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Laws of Logarithms

Q: Can I take the log of a sum, like (x+y)?

There is no law that splits _a(x+y) into separate logs — this is a common and serious error. The laws only apply to products, quotients and powers.

Because a logarithm is just an exponent, every index law has a matching logarithm law — multiplying numbers becomes ADDING their logs, and raising to a power becomes MULTIPLYING by the log. These three laws are what let you solve $a^{x} = b$ for $x$ exactly.

30 min Video by TLMaths Exponentials and Logarithms

The big picture

Logarithms were introduced as "the exponent" — but their real power lies in three laws that mirror the index laws exactly. Every messy product, quotient or power inside a logarithm can be dismantled into simple sums, differences and multiples of logarithms. Combined with taking logs of both sides of an equation, this is the single technique that solves ANY exponential equation, not just the ones with a nice integer answer.

A-Level Maths F4-01: Introducing the Laws of LogarithmsWatch the walkthrough, then work through the notes below.

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What you'll be able to do

State and apply the addition, subtraction and power laws of logarithms
Combine multiple logarithms into a single logarithm, and vice versa
Solve exponential equations by taking logarithms of both sides
Use the change-of-base formula

The three laws

Each law of logarithms mirrors an index law, because $lo g_{a}$ undoes $a^{(\cdot)}$ . Multiplying inside becomes adding outside; dividing inside becomes subtracting outside; a power inside becomes a multiplier outside.

lo g_{a} (x y) = lo g_{a} x + lo g_{a} y

Product law — multiplication becomes addition.

lo g_{a} (\frac{x}{y}) = lo g_{a} x - lo g_{a} y

Quotient law — division becomes subtraction.

lo g_{a} (x^{k}) = k lo g_{a} x

Power law — an exponent becomes a multiplier.

12 = 3 \times 4

, so use the product law.

lo g_{2} 12 = lo g_{2} 3 + lo g_{2} 4

Answer

lo g_{2} 12 = lo g_{2} 3 + lo g_{2} 4

Tip — Two extra useful facts follow directly: $lo g_{a} a = 1$ (since $a^{1} = a$ ) and $lo g_{a} 1 = 0$ (since $a^{0} = 1$ ) — memorise both, they appear constantly.

Solving exponential equations

When an unknown is trapped in an exponent and cannot be matched to a nice power, take logarithms of BOTH sides. The power law then brings the exponent down as a multiplier, turning the equation into ordinary algebra.

1Take

lo g

(or

ln

) of both sides:

lo g (3^{x}) = lo g (20)

2Apply the power law:

x lo g 3 = lo g 20

3Divide:

x = \frac{lo g 20}{lo g 3} = \frac{1.301}{0.477}

Answer

x \approx 2.73

Why this works

It does not matter which base you choose for the logarithm (log or ln) — the ratio $\frac{lo g 20}{lo g 3}$ gives the same value whichever consistent base you use throughout, since the base cancels out.

Change of base

Sometimes you need a logarithm in a base your calculator does not have a button for. The $change-of-base formula$ rewrites any logarithm using a base you can compute (usually 10 or $e$ ).

lo g_{a} b = \frac{lo g _{c} b}{lo g _{c} a}

Rewrite in any convenient base

c

lo g_{5} 30 = \frac{lo g 30}{lo g 5} = \frac{1.477}{0.699}

Answer

\approx 2.11

Think like an examiner

Watch for the classic trap: $lo g_{a} (x + y) \neq = lo g_{a} x + lo g_{a} y$ — there is NO law for the log of a sum, only for a product.
When solving $a^{x} = b$ , take logs of BOTH sides first, then bring the exponent down using the power law.
For change of base, any consistent base works — pick base 10 or $e$ since your calculator has both.
Use $lo g_{a} a = 1$ and $lo g_{a} 1 = 0$ to simplify expressions quickly.

Common misconceptions

Writing

lo g_{a} (x + y) = lo g_{a} x + lo g_{a} y

There is no law for a SUM inside a log — only for a PRODUCT.

lo g_{a} (x y) = lo g_{a} x + lo g_{a} y

, but addition inside has no simplification.

Forgetting to bring the exponent down when solving

a^{x} = b

After taking logs, use the power law:

lo g (a^{x}) = x lo g a

, then divide to isolate

x

Thinking

(lo g_{a} x)^{2} = lo g_{a} (x^{2})

These are different:

lo g_{a} (x^{2}) = 2 lo g_{a} x

(power law), but squaring the whole logarithm is a separate operation entirely.

Laws of logarithms

lo g_{a} (x y) = lo g_{a} x + lo g_{a} y

Product law.

lo g_{a} (\frac{x}{y}) = lo g_{a} x - lo g_{a} y

Quotient law.

lo g_{a} (x^{k}) = k lo g_{a} x

Power law.

lo g_{a} b = \frac{lo g _{c} b}{lo g _{c} a}

Change of base.

Stretch yourself

Solve $5^{2 x + 1} = 3^{x}$ , giving your answer to 3 significant figures.

Hint — Take logs of both sides first, then use the power law on each side separately before collecting the $x$ terms.

Questions students ask

Why does multiplying inside a log become adding outside?

Because logs are exponents, and multiplying two powers of the same base ADDS their exponents (

a^{m} \times a^{n} = a^{m + n}

) — the product law of logs is exactly this fact rewritten.

Can I take the log of a sum, like

lo g (x + y)

There is no law that splits

lo g_{a} (x + y)

into separate logs — this is a common and serious error. The laws only apply to products, quotients and powers.

How do I solve an equation like

2^{x} = 7

Take logarithms of both sides, then use the power law to bring the exponent

x

down as a multiplier:

x lo g 2 = lo g 7

, so

x = \frac{lo g 7}{lo g 2}

Key takeaways

Product law: $lo g_{a} (x y) = lo g_{a} x + lo g_{a} y$ .
Quotient law: $lo g_{a} (x / y) = lo g_{a} x - lo g_{a} y$ .
Power law: $lo g_{a} (x^{k}) = k lo g_{a} x$ .
Solve $a^{x} = b$ by taking logs of both sides and using the power law.
Change of base: $lo g_{a} b = \frac{lo g _{c} b}{lo g _{c} a}$ .

How this fits the course

Build on

Logarithms

Index Laws

Leads to

Exponential Models and Log-Linear Data

Test yourself

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Laws of Logarithms

a^{x} = b

for

x

exactly.

30 min Video by TLMaths Exponentials and Logarithms