) versus merely available, presenting answers in exact form
AO3 (problem-solving)	0–10%	Multi-stage modelling problems where change of base is one step inside a longer argument; rare in section 6 stand-alone but appears in synoptic Paper 2 questions

Examiner-rewarded phrasing: "by the change-of-base formula"; "choosing base 2 because $4 = 2^2$ converts cleanly"; "since $\log_a a = 1$ and $\log_a a^n = n$ ". Phrases that lose marks: "let $\log_4 x = y$ " without re-establishing the relationship to $\log_2 x$ ; quoting the formula but using it with numerator and denominator inverted; treating $\log_b a$ and $\log_a b$ as equal (they are reciprocals: $\log_b a = 1/\log_a b$ ).

A specific Edexcel pattern to watch: when the question states "exact value", the candidate must avoid evaluating $\ln 2$ or $\ln 4$ as decimals at any stage — keep symbolic logarithms intact and only evaluate when the final cancellation makes the answer rational.

Grade-band model answers

3-mark question

Question: Use change of base to evaluate $\log_5 12$ to 3 significant figures.

Grade C response (~210 words):

Using the change-of-base formula:

$\log_5 12 = \dfrac{\ln 12}{\ln 5}$

On a calculator: $\ln 12 \approx 2.4849$ and $\ln 5 \approx 1.6094$ .

So $\log_5 12 \approx \dfrac{2.4849}{1.6094} \approx 1.5439$ .

To 3 significant figures, $\log_5 12 \approx 1.54$ .

Examiner commentary: Full marks (3/3). The candidate correctly applies the change-of-base formula with natural logarithms, reads two intermediate values to 4 decimal places (one more than the requested final precision, which is the right working margin), and rounds correctly to 3sf. The choice of $\ln$ over $\log_{10}$ is incidental — both give the same final value because the two scaling factors cancel inside the ratio. A subtle point worth noting: had the candidate read $\ln 12$ and $\ln 5$ to only 3sf each, accumulated rounding error could have shifted the final answer to 1.53 or 1.55, costing the A1.

Grade A response (~250 words):*

By the change-of-base formula:

$\log_5 12 = \dfrac{\ln 12}{\ln 5}$

(I use natural logarithms for calculator compatibility; $\log_{10}$ would give the same final value because the constant factor $\ln 10$ cancels in the ratio.)

Computing each value to 4 decimal places to retain accuracy through rounding:

$\ln 12 = 2.4849$
$\ln 5 = 1.6094$

Therefore:

$\log_5 12 = \dfrac{2.4849}{1.6094} = 1.5440\ldots$

To 3 significant figures, $\log_5 12 \approx 1.54$ .

Sense check: $5^{1.54} \approx 5 \cdot 5^{0.54} \approx 5 \cdot 2.41 \approx 12.05$ , confirming the answer is consistent with the original equation $5^x = 12$ .

Examiner commentary: Full marks (3/3). Beyond procedural correctness, the candidate (i) names the change-of-base formula explicitly, (ii) justifies the choice of $\ln$ over $\log_{10}$ , (iii) carries one extra significant figure through intermediate working before rounding, and (iv) performs a back-substitution sense-check. The sense-check is the marker of A* discipline: it costs ten seconds and catches both calculator-input errors and conceptual misreads (e.g. computing $\log_{12} 5$ by mistake). Examiner reports repeatedly highlight this habit as separating top-band candidates from the merely competent.

6-mark question

Question: Solve $\log_2 x = \log_4(x + 6)$ , giving $x$ as an exact value.

Grade B response (~285 words):

Use change of base to convert $\log_4(x+6)$ to base 2:

$\log_4(x+6) = \dfrac{\log_2(x+6)}{\log_2 4} = \dfrac{\log_2(x+6)}{2}$ .

So the equation becomes:

$\log_2 x = \dfrac{\log_2(x+6)}{2}$

Multiply both sides by 2:

$2 \log_2 x = \log_2(x+6)$

Change of Base

Change of Base

The Change of Base Formula

Proof

Using the Formula with a Calculator

The Reciprocal Relationship

Solving Equations Using Change of Base

Proving Logarithmic Identities

Converting Between ln and log₁₀

Summary

A-Level Deep Dive: Change of Base

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Synoptic links

Mark-scheme literacy

Grade-band model answers

3-mark question

6-mark question

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