Change of Base Formula: Definition, Proof, Examples and Practice

The change of base formula is one of the most practically useful tools in logarithm algebra. Calculators only have buttons for log (base 10) and ln (base e) — but logarithms appear with all kinds of bases in mathematics. The formula converts any logarithm into one that a calculator can evaluate, and it is used throughout the logarithm units of MCR3U and MHF4U as well as in competition problems where logarithms appear in unexpected bases.

This guide explains the change of base formula precisely, proves it, works through examples at every level of difficulty, and shows where it appears in the Ontario curriculum and Canadian competitions.

What Is the Change of Base Formula?

For any positive numbers $a$ a and $x$ x with $a \neq 1$ a=1, and any valid base $c$ c with $c > 0$ c>0, $c \neq 1$ c=1: $\log_a(x) = \frac{\log_c(x)}{\log_c(a)}$ loga(x)=logc(a)logc(x)

The most common forms in practice: $\log_a(x) = \frac{\log(x)}{\log(a)} = \frac{\ln(x)}{\ln(a)}$ loga(x)=log(a)log(x)=ln(a)ln(x)

where $\log$ log denotes $\log_{10}$ log10 (common logarithm) and $\ln$ ln denotes $\log_e$ loge (natural logarithm).

In plain English: To evaluate $\log_a(x)$ loga(x) using a standard calculator, divide $\log(x)$ log(x) by $\log(a)$ log(a) (or $\ln(x)$ ln(x) by $\ln(a)$ ln(a) — both give the same result).

A first example:

$\log_2(8)$ log2(8): we know $2^3 = 8$ 23=8, so $\log_2(8) = 3$ log2(8)=3.

Using the change of base formula: $\log_2(8) = \frac{\log(8)}{\log(2)} = \frac{0.90309…}{0.30103…} = 3$ log2(8)=log(2)log(8)=0.30103…0.90309…=3. ✓

The formula gives the exact same answer — it is not an approximation, it is an identity.

Proof

The proof follows directly from the definition of logarithm and basic exponent properties.

Claim: $\log_a(x) = \dfrac{\log_c(x)}{\log_c(a)}$ loga(x)=logc(a)logc(x)

Proof:

Let $y = \log_a(x)$ y=loga(x). By the definition of logarithm, this means: $a^y = x$ ay=x

Apply $\log_c$ logc to both sides: $\log_c(a^y) = \log_c(x)$ logc(ay)=logc(x)

By the power law of logarithms ( $\log_c(a^y) = y \log_c(a)$ logc(ay)=ylogc(a)): $y \log_c(a) = \log_c(x)$ ylogc(a)=logc(x)

Divide both sides by $\log_c(a)$ logc(a) (valid since $a \neq 1$ a=1 means $\log_c(a) \neq 0$ logc(a)=0): $y = \frac{\log_c(x)}{\log_c(a)}$ y=logc(a)logc(x)

Substituting back $y = \log_a(x)$ y=loga(x): $\log_a(x) = \frac{\log_c(x)}{\log_c(a)} \quad \square$ loga(x)=logc(a)logc(x)□

Why the proof matters for students:

The proof uses only the definition of logarithm and the power law — both of which students learn in MCR3U Unit 4. Knowing the proof means the formula never needs to be memorised separately — it can be re-derived in under 30 seconds. This is the standard of understanding that MCR3U and MHF4U reward.

The Logarithm Laws: Context

The change of base formula works alongside the standard logarithm laws. For completeness:

Law	Formula	Example
Product	$\log_b(MN) = \log_b M + \log_b N$ logb(MN)=logbM+logbN	$\log_2(8 \times 4) = \log_2 8 + \log_2 4 = 3 + 2 = 5$ log2(8×4)=log28+log24=3+2=5
Quotient	$\log_b(M/N) = \log_b M – \log_b N$ logb(M/N)=logbM−logbN	$\log_3(81/9) = \log_3 81 – \log_3 9 = 4 – 2 = 2$ log3(81/9)=log381−log39=4−2=2
Power	$\log_b(M^p) = p\log_b M$ logb(Mp)=plogbM	$\log_2(8^2) = 2\log_2 8 = 6$ log2(82)=2log28=6
Change of base	$\log_a(x) = \log_c(x)/\log_c(a)$ loga(x)=logc(x)/logc(a)	$\log_4(32) = \log(32)/\log(4)$ log4(32)=log(32)/log(4)
Special values	$\log_b(b) = 1$ logb(b)=1; $\log_b(1) = 0$ logb(1)=0	$\log_5(5) = 1$ log5(5)=1; $\log_7(1) = 0$ log7(1)=0

The change of base formula is distinguished from the others because it changes the base of the logarithm itself — not just the argument. This makes it the bridge between different bases, and the tool that makes any logarithm calculator-evaluable.

Change of Base Formula: Worked Examples

Example 1 — Direct Calculator Evaluation (MCR3U Level)

Evaluate log⁡3(50)\log_3(50) log3(50) to 4 decimal places. $\log_3(50) = \frac{\log(50)}{\log(3)} = \frac{1.69897…}{0.47712…} = 3.5609…$ log3(50)=log(3)log(50)=0.47712…1.69897…=3.5609…

Answer: log⁡3(50)≈3.5609\log_3(50) \approx 3.5609 log3(50)≈3.5609

Check: $3^{3.5609} \approx 50$ 33.5609≈50. ✓

Example 2 — Exact Values Using the Formula

Evaluate log⁡4(32)\log_4(32) log4(32) without a calculator. $\log_4(32) = \frac{\log(32)}{\log(4)} = \frac{\log(2^5)}{\log(2^2)} = \frac{5\log 2}{2\log 2} = \frac{5}{2}$ log4(32)=log(4)log(32)=log(22)log(25)=2log25log2=25

Answer: log⁡4(32)=52\log_4(32) = \frac{5}{2} log4(32)=25

This works because both 32 and 4 are powers of 2 — the change of base formula with base 10 (or any base) reduces to a ratio of powers of 2.

Key insight: When both the argument and the base can be expressed as powers of the same number, the change of base formula gives an exact rational answer.

Example 3 — Expressing in Terms of a Given Logarithm

If log⁡3(2)=k\log_3(2) = k log3(2)=k, express log⁡9(8)\log_9(8) log9(8) in terms of kk k. $\log_9(8) = \frac{\log(8)}{\log(9)} = \frac{\log(2^3)}{\log(3^2)} = \frac{3\log 2}{2\log 3}$ log9(8)=log(9)log(8)=log(32)log(23)=2log33log2

Now use the change of base formula: $\log_3(2) = \frac{\log 2}{\log 3} = k$ log3(2)=log3log2=k, so $\frac{\log 2}{\log 3} = k$ log3log2=k. $\log_9(8) = \frac{3\log 2}{2\log 3} = \frac{3}{2} \cdot \frac{\log 2}{\log 3} = \frac{3k}{2}$ log9(8)=2log33log2=23⋅log3log2=23k

Answer: log⁡9(8)=3k2\log_9(8) = \dfrac{3k}{2} log9(8)=23k

This problem type — expressing a logarithm in terms of a given value — is standard in MCR3U and MHF4U assessments.

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Example 4 — Solving a Logarithmic Equation

Solve log⁡x(81)=4\log_x(81) = 4 logx(81)=4.

By definition: $x^4 = 81 = 3^4$ x4=81=34, so $x = 3$ x=3.

Using the change of base formula as a check:

$\log_x(81) = \frac{\log(81)}{\log(x)} = 4 \implies \log(x) = \frac{\log(81)}{4} = \frac{4\log 3}{4} = \log 3 \implies x = 3$ logx(81)=log(x)log(81)=4⟹log(x)=4log(81)=44log3=log3⟹x=3. ✓

Example 5 — Comparing Logarithms (MHF4U Level)

Without a calculator, determine which is larger: log⁡3(7)\log_3(7) log3(7) or log⁡5(11)\log_5(11) log5(11).

Both expressions are between 1 and 2 (since $3^1 = 3 < 7 < 9 = 3^2$ 31=3<7<9=32 and $5^1 = 5 < 11 < 25 = 5^2$ 51=5<11<25=52).

To compare precisely, use the change of base formula with base 10: $\log_3(7) = \frac{\log 7}{\log 3} \approx \frac{0.8451}{0.4771} \approx 1.7712$ log3(7)=log3log7≈0.47710.8451≈1.7712 $\log_5(11) = \frac{\log 11}{\log 5} \approx \frac{1.0414}{0.6990} \approx 1.4899$ log5(11)=log5log11≈0.69901.0414≈1.4899

log⁡3(7)>log⁡5(11)\log_3(7) > \log_5(11) log3(7)>log5(11).

Example 6 — Simplifying an Expression (Competition Level)

Simplify: 1log⁡26+1log⁡36\dfrac{1}{\log_2 6} + \dfrac{1}{\log_3 6} log261+log361

Using the change of base formula: $\dfrac{1}{\log_a 6} = \log_6 a$ loga61=log6a (reciprocal property). $\frac{1}{\log_2 6} + \frac{1}{\log_3 6} = \log_6 2 + \log_6 3 = \log_6(2 \times 3) = \log_6 6 = 1$ log261+log361=log62+log63=log6(2×3)=log66=1

Answer: 1

This problem combines the change of base formula with the product rule and is the kind of elegant simplification that appears in Euclid Part A and AMC 10 problems.

Example 7 — The Reciprocal Property

A direct consequence of the change of base formula: $\log_a(b) = \frac{1}{\log_b(a)}$ loga(b)=logb(a)1

Proof: $\log_a(b) = \dfrac{\log b}{\log a}$ loga(b)=logalogb and $\log_b(a) = \dfrac{\log a}{\log b}$ logb(a)=logbloga. These are reciprocals. $\square$ □

This property is frequently used in competition problems — whenever a sum of reciprocals of logarithms appears, the change of base formula converts it into a sum of logarithms with a common base.

Natural Logarithms

The change of base formula also explains the relationship between $\log$ log (base 10) and $\ln$ ln (base $e$ e): $\ln(x) = \frac{\log(x)}{\log(e)} = \frac{\log(x)}{0.43429…}$ ln(x)=log(e)log(x)=0.43429…log(x)

Or equivalently: $\log(x) = \frac{\ln(x)}{\ln(10)} = \frac{\ln(x)}{2.30259…}$ log(x)=ln(10)ln(x)=2.30259…ln(x)

These conversion factors appear in applied mathematics and engineering — but for pure mathematics competitions, the formula is most useful for converting between arbitrary bases, not specifically between base 10 and base $e$ e.

Where the Formula Appears in Contests and Curriculum

MCR3U (Ontario Grade 11, Unit 4: Logarithmic Functions): The change of base formula is explicitly taught as part of the logarithm unit and is tested on unit assessments and final exams. The most common problem types: evaluating non-standard bases on a calculator (Example 1), finding exact values when both argument and base share a common factor (Example 2), and expressing logarithms in terms of given values (Example 3). See our MCR3U complete guide.

MHF4U (Ontario Grade 12): MHF4U extends logarithm work with more complex expressions, logarithmic equations requiring the change of base formula, and applications in exponential modelling. The reciprocal property (Example 7) and the comparison of logarithms in different bases (Example 5) are MHF4U-level applications. See our MHF4U Advanced Functions guide.

Math 30-1 (Alberta Grade 12): The change of base formula is part of the Math 30-1 logarithm and exponential functions unit and is explicitly tested on the Alberta Diploma Examination. See our Math 30-1 complete guide and Math 30-1 formula sheet.

Euclid Contest and AMC 10/12:The change of base formula enables elegant solutions to logarithm simplification problems (like Example 6) that would be slow or impossible by direct computation. The reciprocal property in particular — $\frac{1}{\log_a b} = \log_b a$ logab1=logba — appears frequently in competition logarithm problems as a simplification tool. See our Euclid math contest guide.

For the full competition landscape, see our math competitions in Canada guide.

Logarithms appear throughout MCR3U, MHF4U and Math 30-1. A free online assessment identifies whether your child can evaluate, simplify and solve logarithmic expressions independently—before gaps affect their exam performance.

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Common Mistakes with the Formula

Mistake 1: Inverting the fraction. $\log_a(x) = \dfrac{\log x}{\log a}$ loga(x)=logalogx, not $\dfrac{\log a}{\log x}$ logxloga. The argument ( $x$ x) goes in the numerator; the base ( $a$ a) goes in the denominator. A useful mnemonic: “argument over base” — the thing you are taking the log of (the argument) over the base you are converting from.

Mistake 2: Using different bases in numerator and denominator. $\log_a(x) = \dfrac{\log_c(x)}{\log_c(a)}$ loga(x)=logc(a)logc(x) requires the same base $c$ c in both the numerator and denominator. Mixing $\log_{10}$ log10 in the numerator and $\log_e$ loge in the denominator does not give $\log_a(x)$ loga(x) — it gives a different quantity entirely.

Mistake 3: Applying the formula when an exact value is available. $\log_4(16)$ log4(16) can be evaluated exactly as 2 (since $4^2 = 16$ 42=16) without the change of base formula. Reaching for the formula first — and getting a decimal approximation — when an exact integer answer is available is wasteful and produces rounding errors.

Mistake 4: Forgetting the domain restriction.The formula requires $x > 0$ x>0 and $a > 0$ a>0, $a \neq 1$ a=1 (and $c > 0$ c>0, $c \neq 1$ c=1). When solving equations using the formula, check that solutions satisfy these conditions.

Mistake 5: Misapplying to sums of logarithms. $\log_a(x) + \log_a(y) \neq \dfrac{\log x + \log y}{\log a + \log a}$ loga(x)+loga(y)=loga+logalogx+logy. The change of base formula applies to each individual logarithm separately: $\log_a(x) + \log_a(y) = \dfrac{\log x}{\log a} + \dfrac{\log y}{\log a}$ loga(x)+loga(y)=logalogx+logalogy. Or more usefully: use the product law first — $\log_a(x) + \log_a(y) = \log_a(xy)$ loga(x)+loga(y)=loga(xy) — then change base if needed.

Practice Problems

Set A — Direct evaluation

Use the formula to evaluate each (to 4 decimal places where appropriate):

$\log_2(50)$ log2(50)
$\log_3(20)$ log3(20)
$\log_5(100)$ log5(100)
$\log_7(200)$ log7(200)
$\log_4(64)$ log4(64) (exact value)
$\log_9(27)$ log9(27) (exact value)

Set B — Exact values and expressions

If $\log_2(3) = p$ log2(3)=p, express $\log_4(9)$ log4(9) in terms of $p$ p.
If $\log_5(2) = k$ log5(2)=k, express $\log_{25}(8)$ log25(8) in terms of $k$ k.
Simplify: $\dfrac{1}{\log_3 12} + \dfrac{1}{\log_4 12}$ log3121+log4121
Simplify: $\log_2(3) \cdot \log_3(4) \cdot \log_4(8)$ log2(3)⋅log3(4)⋅log4(8)

Set C — Equations and inequalities

Solve: $\log_x(125) = 3$ logx(125)=3
Solve: $\log_2(x) = \log_4(x+6)$ log2(x)=log4(x+6)
Solve: $\log_3(x) + \log_9(x) = 4$ log3(x)+log9(x)=4
For which values of $x$ x is $\log_2(x) > \log_3(x)$ log2(x)>log3(x)?

Answers:

Set A:

$\log(50)/\log(2) \approx 1.6990/0.3010 \approx$ log(50)/log(2)≈1.6990/0.3010≈ 5.6439
$\log(20)/\log(3) \approx 1.3010/0.4771 \approx$ log(20)/log(3)≈1.3010/0.4771≈ 2.7268
$\log(100)/\log(5) = 2/0.6990 \approx$ log(100)/log(5)=2/0.6990≈ 2.8614
$\log(200)/\log(7) \approx 2.3010/0.8451 \approx$ log(200)/log(7)≈2.3010/0.8451≈ 2.7227
$\log_4(64) = \log(2^6)/\log(2^2) = 6\log 2/2\log 2 =$ log4(64)=log(26)/log(22)=6log2/2log2= 3 (exact)
$\log_9(27) = \log(3^3)/\log(3^2) = 3\log 3/2\log 3 =$ log9(27)=log(33)/log(32)=3log3/2log3= 3/2 (exact)

Set B: 7) $\log_4(9) = \log(3^2)/\log(2^2) = 2\log 3/2\log 2 = \log_2(3) =$ log4(9)=log(32)/log(22)=2log3/2log2=log2(3)= p8) $\log_{25}(8) = \log(2^3)/\log(5^2) = 3\log 2/2\log 5 = (3/2)(\log 2/\log 5) = (3/2)\log_5(2) =$ log25(8)=log(23)/log(52)=3log2/2log5=(3/2)(log2/log5)=(3/2)log5(2)= 3k/29) $\frac{1}{\log_3 12} + \frac{1}{\log_4 12} = \log_{12}3 + \log_{12}4 = \log_{12}(3 \times 4) = \log_{12}12 =$ log3121+log4121=log123+log124=log12(3×4)=log1212= 110) $\log_2(3) \cdot \log_3(4) \cdot \log_4(8) = \frac{\log 3}{\log 2} \cdot \frac{\log 4}{\log 3} \cdot \frac{\log 8}{\log 4} = \frac{\log 8}{\log 2} = \frac{3\log 2}{\log 2} =$ log2(3)⋅log3(4)⋅log4(8)=log2log3⋅log3log4⋅log4log8=log2log8=log23log2= 3

Set C: 11) $x^3 = 125 = 5^3$ x3=125=53, so $x =$ x= 512) Change base: $\log_4(x+6) = \frac{\log_2(x+6)}{\log_2 4} = \frac{\log_2(x+6)}{2}$ log4(x+6)=log24log2(x+6)=2log2(x+6). So $\log_2 x = \frac{\log_2(x+6)}{2}$ log2x=2log2(x+6). Let $u = \log_2 x$ u=log2x: $2u = \log_2(x+6)$ 2u=log2(x+6), so $x^2 = x + 6$ x2=x+6, giving $x^2 – x – 6 = 0$ x2−x−6=0, $(x-3)(x+2) = 0$ (x−3)(x+2)=0. Check: $x > 0$ x>0 and $x + 6 > 0$ x+6>0, so $x =$ x= 313) $\log_9(x) = \frac{\log_3 x}{2}$ log9(x)=2log3x. So $\log_3 x + \frac{\log_3 x}{2} = 4$ log3x+2log3x=4, giving $\frac{3}{2}\log_3 x = 4$ 23log3x=4, $\log_3 x = 8/3$ log3x=8/3, $x = 3^{8/3} =$ x=38/3= 383=65613\sqrt[3]{3^8} = \sqrt[3]{6561} 338=3656114) $\log_2(x) > \log_3(x)$ log2(x)>log3(x): for $x > 1$ x>1, both are positive; $\frac{\log x}{\log 2} > \frac{\log x}{\log 3}$ log2logx>log3logx iff $\frac{1}{\log 2} > \frac{1}{\log 3}$ log21>log31 iff $\log 3 > \log 2$ log3>log2 ✓. For $0 < x < 1$ 0<x<1, both are negative; inequality reverses. Therefore: $\log_2(x) > \log_3(x)$ log2(x)>log3(x) when x>1x > 1 x>1.

Frequently Asked Questions

What is the change of base formula?

loga(x)=logc(a)logc(x) for any valid base $c$ c. Most commonly used with $c = 10$ c=10 (common log) or $c = e$ c=e (natural log): $\log_a(x) = \dfrac{\log x}{\log a} = \dfrac{\ln x}{\ln a}$ loga(x)=logalogx=lnalnx.

Why do we need the change of base formula?

Calculators only compute $\log_{10}$ log10 and $\ln$ ln directly. The change of base formula converts any other base into one of these, allowing calculation. It is also used algebraically to simplify expressions and compare logarithms with different bases.

Does it matter which base you use?

No — as long as you use the same base in both numerator and denominator, the result is the same. $\dfrac{\log x}{\log a} = \dfrac{\ln x}{\ln a} = \dfrac{\log_2 x}{\log_2 a}$ logalogx=lnalnx=log2alog2x all give identical values.

What is the reciprocal property of logarithms?

loga(b)=logb(a)1. This follows directly from the change of base formula and is frequently useful in competition mathematics.

Is the change of base formula on the Math 30-1 formula sheet?

No — the Alberta Diploma Examination does not provide a formula sheet. All formulas including the change of base formula must be memorised. See our Math 30-1 formula sheet guide.

How do I remember the change of base formula?

“Log of the argument over log of the base” — $\log_a(x) = \frac{\log x}{\log a}$ loga(x)=logalogx. The argument ( $x$ x, the thing inside) goes on top; the base ( $a$ a, the subscript) goes on the bottom.

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Change of Base Formula Explained: Definition, Proof and Examples