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Logarithms Class 9 ICSE Important Questions with Answers

The 13 questions below are Logarithms Class 9 ICSE important questions, each with a full step-by-step answer: extra questions beyond the textbook exercises, covering 5 topics of ICSE Class 9 Maths in the question formats used in the exam.

By The Classmate AI Editorial Team

Reviewed by Classmate AI Team · 17 October 2026

Last updated 25 October 2026

13
Questions
5
Topics
3
Key concepts
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With answers
Quick answer

ICSE Class 9 Logarithms questions convert between exponential and logarithmic form, use the laws log⁡a(mn)=log⁡am+log⁡an\log_a(mn)=\log_a m+\log_a nlog _a(mn)=log _a m+log _a n, log⁡amn=log⁡am−log⁡an\log_a\dfrac{m}{n}=\log_a m-\log_a nlog _am/n=log _a m-log _a n and log⁡amp=plog⁡am\log_a m^p=p\log_a mlog _a m^p=p log _a m to expand, combine and evaluate expressions, find values from given logs such as log⁡2\log 2log 2 and log⁡3\log 3log 3, and solve simple logarithmic equations.

Syllabus reference (2026–27): ICSE 2028 Mathematics syllabus (PDF) · CISCE (Council for the Indian School Certificate Examinations)

About Logarithms

In the ICSE Class 9 Maths chapter Logarithms you interpret log⁡aN=x\log_a N=xlog _a N=x as ax=Na^x=Na^x=N, apply the product, quotient, and power laws of logarithms, work with common logarithms (base 101010), and solve equations involving logarithms. The chapter links indices and logarithms as inverse ideas.

Log-exponential conversionLaws of logarithmsCommon logarithms (base 10)Expanding and combining log expressionsSolving logarithmic equations

Key concepts & formulas

Definition

log⁡aN=x  ⟺  ax=N\log_a N=x\iff a^x=Nlog _a N=x⇔ a^x=N, for a>0, a≠1, N>0a>0,\ a\neq1,\ N>0a>0, a≠1, N>0. Base 101010 logarithms are common logarithms, written log⁡N\log Nlog N.

Laws of logarithms

log⁡a(mn)=log⁡am+log⁡an\log_a(mn)=\log_a m+\log_a nlog _a(mn)=log _a m+log _a n; log⁡amn=log⁡am−log⁡an\log_a\dfrac{m}{n}=\log_a m-\log_a nlog _am/n=log _a m-log _a n; log⁡amp=plog⁡am\log_a m^p=p\log_a mlog _a m^p=p log _a m.

Special values

log⁡a1=0\log_a 1=0log _a 1=0 and log⁡aa=1\log_a a=1log _a a=1. The logarithm of zero or of a negative number is not defined, so check that every argument is positive when solving an equation.

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Logarithms Questions and Answers

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MCQs with answers (1 mark)

Q1MCQEasy1 mark

The value of log⁡232\log_2 32log₂ 32 is:

  1. (a)

    555

  2. (b)

    444

  3. (c)

    666

  4. (d)

    161616

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Answer: (a) 555.

32=2532=2^532=2⁵, so log⁡232=5\log_2 32=5log₂ 32=5.

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Q2MCQEasy1 mark

In exponential form, log⁡381=4\log_3 81=4log₃ 81=4 is written as:

  1. (a)

    34=813^4=813⁴=81

  2. (b)

    43=814^3=814³=81

  3. (c)

    813=481^3=481³=4

  4. (d)

    381=43^{81}=43⁸¹=4

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Answer: (a) 34=813^4=813⁴=81.

By definition log⁡aN=x  ⟺  ax=N\log_a N=x\iff a^x=Nlog _a N=x⇔ a^x=N, so log⁡381=4\log_3 81=4log₃ 81=4 means 34=813^4=813⁴=81.

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Q3MCQModerate1 mark

The value of log⁡8+log⁡125\log 8+\log 125log 8+log 125 (base 101010) is:

  1. (a)

    333

  2. (b)

    222

  3. (c)

    111

  4. (d)

    log⁡133\log 133log 133

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Answer: (a) 333.

log⁡8+log⁡125=log⁡(8×125)=log⁡1000=log⁡103=3\log 8+\log 125=\log(8\times125)=\log 1000=\log 10^3=3log 8+log 125=log (8×125)=log 1000=log 10³=3.

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Q4MCQHOTS1 mark

If log⁡102=0.3010\log_{10}2=0.3010log₁₀2=0.3010, then log⁡105\log_{10}5log₁₀5 equals:

  1. (a)

    0.69900.69900.6990

  2. (b)

    0.30100.30100.3010

  3. (c)

    0.60200.60200.6020

  4. (d)

    0.15050.15050.1505

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Answer: (a) 0.69900.69900.6990.

log⁡5=log⁡102=log⁡10−log⁡2=1−0.3010=0.6990\log 5=\log\dfrac{10}{2}=\log 10-\log 2=1-0.3010=0.6990log 5=log 10/2=log 10-log 2=1-0.3010=0.6990.

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonModerate1 mark

Assertion (A): log⁡a1=0\log_a 1=0log _a 1=0 for every valid base aaa.

Reason (R): Any non-zero number raised to the power 000 equals 111.

  1. (a)

    Both A and R are true and R is the correct explanation of A

  2. (b)

    Both A and R are true but R is not the correct explanation of A

  3. (c)

    A is true but R is false

  4. (d)

    A is false but R is true

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Answer: (a) Since a0=1a^0=1a⁰=1, by definition log⁡a1=0\log_a 1=0log _a 1=0. R gives exactly the reason A holds, so R correctly explains A.

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Very short answer questions (2 marks)

Q6Very ShortEasy2 marks

Evaluate log⁡381\log_{\sqrt3}81log _√381.

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Write 3=31/2\sqrt3=3^{1/2}√3=3^1/2 and 81=3481=3^481=3⁴. Let log⁡381=x\log_{\sqrt3}81=xlog _√381=x:

(31/2)x=34⇒3x/2=34⇒x2=4⇒x=8.(3^{1/2})^{x}=3^{4}\Rightarrow 3^{x/2}=3^{4}\Rightarrow \frac{x}{2}=4\Rightarrow x=8.(3^1/2)^x=3⁴⇒ 3^x/2=3⁴⇒ x/2=4⇒ x=8.

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Q7Very ShortModerate2 marks

Expand log⁡a3b2c\log\dfrac{a^3b^2}{\sqrt c}log a³b²/√c using the laws of logarithms.

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log⁡a3b2c=log⁡a3+log⁡b2−log⁡c12=3log⁡a+2log⁡b−12log⁡c.\log\frac{a^3b^2}{\sqrt c}=\log a^3+\log b^2-\log c^{\frac12}=3\log a+2\log b-\tfrac12\log c.log a³b²/√c=log a³+log b²-log c^1/2=3 log a+2 log b-1/2 log c.
(Product law for the numerator, quotient law for the division, and power law for each power, with c=c12\sqrt c=c^{\frac12}√c=c^1/2.)

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Short answer questions (3 marks)

Q8Short AnswerModerate3 marks

Express as a single logarithm and simplify: 2log⁡3+log⁡5−log⁡452\log 3+\log 5-\log 452 log 3+log 5-log 45.

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Use the power law then combine:

2log⁡3=log⁡32=log⁡9.2\log 3=\log 3^2=\log 9.2 log 3=log 3²=log 9.

So the expression becomes

log⁡9+log⁡5−log⁡45=log⁡9×545=log⁡4545=log⁡1=0.\log 9+\log 5-\log 45=\log\frac{9\times5}{45}=\log\frac{45}{45}=\log 1=0.log 9+log 5-log 45=log 9×5/45=log 45/45=log 1=0.

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Q9Short AnswerModerate3 marks

If log⁡(a+b3)=12(log⁡a+log⁡b)\log\left(\dfrac{a+b}{3}\right)=\dfrac12(\log a+\log b)log ((a+b)/3)=1/2(log a+log b), prove that a2+b2=7aba^2+b^2=7aba²+b²=7ab.

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The right side is 12log⁡(ab)=log⁡(ab)1/2=log⁡ab\dfrac12\log(ab)=\log(ab)^{1/2}=\log\sqrt{ab}1/2 log (ab)=log (ab)^1/2=log √ab.

So log⁡a+b3=log⁡ab\log\dfrac{a+b}{3}=\log\sqrt{ab}log (a+b)/3=log √ab, giving

a+b3=ab.\frac{a+b}{3}=\sqrt{ab}.(a+b)/3=√ab.

Square both sides:

(a+b)29=ab⇒(a+b)2=9ab.\frac{(a+b)^2}{9}=ab\Rightarrow (a+b)^2=9ab.((a+b)²)/9=ab⇒ (a+b)²=9ab.

a2+2ab+b2=9ab⇒a2+b2=7ab.a^2+2ab+b^2=9ab\Rightarrow a^2+b^2=7ab.a²+2ab+b²=9ab⇒ a²+b²=7ab.

Hence proved.

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Q10Short AnswerEasy3 marks

Given log⁡2=0.3010\log 2=0.3010log 2=0.3010 and log⁡3=0.4771\log 3=0.4771log 3=0.4771, evaluate log⁡6\log 6log 6 and log⁡1.5\log 1.5log 1.5.

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log⁡6\log 6log 6: log⁡6=log⁡(2×3)=log⁡2+log⁡3=0.3010+0.4771=0.7781.\log 6=\log(2\times3)=\log 2+\log 3=0.3010+0.4771=0.7781.log 6=log (2×3)=log 2+log 3=0.3010+0.4771=0.7781.

log⁡1.5\log 1.5log 1.5: log⁡1.5=log⁡32=log⁡3−log⁡2=0.4771−0.3010=0.1761.\log 1.5=\log\dfrac{3}{2}=\log 3-\log 2=0.4771-0.3010=0.1761.log 1.5=log 3/2=log 3-log 2=0.4771-0.3010=0.1761.

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Long answer questions (5 marks)

Q11Long AnswerModerate5 marks

Solve for xxx: log⁡10(x+5)+log⁡10(x−5)=4log⁡102+2log⁡103\log_{10}(x+5)+\log_{10}(x-5)=4\log_{10}2+2\log_{10}3log₁₀(x+5)+log₁₀(x-5)=4 log₁₀2+2 log₁₀3.

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Left side (product law): log⁡10[(x+5)(x−5)]=log⁡10(x2−25)\log_{10}\big[(x+5)(x-5)\big]=\log_{10}(x^2-25)log₁₀[(x+5)(x-5)]=log₁₀(x²-25).

Right side (power law, then product law): 4log⁡102+2log⁡103=log⁡1024+log⁡1032=log⁡1016+log⁡109=log⁡101444\log_{10}2+2\log_{10}3=\log_{10}2^4+\log_{10}3^2=\log_{10}16+\log_{10}9=\log_{10}1444 log₁₀2+2 log₁₀3=log₁₀2⁴+log₁₀3²=log₁₀16+log₁₀9=log₁₀144.

Equating the arguments: x2−25=144⇒x2=169⇒x=±13x^2-25=144\Rightarrow x^2=169\Rightarrow x=\pm13x²-25=144⇒ x²=169⇒ x=±13.

The logarithm of a negative number is not defined, so x−5>0x-5>0x-5>0, i.e. x>5x>5x>5. Rejecting x=−13x=-13x=-13: x=13\boxed{x=13}x=13.
Check: log⁡18+log⁡8=log⁡144\log 18+\log 8=\log 144log 18+log 8=log 144 ✓.

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Q12Long AnswerHOTS5 marks

If log⁡x2=12log⁡y=13log⁡z\log\dfrac{x}{2}=\dfrac12\log y=\dfrac13\log zlog x/2=1/2 log y=1/3 log z, prove that x6=64y3x^6=64y^3x⁶=64y³ and that y3=z2y^3=z^2y³=z².

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Let each expression equal log⁡t\log tlog t (with t>0t>0t>0).

log⁡x2=log⁡t⇒x=2t\log\dfrac{x}{2}=\log t\Rightarrow x=2tlog x/2=log t⇒ x=2t.

12log⁡y=log⁡t⇒log⁡y=2log⁡t=log⁡t2⇒y=t2\tfrac12\log y=\log t\Rightarrow\log y=2\log t=\log t^2\Rightarrow y=t^21/2 log y=log t⇒ log y=2 log t=log t²⇒ y=t².

13log⁡z=log⁡t⇒log⁡z=3log⁡t=log⁡t3⇒z=t3\tfrac13\log z=\log t\Rightarrow\log z=3\log t=\log t^3\Rightarrow z=t^31/3 log z=log t⇒ log z=3 log t=log t³⇒ z=t³.

Then x6=(2t)6=64t6x^6=(2t)^6=64t^6x⁶=(2t)⁶=64t⁶ and 64y3=64(t2)3=64t664y^3=64(t^2)^3=64t^664y³=64(t²)³=64t⁶, so x6=64y3x^6=64y^3x⁶=64y³.

Also y3=(t2)3=t6y^3=(t^2)^3=t^6y³=(t²)³=t⁶ and z2=(t3)2=t6z^2=(t^3)^2=t^6z²=(t³)²=t⁶, so y3=z2y^3=z^2y³=z². Hence proved.

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Case study questions (4 marks)

Q13Case-basedModerate4 marks

Scientists measure sound intensity on a logarithmic scale. The loudness LLL in bels of a sound of intensity III relative to a reference I0I_0I₀ is L=log⁡10II0L=\log_{10}\dfrac{I}{I_0}L=log₁₀I/I₀.

(i) A sound has I=1000 I0I=1000\,I_0I=1000 I₀. Find LLL.

(ii) Another sound has L=2.5L=2.5L=2.5 bels. Express II0\dfrac{I}{I_0}I/I₀ as a power of 101010.

(iii) If the intensity increases 100100100-fold, by how many bels does LLL increase?

(iv) Using log⁡2=0.3010\log 2=0.3010log 2=0.3010, find LLL (to 4 decimals) when I=2 I0I=2\,I_0I=2 I₀.

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(i) L=log⁡101000 I0I0=log⁡101000=log⁡10103=3L=\log_{10}\dfrac{1000\,I_0}{I_0}=\log_{10}1000=\log_{10}10^3=3L=log₁₀(1000 I₀)/I₀=log₁₀1000=log₁₀10³=3 bels.

(ii) 2.5=log⁡10II0⇒II0=102.52.5=\log_{10}\dfrac{I}{I_0}\Rightarrow \dfrac{I}{I_0}=10^{2.5}2.5=log₁₀I/I₀⇒ I/I₀=10^2.5.

(iii) New loudness =log⁡10100II0=log⁡10100+log⁡10II0=2+L=\log_{10}\dfrac{100I}{I_0}=\log_{10}100+\log_{10}\dfrac{I}{I_0}=2+L=log₁₀100I/I₀=log₁₀100+log₁₀I/I₀=2+L. So LLL increases by 222 bels.

(iv) L=log⁡102=0.3010L=\log_{10}2=0.3010L=log₁₀2=0.3010 bels.

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All ICSE Class 9 Maths Chapters

Rational and Irrational NumbersCompound Interest (Without Using Formula)Compound Interest (Using Formula)ExpansionsFactorisationSimultaneous (Linear) EquationsIndices (Exponents)LogarithmsTriangles (Congruency in Triangles)Isosceles TrianglesInequalitiesMid-point and Its Converse (Including Intercept Theorem)Pythagoras TheoremRectilinear Figures (Quadrilaterals)Construction of PolygonsArea TheoremsCircleStatistics (Classification of Data, Tabulation)Mean and MedianArea and Perimeter of Plane FiguresSolids (Surface Area and Volume of 3-D Solids)Trigonometrical RatiosTrigonometrical Ratios of Standard AnglesSolution of Right TrianglesComplementary AnglesCo-ordinate GeometryGraphical Solution (Simultaneous Linear Equations, Graphically)Distance Formula

Frequently asked questions

  • What are the laws of logarithms for ICSE Class 9?
    log⁡a(mn)=log⁡am+log⁡an\log_a(mn)=\log_a m+\log_a nlog _a(mn)=log _a m+log _a n, log⁡amn=log⁡am−log⁡an\log_a\tfrac mn=\log_a m-\log_a nlog _am/n=log _a m-log _a n and log⁡amp=plog⁡am\log_a m^p=p\log_a mlog _a m^p=p log _a m, with log⁡a1=0\log_a1=0log _a1=0 and log⁡aa=1\log_aa=1log _aa=1. They come straight from the laws of indices because log⁡aN=x\log_aN=xlog _aN=x means ax=Na^x=Na^x=N.
  • How do you find log 5 if log 2 = 0.3010?
    Write 5=1025=\tfrac{10}{2}5=10/2, so log⁡5=log⁡10−log⁡2=1−0.3010=0.6990\log5=\log10-\log2=1-0.3010=0.6990log 5=log 10-log 2=1-0.3010=0.6990. In the same way, log⁡6=log⁡2+log⁡3\log6=\log2+\log3log 6=log 2+log 3 and log⁡1.5=log⁡3−log⁡2\log1.5=\log3-\log2log 1.5=log 3-log 2 when log⁡3=0.4771\log3=0.4771log 3=0.4771 is given. Most questions of this type ask you to build the number from 222, 333 and 101010 using products and quotients, then apply the matching law of logarithms.
  • How do you evaluate log base √3 of 81?
    Write 818181 as a power of 3\sqrt3√3: 81=34=(3)881=3^4=(\sqrt3)^881=3⁴=(√3)⁸. So log⁡381=8\log_{\sqrt3}81=8log _√381=8. Converting both base and number to powers of the same prime is the standard method.
  • How do you solve log(x + 5) + log(x − 5) = 4 log 2 + 2 log 3?
    Combine each side: log⁡(x2−25)=log⁡(16×9)=log⁡144\log(x^2-25)=\log(16\times9)=\log144log (x²-25)=log (16×9)=log 144. So x2−25=144x^2-25=144x²-25=144, x2=169x^2=169x²=169 and x=13x=13x=13. Reject x=−13x=-13x=-13, because log⁡(x−5)\log(x-5)log (x-5) would then be the log of a negative number. Checking the answer in the original equation is part of the method here.
  • Which ICSE Class 9 chapter should I revise before Logarithms?
    Indices (Exponents), Chapter 7, because every law of logarithms is an index law rewritten. Logarithms is Chapter 8, and Triangles (Congruency) follows as Chapter 9.
  • What mistakes do students make in ICSE Logarithms?
    Writing log⁡(m+n)=log⁡m+log⁡n\log(m+n)=\log m+\log nlog (m+n)=log m+log n, which is false, forgetting that the log of zero or a negative number is undefined, and dropping the coefficient when using plog⁡m=log⁡mpp\log m=\log m^pp log m=log m^p. Check every step against the three laws.
  • Can I download Logarithms ICSE Class 9 questions with answers?
    Yes. Enter your email in the free PDF box on this page and all 13 logarithm questions, the sound-intensity case study included, open with worked answers as a printable worksheet.

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