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Next exam on November 14 – 15, 2026:
All concepts

Logarithms and Logarithmic Functions

对数运算与对数函数

Logarithm laws, the change-of-base formula and logarithmic domains, with monotonicity used to compare logarithmic values.

6 real-exam questions · 5 sittings33 in mock papers27 practice

Formulas

  • Laws of Logarithms

    Law
    log⁡a(MN)=log⁡aM+log⁡aN,log⁡aMN=log⁡aM−log⁡aN,log⁡aMn=nlog⁡aM\log_{a}(MN)=\log_{a}M+\log_{a}N,\quad \log_{a}\frac{M}{N}=\log_{a}M-\log_{a}N,\quad \log_{a}M^{n}=n\log_{a}M

    Applies when：a > 0, a ≠ 1, M > 0, N > 0

  • Change of Base

    Identity
    log⁡ab=log⁡cblog⁡ca\log_{a}b=\frac{\log_{c}b}{\log_{c}a}

    Applies when：a > 0, a ≠ 1, b > 0, c > 0, c ≠ 1

  • Exponential–Logarithmic Conversion

    Definition
    ax=N  ⟺  x=log⁡aNa^{x}=N\iff x=\log_{a}N

    Applies when：a > 0, a ≠ 1, N > 0

How the real exam tests it

Questions from real CSCA papers that test this concept. Work each one out before you open the answer.

  1. CSCA Math December 2025 Past Paper
    log⁡232+log⁡1232=\log_2 32+\log_{\frac{1}{2}}32= ( ).
    • A．0
    • B．-5
    • C．5
    • D．10
    Show answer & solution

    Answer: A

    Answer: A log⁡232=5\log_2 32=5 and log⁡1232=−5\log_{\frac{1}{2}}32=-5, so the sum is 0. Choose A.
    See it in the full CSCA Math December 2025 Past Paper
  2. CSCA Math January 2026 Past Paper
    Let a>0a>0 and b>0b>0 be real numbers. Which statement is incorrect?
    • A．ln⁡(ab)=bln⁡a\ln(a^b)=b\ln a
    • B．log⁡ab=ln⁡aln⁡b\log_a b=\frac{\ln a}{\ln b} (a≠1,b≠1a\neq1,b\neq1)
    • C．ln⁡(ab)=ln⁡a−ln⁡b\ln\left(\frac{a}{b}\right)=\ln a-\ln b
    • D．ln⁡(ab)=ln⁡a+ln⁡b\ln(ab)=\ln a+\ln b
    Show answer & solution

    Answer: B

    Answer: B The change-of-base formula is log⁡ab=ln⁡bln⁡a\log_a b=\frac{\ln b}{\ln a} for a>0a>0, a≠1a\neq1, and b>0b>0. Option B reverses the numerator and denominator.
    Take the full CSCA Math January 2026 Past Paper — timed & graded
  3. CSCA Math April 2026 Past Paper
    (log⁡327)(log⁡416)=\left( \log _ { 3 } 27 \right) \left( \log _ { 4 } 16 \right) =
    • A．16\frac { 1 } { 6 }
    • B．23\frac { 2 } { 3 }
    • C．6
    • D．5
    Show answer & solution

    Answer: C

    log⁡327=3\log _ { 3 } 27 = 3 (since 33=273 ^ { 3 } = 27 ); log⁡416=2\log _ { 4 } 16 = 2 (since 42=164 ^ { 2 } = 16 ). Product =3⋅2=6= 3 \cdot 2 = 6.
    Take the full CSCA Math April 2026 Past Paper — timed & graded

Real-exam evidence

Recorded past-paper appearances are listed below; online verification is labelled separately.

6 questions across 5 sittings1 of 5 sittings available to verify free online
  • CSCA Math June 2026 Past Paper2 questionsreading page not yet public
  • CSCA Math April 2026 Past Paper1 questionreading page not yet public
  • CSCA Math March 2026 Past Paper1 questionreading page not yet public
  • CSCA Math January 2026 Past Paper1 questionreading page not yet public
  • CSCA Math December 2025 Past Paper1 questionView question 30 and answer free (full paper)

33 questions across 16 mock papers — our own papers, so a coverage signal rather than real-exam evidence

27 related practice questions availablePractise free: Elementary FunctionPractise free: Inequality

Sitting evidence comes from the CSCA past papers on record; questions whose original paper is missing are excluded, and sprint papers are deliberately not counted.

Glossary

logarithmic function

Related concepts

Exponents and Exponential FunctionsFunction ParityFunction MonotonicityFunction DomainsInverse Functions