Rules of Logarithms

This article presents the rules of logarithms using complete, line-by-line derivations. Every identity is built directly from its exponential origin, without shortcuts, matching the structure of formal handwritten algebra.

1. Definition

We begin with fundamental exponent facts:

a⁰ = 1 ⇒ logₐ(1) = 0
a¹ = a ⇒ logₐ(a) = 1

Say:

aᵐ = p

Then, by definition:

logₐ(p) = m

Raise both sides of aᵐ = p to the power 1/m (with m ≠ 0):

p^(1/m) = a

Therefore:

logₚ(a) = 1/m

Since m = logₐ(p), we obtain:

logₐ(p) = 1 / logₚ(a)

2. Product Rule — Full Derivation

Say:

aᵐ = p and aⁿ = q

Multiply:

aᵐ · aⁿ = p · q

Using index addition:

a^(m+n) = p · q

Taking logarithms:

logₐ(p · q) = m + n

Substitute:

logₐ(p · q) = logₐ(p) + logₐ(q)

3. Quotient Rule — Full Derivation

Say:

aᵐ = p and aⁿ = q

Divide:

aᵐ / aⁿ = p / q

Index subtraction gives:

a^(m−n) = p / q

Taking logarithms:

logₐ(p / q) = m − n

So:

logₐ(p / q) = logₐ(p) − logₐ(q)

4. Power Rule — Full Derivation

Say:

aᵇ = c

Raise both sides to the power d:

(aᵇ)ᵈ = cᵈ

Index multiplication gives:

a^(b·d) = cᵈ

Taking logarithms:

logₐ(cᵈ) = b·d

From aᵇ = c we know b = logₐ(c).

Substitute:

logₐ(cᵈ) = d · logₐ(c)

logₐ(cᵈ) = d · logₐ(c)

5. Change of Base — Full Derivation

Say:

aᵇ = c

Take logarithms base d:

log_d(aᵇ) = log_d(c)

Apply the power rule:

b · log_d(a) = log_d(c)

Solve for b:

b = log_d(c) / log_d(a)

But from the original equation b = logₐ(c).

Equating:

logₐ(c) = log_d(c) / log_d(a)

6. Example — Change of Base in Full Detail

Start with:

2³ = 8

Thus:

log₂(8) = 3

Convert using base-10 logs:

log₁₀(2³) = log₁₀(8)

Apply the power rule:

3 · log₁₀(2) = log₁₀(8)

Solve for the exponent:

3 = log₁₀(8) / log₁₀(2)

log₂(8) = log₁₀(8) / log₁₀(2)

@mathematics.proofs

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