SolveNCERT
SolveNCERT
HomeSolutionsAdvanced MathematicsCh 2
Class 9 · Advanced MathematicsNCERT 2026 Revised Syllabus

Chapter 2: Logarithms

Class IX CBSE Advanced Mathematics

Covers the definition of logarithms, conversion between exponential and logarithmic forms, the laws of logarithms, changing bases, and solving logarithmic equations step by step.

Find complete NCERT solutions for Class 9 Advanced Mathematics, Class IX CBSE Advanced Mathematics Chapter 2. Every question has a clear, step-by-step answer.

Jump to:

Exercise 2.1(3 questions)

1

Write an equivalent logarithmic statement for: (a) $5^3 = 125$ (b) $2^5 = 32$ (c) $7^{-1} = \frac{1}{7}$ (d) $3^{-1/2} = \frac{1}{\sqrt{3}}$.

  1. a.$5^3 = 125$
  2. b.$2^5 = 32$
  3. c.$7^{-1} = \frac{1}{7}$
  4. d.$3^{-1/2} = \frac{1}{\sqrt{3}}$

School Level Solution

Sol.
Solution:If $a^b = c$ then $\log_a c = b$.(a) $5^3 = 125 \Rightarrow \log_5 125 = 3$ (b) $2^5 = 32 \Rightarrow \log_2 32 = 5$ (c) $7^{-1} = \frac{1}{7} \Rightarrow \log_7 \frac{1}{7} = -1$ (d) $3^{-1/2} = \frac{1}{\sqrt{3}} \Rightarrow \log_3 \frac{1}{\sqrt{3}} = -\frac{1}{2}$Answer: $\boxed{\text{(a) }\log_5 125 = 3\text{, (b) }\log_2 32 = 5\text{, (c) }\log_7 \frac{1}{7} = -1\text{, (d) }\log_3 \frac{1}{\sqrt{3}} = -\frac{1}{2}}$.
2

Write an equivalent exponential statement for: (a) $\log_2 16 = 4$ (b) $\log_9 81 = 2$ (c) $\log_5 \sqrt{5} = \frac{1}{2}$ (d) $\log_2 \frac{1}{2} = -1$.

  1. a.$\log_2 16 = 4$
  2. b.$\log_9 81 = 2$
  3. c.$\log_5 \sqrt{5} = \frac{1}{2}$
  4. d.$\log_2 \frac{1}{2} = -1$

School Level Solution

Sol.
Solution:If $\log_a c = b$ then $a^b = c$.(a) $\log_2 16 = 4 \Rightarrow 2^4 = 16$ (b) $\log_9 81 = 2 \Rightarrow 9^2 = 81$ (c) $\log_5 \sqrt{5} = \frac{1}{2} \Rightarrow 5^{1/2} = \sqrt{5}$ (d) $\log_2 \frac{1}{2} = -1 \Rightarrow 2^{-1} = \frac{1}{2}$Answer: $\boxed{\text{(a) }2^4 = 16\text{, (b) }9^2 = 81\text{, (c) }5^{1/2} = \sqrt{5}\text{, (d) }2^{-1} = \frac{1}{2}}$.
3

Find the value of: (a) $\log_{10} 1000$ (b) $\log_6 36$ (c) $\log_2 64$.

  1. a.$\log_{10} 1000$
  2. b.$\log_6 36$
  3. c.$\log_2 64$

School Level Solution

Sol.
Solution:(a) $10^3 = 1000 \Rightarrow \log_{10} 1000 = 3$ (b) $6^2 = 36 \Rightarrow \log_6 36 = 2$ (c) $2^6 = 64 \Rightarrow \log_2 64 = 6$Answer: $\boxed{\text{(a) 3, (b) 2, (c) 6}}$.

Exercise 2.2(5 questions)

1

Express the following as a single logarithm: (a) $\log 2 + 2 \log 7$ (b) $\log_3 8 + \log_3 5 - \log_3 4$ (c) $\log 5 + 2 \log 3 - \log 15$ (d) $2 + 2 \log_5 3$ (e) $3 - \frac{1}{2}\log_3 9$ (f) $1 + 2 \log_4 3 - 3 \log_4 4$.

  1. a.$\log 2 + 2 \log 7$
  2. b.$\log_3 8 + \log_3 5 - \log_3 4$
  3. c.$\log 5 + 2 \log 3 - \log 15$
  4. d.$2 + 2 \log_5 3$
  5. e.$3 - 2 \log_3 9$
  6. f.$1 + 2 \log_4 3 - 3 \log_4 4$

School Level Solution

Sol.
Solution:Use: $\log m + \log n = \log mn$, $\log m - \log n = \log \frac{m}{n}$, $k \log m = \log m^k$.(a) $\log 2 + 2 \log 7 = \log 2 + \log 49 = \log(2 \times 49) = \log 98$(b) $\log_3 8 + \log_3 5 - \log_3 4 = \log_3 \frac{8 \times 5}{4} = \log_3 10$(c) $\log 5 + 2 \log 3 - \log 15 = \log \frac{5 \times 9}{15} = \log 3$(d) $2 + 2 \log_5 3 = \log_5 25 + \log_5 9 = \log_5(25 \times 9) = \log_5 225$(e) $3 - \frac{1}{2}\log_3 9 = 3 - 1 = 2 = \log_3 9$(f) $1 + 2 \log_4 3 - 3 \log_4 4 = \log_4 4 + \log_4 9 - \log_4 64 = \log_4 \frac{4 \times 9}{64} = \log_4 \frac{9}{16}$Answer: $\boxed{\text{(a) }\log 98\text{, (b) }\log_3 10\text{, (c) }\log 3\text{, (d) }\log_5 225\text{, (e) }\log_3 \frac{1}{3} = -1\text{, (f) }\log_4 \frac{9}{16}}$.
2

Find the exact value of: (a) $\log_{11} 121$ (b) $\log_7 1$ (c) $\log_5 625$ (d) $\log_8 8$ (e) $\log 1000$.

  1. a.$\log_{11} 121$
  2. b.$\log_7 1$
  3. c.$\log_5 625$
  4. d.$\log_8 8$
  5. e.$\log 1000$

School Level Solution

Sol.
Solution:(a) $11^2 = 121 \Rightarrow \log_{11} 121 = 2$ (b) $7^0 = 1 \Rightarrow \log_7 1 = 0$ (c) $5^4 = 625 \Rightarrow \log_5 625 = 4$ (d) $8^1 = 8 \Rightarrow \log_8 8 = 1$ (e) $10^3 = 1000 \Rightarrow \log 1000 = 3$Answer: $\boxed{\text{(a) 2, (b) 0, (c) 4, (d) 1, (e) 3}}$.
3
Hard

If $\log_2 3 = p$ and $\log_2 5 = q$, write in terms of $p$ and $q$: (a) $\log_2 15$ (b) $\log_2 45$ (c) $\log_2 \frac{5}{3}$ (d) $\log_2 10$.

  1. a.$\log_2 15$
  2. b.$\log_2 45$
  3. c.$\log_2 \frac{5}{3}$
  4. d.$\log_2 10$

School Level Solution

Sol.
Solution:(a) $\log_2 15 = \log_2 3 + \log_2 5 = p + q$ (b) $\log_2 45 = \log_2(9 \times 5) = 2 \log_2 3 + \log_2 5 = 2p + q$ (c) $\log_2 \frac{5}{3} = \log_2 5 - \log_2 3 = q - p$ (d) $\log_2 10 = \log_2 2 + \log_2 5 = 1 + q$Answer: $\boxed{\text{(a) }p + q\text{, (b) }2p + q\text{, (c) }q - p\text{, (d) }1 + q}$.
4
Hard

Which of the following are true? (a) If $2^{x+1} = 3^{x+2}$ then $x+1 = x+2$. (b) $\log(x+1) = \log x$. (c) $\log_b b^3 = 3$. (d) Logarithm to base 1 is not defined.

  1. a.If $2^{x+1} = 3^{x+2}$ then $x+1 = x+2$
  2. b.$\log(x+1) = \log x$
  3. c.$\log_b b^3 = 3$
  4. d.Logarithm to base 1 is not defined

School Level Solution

Sol.
Solution:(a) False — bases are different, so the exponents cannot simply be equated. (b) False in general — $\log(x+1) = \log x$ only if $x+1 = x$, impossible. (c) True — $\log_b b^3 = 3 \log_b b = 3(1) = 3$. (d) True — base must be $> 0$ and $\neq 1$.Answer: $\boxed{\text{(c) and (d) are True}}$.
5
Hard

Given $\log_{2026} x - \log_{2026} y = a$, $\log_{2026} y - \log_{2026} z = b$, $\log_{2026} z - \log_{2026} x = c$ (note $a + b + c = 0$). Let $E = \left(\frac{x}{y}\right)^{b-c} \cdot \left(\frac{y}{z}\right)^{c-a} \cdot \left(\frac{z}{x}\right)^{a-b}$. Find $E$.

School Level Solution

Sol.
Solution:Taking $\log_{2026}$ of $E$: $$\log_{2026} E = a(b-c) + b(c-a) + c(a-b)$$ $$= ab - ac + bc - ab + ac - bc = 0$$Since $\log_{2026} E = 0$: $$E = 1$$Answer: $\boxed{E = 1}$.

Exercise 2.3(8 questions)

1

Express the following in logarithmic form: (a) $5^4 = 625$ (b) $10^{-2} = 0.01$ (c) $7^0 = 1$ (d) $8^1 = 8$.

  1. a.$5^4 = 625$
  2. b.$10^{-2} = 0.01$
  3. c.$7^0 = 1$
  4. d.$8^1 = 8$

School Level Solution

Sol.
Solution:(a) $5^4 = 625 \Rightarrow \log_5 625 = 4$ (b) $10^{-2} = 0.01 \Rightarrow \log_{10} 0.01 = -2$ (c) $7^0 = 1 \Rightarrow \log_7 1 = 0$ (d) $8^1 = 8 \Rightarrow \log_8 8 = 1$Answer: $\boxed{\text{(a) }\log_5 625 = 4\text{, (b) }\log_{10} 0.01 = -2\text{, (c) }\log_7 1 = 0\text{, (d) }\log_8 8 = 1}$}.
2

Evaluate: $\log_2 16 + \log_2 4$.

School Level Solution

Sol.
Solution:$$\log_2 16 + \log_2 4 = \log_2(16 \times 4) = \log_2 64 = 6$$Answer: $\boxed{\text{6}}$.
3

Evaluate: (a) $\log_2 256$ (b) $\log_4 16$ (c) $\log_5 125$ (d) $\log_{10} 0.001$.

  1. a.$\log_2 256$
  2. b.$\log_4 16$
  3. c.$\log_5 125$
  4. d.$\log_{10} 0.001$

School Level Solution

Sol.
Solution:(a) $2^8 = 256 \Rightarrow \log_2 256 = 8$ (b) $4^2 = 16 \Rightarrow \log_4 16 = 2$ (c) $5^3 = 125 \Rightarrow \log_5 125 = 3$ (d) $10^{-3} = 0.001 \Rightarrow \log_{10} 0.001 = -3$Answer: $\boxed{\text{(a) 8, (b) 2, (c) 3, (d) -3}}$.
4
Hard

If $\log_2 7 = p$ and $\log_2 3 = q$, write in terms of $p$ and $q$: (a) $\log_2 21$ (b) $\log_2 49$ (c) $\log_2 \frac{7}{3}$ (d) $\log_2 63$.

  1. a.$\log_2 21$
  2. b.$\log_2 49$
  3. c.$\log_2 \frac{7}{3}$
  4. d.$\log_2 63$

School Level Solution

Sol.
Solution:(a) $\log_2 21 = \log_2(3 \times 7) = q + p$ (b) $\log_2 49 = \log_2 7^2 = 2 \log_2 7 = 2p$ (c) $\log_2 \frac{7}{3} = \log_2 7 - \log_2 3 = p - q$ (d) $\log_2 63 = \log_2(9 \times 7) = 2 \log_2 3 + \log_2 7 = 2q + p$Answer: $\boxed{\text{(a) }p + q\text{, (b) }2p\text{, (c) }p - q\text{, (d) }p + 2q}$.
5

Real world applications: (a) Evaluate $2.5 \times \log_{10}(100)$. (b) If a solution has pH 6 and another pH 3, how many times more acidic is the second? (c) If two earthquakes measure 9 and 4 on the Richter scale, how many times stronger is the first?

  1. a.$2.5 \times \log_{10}(100)$
  2. b.pH difference 6 to 3
  3. c.Magnitude 9 vs 4

School Level Solution

Sol.
Solution:(a) $2.5 \times \log_{10}(100) = 2.5 \times 2 = 5$(b) pH difference $= 6 - 3 = 3$; ratio $= 10^3 = 1000$ times more acidic.(c) Magnitude difference $= 9 - 4 = 5$; ratio $= 10^5 = 100000$ times stronger.Answer: $\boxed{\text{(a) 5, (b) 1000 times more acidic, (c) 100000 times stronger}}$.
6
Hard

True or False (with reasoning): (a) $\frac{1}{3} \log_b x = \sqrt[3]{x}$ for $x > 0$. (b) $\log_8 e = \frac{1}{\ln 8}$. (c) The logarithm of a negative number is defined. (d) $\log_b(M + N) = \log_b M + \log_b N$. (e) The base of a logarithm can be any real number.

  1. a.$\frac{1}{3} \log_b x = \sqrt[3]{x}$ for $x > 0$
  2. b.$\log_8 e = \frac{1}{\ln 8}$
  3. c.Logarithm of a negative number is defined
  4. d.$\log_b(M + N) = \log_b M + \log_b N$
  5. e.Base can be any real number

School Level Solution

Sol.
Solution:(a) False — the correct relation is $\log_b x^{1/3} = \frac{1}{3} \log_b x$; a logarithm can never equal a root of $x$ itself.(b) True — by change of base, $\log_8 e = \frac{\ln e}{\ln 8} = \frac{1}{\ln 8}$, so $\frac{1}{\log_8 e} = \ln 8$.(c) False — logarithm of a negative number is not defined. (d) False — log of a sum is not the sum of logs. (e) False — base must be $> 0$ and $\neq 1$.Answer: $\boxed{\text{(b) is True; (a), (c), (d), (e) are False}}$.
7
Hard

Which is the greatest integer that is less than the number $\log_4 9 + \log_9 28$? (Do not use calculator)

School Level Solution

Sol.
Solution:$\log_4 9 = \frac{\log_2 9}{\log_2 4} = \frac{2 \log_2 3}{2} = \log_2 3$.Since $2^{1.5} = 2\sqrt{2} \approx 2.83 < 3$, we get $\log_2 3 > 1.5$ and also $\log_2 3 < 2$.Similarly $\log_9 28 = \frac{\log_2 28}{2 \log_2 3}$. Since $9^{1.5} = 27 < 28$, $\log_9 28 > 1.5$, and $\log_9 28 < 2$.So the sum lies strictly between 3 and 4 (numerically about 3.10).Answer: $\boxed{\text{3}}$.
8
Hard

Evaluate the value of $(x + 5y)$, where $x = \log_{1.43}\left(\frac{43}{30}\right)$ and $y = \left(\frac{1}{2}\right)^{\log_2 5}$.

School Level Solution

Sol.
Solution:$$x = \log_{1.43} \left(\frac{43}{30}\right) = 1$$$$y = \left(\frac{1}{2}\right)^{\log_2 5} = 5^{-1} = \frac{1}{5}$$Therefore, $x + 5y = 1 + 5(1/5) = 2$.Answer: $\boxed{x + 5y = 2}$.

Exercise 2.4(5 questions)

1

Solve for $x$: (a) $\log_3(2x - 5) = 2$ (b) $\log_7(3x) + \log_7 2 = \log_7 24$ (c) $\log_5(x+3) - \log_5(x-1) = 1$ (d) $\log_2(x^2 - 7) = 3$.

  1. a.$\log_3(2x - 5) = 2$
  2. b.$\log_7(3x) + \log_7 2 = \log_7 24$
  3. c.$\log_5(x+3) - \log_5(x-1) = 1$
  4. d.$\log_2(x^2 - 7) = 3$

School Level Solution

Sol.
Solution:(a) $\log_3(2x - 5) = 2 \Rightarrow 2x - 5 = 3^2 = 9 \Rightarrow 2x = 14 \Rightarrow x = 7$(b) $\log_7(3x) + \log_7 2 = \log_7 24 \Rightarrow \log_7(6x) = \log_7 24 \Rightarrow 6x = 24 \Rightarrow x = 4$(c) $\log_5(x+3) - \log_5(x-1) = 1 \Rightarrow \frac{x+3}{x-1} = 5 \Rightarrow x + 3 = 5x - 5 \Rightarrow 8 = 4x \Rightarrow x = 2$(d) $\log_2(x^2 - 7) = 3 \Rightarrow x^2 - 7 = 8 \Rightarrow x^2 = 15 \Rightarrow x = \pm\sqrt{15}$ (both keep $x^2 - 7 = 8 > 0$, so both are valid).Answer: $\boxed{\text{(a) }x = 7\text{, (b) }x = 4\text{, (c) }x = 2\text{, (d) }x = \pm\sqrt{15}}$.
2
Hard

Solve for $x$: (a) $\log_2(x-3) + \log_2(x+1) = 5$ (b) $2 \log_4 x = \log_4(5x - 4)$ (c) $\log_5(x+2) + \log_5(x-2) = 1$ (d) $\log_{10}(x-2) + \log_{10}(x+1) = 1$.

  1. a.$\log_2(x-3) + \log_2(x+1) = 5$
  2. b.$2 \log_4 x = \log_4(5x - 4)$
  3. c.$\log_5(x+2) + \log_5(x-2) = 1$
  4. d.$\log_{10}(x-2) + \log_{10}(x+1) = 1$

School Level Solution

Sol.
Solution:(a) $\log_2(x-3) + \log_2(x+1) = 5 \Rightarrow (x-3)(x+1) = 2^5 = 32$ $$x^2 - 2x - 3 = 32 \Rightarrow x^2 - 2x - 35 = 0 \Rightarrow (x-7)(x+5) = 0$$ Domain requires $x > 3$, so $x = -5$ is rejected. $x = 7$.(b) $2 \log_4 x = \log_4(5x - 4) \Rightarrow x^2 = 5x - 4 \Rightarrow x^2 - 5x + 4 = 0 \Rightarrow (x-4)(x-1) = 0$ Both satisfy $x > 0$ and $5x - 4 > 0$. $x = 1, 4$.(c) $(x+2)(x-2) = 5 \Rightarrow x^2 - 4 = 5 \Rightarrow x^2 = 9 \Rightarrow x = \pm 3$ Domain requires $x > 2$, so $x = -3$ is rejected. $x = 3$.(d) $(x-2)(x+1) = 10 \Rightarrow x^2 - x - 12 = 0 \Rightarrow (x-4)(x+3) = 0$ Domain requires $x > 2$, so $x = -3$ is rejected. $x = 4$.Answer: $\boxed{\text{(a) }x = 7\text{, (b) }x = 1, 4\text{, (c) }x = 3\text{, (d) }x = 4}$.
3
Hard

Solve for $x$: (a) $\log_x(3x + 10) = 2$, $x > 0$, $x \neq 1$ (b) $(\log_3 x)^2 - 4\log_3 x + 3 = 0$ (c) $(\log_2 x)^2 + 3 \log_2 x - 10 = 0$ (d) $x^{\log_{10} x} = 1000x^2$.

  1. a.$\log_x(3x + 10) = 2$, $x > 0$, $x \neq 1$
  2. b.$(\log_3 x)^2 - 4\log_3 x + 3 = 0$
  3. c.$(\log_2 x)^2 + 3 \log_2 x - 10 = 0$
  4. d.$x^{\log_{10} x} = 1000x^2$

School Level Solution

Sol.
Solution:(a) $\log_x(3x + 10) = 2 \Rightarrow 3x + 10 = x^2 \Rightarrow x^2 - 3x - 10 = 0 \Rightarrow (x-5)(x+2) = 0$ Since $x > 0$, $x = -2$ is rejected. $x = 5$.(b) Let $y = \log_3 x$: $y^2 - 4y + 3 = 0 \Rightarrow (y-1)(y-3) = 0 \Rightarrow y = 1, 3$ $$x = 3, 27$$(c) Let $y = \log_2 x$: $y^2 + 3y - 10 = 0 \Rightarrow (y+5)(y-2) = 0 \Rightarrow y = -5, 2$ $$x = \frac{1}{32}, 4$$(d) Taking $\log_{10}$ both sides, let $t = \log_{10} x$: $$t \cdot t = \log_{10} 1000 + 2t \Rightarrow t^2 = 3 + 2t \Rightarrow t^2 - 2t - 3 = 0 \Rightarrow (t-3)(t+1) = 0$$ $$t = 3 \Rightarrow x = 1000; \quad t = -1 \Rightarrow x = 0.1$$Answer: $\boxed{\text{(a) }x = 5\text{, (b) }x = 3, 27\text{, (c) }x = \frac{1}{32}, 4\text{, (d) }x = 1000, 0.1}$.
4
Hard

Solve for $x$: (a) $\log_3(x^2 - 1) = \log_3(2x - 1)$ (b) $\log_x 5 - \log_x 2 = \log_x \sqrt{x}$ (c) $\log_2 x + \frac{1}{\log_x 2} = 4$ (d) $\log_3(3+x) + \log_3(8-x) - \log_3(9x-8) = 2 - \log_3 9$ (e) $\log_{10}[\log_2(\log_3 9)] = 5x$.

  1. a.$\log_3(x^2 - 1) = \log_3(2x - 1)$
  2. b.$\log_x 5 - \log_x 2 = \log_x \sqrt{x}$
  3. c.$\log_2 x + \frac{1}{\log_x 2} = 4$
  4. d.$\log_3(3+x) + \log_3(8-x) - \log_3(9x-8) = 2 - \log_3 9$
  5. e.$\log_{10}[\log_2(\log_3 9)] = 5x$

School Level Solution

Sol.
Solution:(a) $x^2 - 1 = 2x - 1 \Rightarrow x^2 - 2x = 0 \Rightarrow x(x-2) = 0$ $x = 0$ fails the domain ($x^2 - 1 > 0$ needs $|x| > 1$); $x = 2$ satisfies both domains. $x = 2$.(b) $\log_x \frac{5}{2} = \log_x x^{1/2} \Rightarrow \frac{5}{2} = x^{1/2} \Rightarrow x = \frac{25}{4}$(c) Since $\frac{1}{\log_x 2} = \log_2 x$, the equation becomes $2\log_2 x = 4$. Thus $\log_2 x = 2$, so $x = 4$. (d) RHS $= 2 - 2 = 0$ since $\log_3 9 = 2$. $$\log_3 \frac{(3+x)(8-x)}{9x-8} = 0 \Rightarrow \frac{(3+x)(8-x)}{9x-8} = 1$$ $$(3+x)(8-x) = 9x - 8 \Rightarrow 24 + 5x - x^2 = 9x - 8 \Rightarrow x^2 + 4x - 32 = 0 \Rightarrow (x+8)(x-4) = 0$$ Domain: $-3 < x < 8$ and $x > \frac{8}{9}$, so $x = -8$ rejected. $x = 4$.(e) $\log_3 9 = 2$, $\log_2 2 = 1$, $\log_{10} 1 = 0$: $$0 = 5x \Rightarrow x = 0$$Answer: $\boxed{\text{(a) }x = 2\text{, (b) }x = \frac{25}{4}\text{, (c) }x = 4\text{, (d) }x = 4\text{, (e) }x = 0}$.
5
Hard

Let $x = \log \frac{1}{2} + \log \frac{2}{3} + \log \frac{3}{4} + \dots + \log \frac{99}{100}$ (base 10). Find the value of $(x+1)(x+2)(x+3)\dots(x+99)$.

School Level Solution

Sol.
Solution:This telescopes: $$x = \log\left(\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdots \frac{99}{100}\right) = \log \frac{1}{100} = -2$$So: $$(x+1)(x+2)(x+3)\dots(x+99) = (-1)(0)(1)\dots(97)$$Since one factor $(x + 2 = 0)$ is zero, the whole product is zero.Answer: $\boxed{\text{0}}$.

Other Chapters

123456