DWG NO. 3.4 — Lesson 4 of 6

Properties of Logarithms

Unit 3: Exponential and Logarithmic Functions · ~20 min

Objective Expand and condense logarithmic expressions using the product, quotient, and power rules, and evaluate any-base logarithms with the change-of-base formula.

Because a logarithm is an exponent (lo\(g_{b}\)(x) asks "what power of b gives x?"), the properties of logarithms are really the laws of exponents from Algebra II, translated into log language. Each rule below turns a hard computation with logs into an easier one with arithmetic.

The three core rules

Product rule: lo\(g_{b}\)(MN) = lo\(g_{b}\)(M) + lo\(g_{b}\)(N) — the log of a product is the sum of the logs.
Quotient rule: lo\(g_{b}\)(\(\frac{M}{N}\)) = lo\(g_{b}\)(M) − lo\(g_{b}\)(N) — the log of a quotient is the difference of the logs.
Power rule: lo\(g_{b}\)(\(M^{p}\)) = p·lo\(g_{b}\)(M) — an exponent inside a log becomes a multiplier out front.

All three require M and N to be positive, since logs of non-positive numbers are undefined. Two extra identities are worth memorizing alongside them: lo\(g_{b}\)(1) = 0 and lo\(g_{b}\)(b) = 1 — both follow directly from b⁰ = 1 and b¹ = b.

Change of base

Calculators only compute log (base 10) and ln (base e) directly. To evaluate a logarithm in any other base, rewrite it as a ratio of two logs in a base your calculator knows:

lo\(g_{b}\)(M) = \(\frac{log(M)}{log(b)}\) = \(\frac{ln(M)}{ln(b)}\)

y = log₂(4x) = log₂(4) + log₂x = log₂x + 2 — the product rule turns a factor inside the log into a vertical shift

Worked Example 1 · Expanding a logarithm

ProblemExpand lo\(g_{b}\)(\(\frac{x^{3}y}{z}\)) fully.

1Quotient rule first: lo\(g_{b}\)(x³y) − lo\(g_{b}\)(z).

2Product rule on the first term: lo\(g_{b}\)(x³) + lo\(g_{b}\)(y) − lo\(g_{b}\)(z).

3Power rule on lo\(g_{b}\)(x³): 3·lo\(g_{b}\)(x).

3·lo\(g_{b}\)(x) + lo\(g_{b}\)(y) − lo\(g_{b}\)(z)

Worked Example 2 · Condensing a logarithm

ProblemCondense 2·ln(x) + ln(y) − 3·ln(z) into a single logarithm.

1Power rule in reverse: 2·ln(x) = ln(x²) and 3·ln(z) = ln(z³).

2Now combine: ln(x²) + ln(y) = ln(x²y) by the product rule; then ln(x²y) − ln(z³) = ln(\(\frac{x^{2}y}{z^{3}}\)) by the quotient rule.

ln(\(\frac{x^{2}y}{z^{3}}\))

Worked Example 3 · Change of base

ProblemEvaluate lo\(g_{7}\)(50) to three decimal places.

1Rewrite with common logs: lo\(g_{7}\)(50) = \(\frac{log(50)}{log(7)}\).

2log(50) ≈ 1.699, log(7) ≈ 0.845. Divide: 1.699 ÷ 0.845 ≈ 2.011.

lo\(g_{7}\)(50) ≈ 2.011

Guided practice

1.Expand log(\(\frac{a}{b^{2}}\)) fully.
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Quotient rule: log(a) − log(b²); power rule on the second term: log(a) − 2·log(b).
2.Condense 3·log(x) − log(y) into a single logarithm.
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Power rule: 3·log(x) = log(x³). Quotient rule: log(\(\frac{x^{3}}{y}\)).
3.Use lo\(g_{b}\)(2) ≈ 0.301 and lo\(g_{b}\)(3) ≈ 0.477 (both base 10) to find log(6).
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6 = 2 · 3, so by the product rule, log(6) = log(2) + log(3) ≈ 0.301 + 0.477 = 0.778.
4.Evaluate lo\(g_{3}\)(40) to three decimal places using change of base.
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lo\(g_{3}\)(40) = \(\frac{log(40)}{log(3)}\) ≈ \(\frac{1.602}{0.477}\) ≈ 3.358.
5.Explain why lo\(g_{b}\)(M) + lo\(g_{b}\)(N) cannot simply be rewritten as lo\(g_{b}\)(M + N).
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The product rule combines logs of a product, lo\(g_{b}\)(MN), not a sum. lo\(g_{b}\)(M) + lo\(g_{b}\)(N) = lo\(g_{b}\)(MN), which is generally a different value from lo\(g_{b}\)(M + N) — for example lo\(g_{2}\)(4) + lo\(g_{2}\)(4) = 2 + 2 = 4, but lo\(g_{2}\)(4+4) = lo\(g_{2}\)(8) = 3.

← 3.3 Logarithmic Functions as Inverses of Exponentials Unit Overview 3.5 Solving Exponential and Logarithmic Equations →