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4.4 Laws of Logarithms

Focus on log⁡\log product law and quotient law.

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Ideas in this lesson

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Lesson brief and rules

4.4 Laws of Logarithms

For positive inputs, log⁡b(MN)=log⁡bM+log⁡bN\log _b(MN)=\log _b M+\log _b N.

Worked example 11
01 / For b>0, b≠1, A>0, B>0b>0,\ b\ne1,\ A>0,\ B>0, prove the product, quotient and real-power laws from the exponential definition, rather than assuming them. Then explain why log⁡b(A+B)\log_b(A+B) has no corresponding sum law. Written work is required; selecting an answer alone is not a complete response. Complete the written work or drawing first. The selected answer checks the result only; your reasoning or drawing needs teacher/self-review.
02 / Set u=log⁡bA, v=log⁡bBu=\log_b A,\ v=\log_b B, so A=bu, B=bvA=b^{u},\ B=b^{v}.
03 / The exponential law AB=bu+vA B=b^{u+v} gives log⁡b(AB)=u+v=log⁡bA+log⁡bB\log_b(A B)=u+v=\log_b A+\log_b B.
04 / Similarly AB=bu−v\dfrac{A}{B}=b^{u-v} gives log⁡bAB=log⁡bA−log⁡bB\log_b\dfrac{A}{B}=\log_b A-\log_b B.
05 / For real rr, positivity of AA makes Ar=(bu)r=bruA^{r}=(b^{u})^{r}=b^{r u} well-defined, hence log⁡b(Ar)=rlog⁡bA\log_b(A^{r})=r\log_b A.
06 / A trial for a counterexample to an additive law is b=2, A=B=2b=2,\ A=B=2: log⁡2(2+2)=2\log_2(2+2)=2 but log⁡22+log⁡22=2\log_2 2+\log_2 2=2 happens to agree, so it is not a counterexample. Instead use A=B=4A=B=4: log⁡28=3≠4=log⁡24+log⁡24\log_2 8=3\ne4=\log_2 4+\log_2 4.
Worked example 2 · Prove log laws and diagnose false identities

A student claims log⁡2(A−B)=log⁡2A−log⁡2B\log_2(A-B)=\log_2A-\log_2B. Give a counterexample with every logarithm defined, then state the correct subtraction-of-logs law. Written work is required; selecting an answer alone is not a complete response. Complete the written work or drawing first. The selected answer checks the result only; your reasoning or drawing needs teacher/self-review.

Choose A=8, B=2A=8,\ B=2. All arguments are positive.
The left side is log⁡26\log_{2}6 and the right side is 3−1=23-1=2.
Since 6≠46\ne4, these are unequal.
The correct identity is log⁡bA−log⁡bB=log⁡bAB\log_b A-\log_b B=\log_b\dfrac{A}{B} for positive arguments and a valid base.
Written-work review:

15 questions with a worked answer for each.Start practice ↗
Textbook reference: Stewart, Redlin & Watson, Precalculus: Mathematics for Calculus, 7th ed. (Cengage, 2016). The explanations and practice here are original, organized by textbook section.