Mathematics

Class 92025–26 SNC36 Questions SolvedEnglish Medium

Exercise 2.4 — Class 9 Mathematics Solutions

All 36 questions below are worked step by step — every step clearly worked out with mathematical reasons — for the Punjab Board 9th class Mathematics textbook.

Exercise 2.4

8 Questions (28 Sub-parts)

Every solution is shown in full. Switch to Self-Test Mode to hide them and practice first.

Q1Question 1

Medium

Without using calculator, evaluate the following:

Q1Part (i)

EasyApproved by Miss Aniya • Aug 24, 2026

$\log_{2} 18 - \log_{2} 9$

Hint

Apply the quotient law: $\log_{a} m - \log_{a} n = \log_{a} (\frac{m}{n})$ .

Solution

Apply the quotient law of logarithms.
$\begin{matrix} \log_{2} 18 - \log_{2} 9 & = \log_{2} (\frac{18}{9}) \end{matrix}$
Simplify the fraction.
$\begin{matrix} = \log_{2} (2) \end{matrix}$
Evaluate using base identity $\log_{a} a = 1$ .
$\begin{matrix} = 1 \end{matrix}$

Answer

$1$

Q1Part (ii)

EasyApproved by Miss Aniya • Aug 24, 2026

$\log_{2} 64 + \log_{2} 2$

Hint

Express $64 = 2^{6}$ and apply the power law $\log_{a} (m^{n}) = n \log_{a} m$ .

Solution

Express arguments as powers of base $2$ .

Since $64 = 2^{6}$ :
$\begin{matrix} \log_{2} 64 + \log_{2} 2 & = \log_{2} (2^{6}) + \log_{2} 2 \end{matrix}$
Apply the power law of logarithms.
$\begin{matrix} = 6 \log_{2} 2 + \log_{2} 2 \end{matrix}$
Substitute the base identity $\log_{2} 2 = 1$ .
$\begin{matrix} = 6 (1) + 1 \\ = 6 + 1 \\ = 7 \end{matrix}$

Answer

$7$

Q1Part (iii)

MediumApproved by Miss Aniya • Aug 24, 2026

$\frac{1}{3} \log_{3} 8 - \log_{3} 18$

Hint

Bring $\frac{1}{3}$ inside: $8^{\frac{1}{3}} = (2^{3})^{\frac{1}{3}} = 2$ . Then apply quotient law: $\log_{3} (\frac{2}{18}) = \log_{3} (\frac{1}{9}) = - 2$ .

Solution

Apply the power law to the first term.
$\begin{matrix} \frac{1}{3} \log_{3} 8 - \log_{3} 18 & = \log_{3} (8^{1 / 3}) - \log_{3} 18 \end{matrix}$
Simplify $8^{1 / 3} = (2^{3})^{1 / 3} = 2$ .
$\begin{matrix} = \log_{3} 2 - \log_{3} 18 \end{matrix}$
Apply the quotient law of logarithms.
$\begin{matrix} = \log_{3} (\frac{2}{18}) \\ = \log_{3} (\frac{1}{9}) \end{matrix}$
Express as a power of base $3$ and evaluate.
$\begin{matrix} = \log_{3} (3^{- 2}) \\ = - 2 \log_{3} 3 \\ = - 2 (1) \\ = - 2 \end{matrix}$

Answer

$- 2$

Q1Part (iv)

EasyApproved by Miss Aniya • Aug 24, 2026

$2 \log 2 + \log 25$

Hint

Move coefficient to exponent: $2^{2} = 4$ . Then apply product law: $\log (4 \times 25) = \log (100) = 2$ .

Solution

Apply the power law to the first term.
$\begin{matrix} 2 \log 2 + \log 25 & = \log (2^{2}) + \log 25 \\ = \log 4 + \log 25 \end{matrix}$
Apply the product law of logarithms.
$\begin{matrix} = \log (4 \times 25) \\ = \log 100 \end{matrix}$
Evaluate base 10 logarithm.
$\begin{matrix} = \log_{10} (10^{2}) \\ = 2 \log_{10} 10 \\ = 2 (1) \\ = 2 \end{matrix}$

Answer

$2$

Q1Part (v)

MediumApproved by Miss Aniya • Aug 24, 2026

$\frac{1}{3} \log_{4} 64 + 2 \log_{5} 25$

Hint

Write $64 = 4^{3}$ and $25 = 5^{2}$ . Simplify each logarithm using $\log_{a} (a^{n}) = n$ .

Solution

Express arguments as powers of their bases.

Since $64 = 4^{3}$ and $25 = 5^{2}$ :
$\begin{matrix} \frac{1}{3} \log_{4} 64 + 2 \log_{5} 25 & = \frac{1}{3} \log_{4} (4^{3}) + 2 \log_{5} (5^{2}) \end{matrix}$
Apply the power law to each term.
$\begin{matrix} = \frac{1}{3} (3 \log_{4} 4) + 2 (2 \log_{5} 5) \\ = 1 \log_{4} 4 + 4 \log_{5} 5 \end{matrix}$
Substitute $\log_{4} 4 = 1$ and $\log_{5} 5 = 1$ .
$\begin{matrix} = 1 (1) + 4 (1) \\ = 1 + 4 \\ = 5 \end{matrix}$

Answer

$5$

Q1Part (vi)

MediumApproved by Miss Aniya • Aug 24, 2026

$\log_{3} 12 + \log_{3} 0.25$

Hint

Convert $0.25 = \frac{1}{4}$ , then apply product law: $\log_{3} (12 \times \frac{1}{4}) = \log_{3} (3) = 1$ .

Solution

Convert decimal to fraction.

Since $0.25 = \frac{1}{4}$ :
$\begin{matrix} \log_{3} 12 + \log_{3} 0.25 & = \log_{3} 12 + \log_{3} (\frac{1}{4}) \end{matrix}$
Apply the product law of logarithms.
$\begin{matrix} = \log_{3} (12 \times \frac{1}{4}) \\ = \log_{3} (3) \end{matrix}$
Apply the base identity.
$\begin{matrix} = 1 \end{matrix}$

Answer

$1$

Q2Question 2

Medium

Write the following as a single logarithm:

Q2Part (i)

EasyApproved by Miss Aniya • Aug 24, 2026

$\frac{1}{2} \log 25 + 2 \log 3$

Hint

Use power law: $25^{\frac{1}{2}} = 5$ and $3^{2} = 9$ . Then apply product law: $\log (5 \times 9) = \log 45$ .

Solution

Apply the power law of logarithms.
$\begin{matrix} \frac{1}{2} \log 25 + 2 \log 3 & = \log (25^{\frac{1}{2}}) + \log (3^{2}) \end{matrix}$
Simplify powers.
$\begin{matrix} = \log 5 + \log 9 \end{matrix}$
Apply the product law of logarithms.
$\begin{matrix} = \log (5 \times 9) \\ = \log 45 \end{matrix}$

Answer

$\log 45$

Q2Part (ii)

EasyApproved by Miss Aniya • Aug 24, 2026

$\log 9 - \log \frac{1}{3}$

Hint

Apply quotient law: $\log (\frac{9}{\frac{1}{3}}) = \log (9 \times 3) = \log 27$ .

Solution

Apply the quotient law of logarithms.
$\begin{matrix} \log 9 - \log (\frac{1}{3}) & = \log (\frac{9}{\frac{1}{3}}) \end{matrix}$
Simplify the fraction.
$\begin{matrix} = \log (9 \times 3) \\ = \log 27 \end{matrix}$

Answer

$\log 27$

Q2Part (iii)

HardApproved by Miss Aniya • Aug 24, 2026

$\log_{5} b^{2} \cdot \log_{a} 5^{3}$

Hint

Apply power laws and change of base: $(2 \log_{5} b) (3 \log_{a} 5) = 6 (\frac{\log b}{\log 5} \cdot \frac{\log 5}{\log a}) = 6 \log_{a} b = \log_{a} (b^{6})$ .

Solution

Apply the power law of logarithms.
$\begin{matrix} \log_{5} b^{2} \cdot \log_{a} 5^{3} & = (2 \log_{5} b) \cdot (3 \log_{a} 5) \\ = 6 \cdot (\log_{5} b \cdot \log_{a} 5) \end{matrix}$
Apply change of base rule $\log_{x} y = \frac{\log y}{\log x}$ .
$\begin{matrix} = 6 \cdot (\frac{\log b}{\log 5} \cdot \frac{\log 5}{\log a}) \\ = 6 \cdot \frac{\log b}{\log a} \\ = 6 \log_{a} b \end{matrix}$
Express as a single logarithm.
$\begin{matrix} = \log_{a} (b^{6}) \end{matrix}$

Answer

$\log_{a} (b^{6})$

Q2Part (iv)

EasyApproved by Miss Aniya • Aug 24, 2026

$2 \log_{3} x + \log_{3} y$

Hint

Apply power law to get $\log_{3} (x^{2})$ , then product law: $\log_{3} (x^{2} y)$ .

Solution

Apply the power law of logarithms.
$\begin{matrix} 2 \log_{3} x + \log_{3} y & = \log_{3} (x^{2}) + \log_{3} y \end{matrix}$
Apply the product law of logarithms.
$\begin{matrix} = \log_{3} (x^{2} y) \end{matrix}$

Answer

$\log_{3} (x^{2} y)$

Q2Part (v)

MediumApproved by Miss Aniya • Aug 24, 2026

$4 \log_{5} x - \log_{5} y + \log_{5} z$

Hint

Combine positive terms in numerator and negative term in denominator: $\log_{5} (\frac{x^{4} z}{y})$ .

Solution

Apply the power law of logarithms.
$\begin{matrix} 4 \log_{5} x - \log_{5} y + \log_{5} z & = \log_{5} (x^{4}) - \log_{5} y + \log_{5} z \end{matrix}$
Group positive terms and apply product law.
$\begin{matrix} = [\log_{5} (x^{4}) + \log_{5} z] - \log_{5} y \\ = \log_{5} (x^{4} z) - \log_{5} y \end{matrix}$
Apply the quotient law of logarithms.
$\begin{matrix} = \log_{5} (\frac{x^{4} z}{y}) \end{matrix}$

Answer

$\log_{5} (\frac{x^{4} z}{y})$

Q2Part (vi)

MediumApproved by Miss Aniya • Aug 24, 2026

$2 \ln a + 3 \ln b - 4 \ln c$

Hint

Apply power laws, then product and quotient laws: $\ln (\frac{a^{2} b^{3}}{c^{4}})$ .

Solution

Apply the power law of logarithms.
$\begin{matrix} 2 \ln a + 3 \ln b - 4 \ln c & = \ln (a^{2}) + \ln (b^{3}) - \ln (c^{4}) \end{matrix}$
Combine terms using product and quotient laws.
$\begin{matrix} = \ln (a^{2} b^{3}) - \ln (c^{4}) \\ = \ln (\frac{a^{2} b^{3}}{c^{4}}) \end{matrix}$

Answer

$\ln (\frac{a^{2} b^{3}}{c^{4}})$

Q3Question 3

Medium

Expand the following using laws of logarithms:

Q3Part (i)

EasyApproved by Wasif • Aug 25, 2026

$\log (\frac{11}{5})$

Hint

Apply the quotient law: $\log (\frac{m}{n}) = \log m - \log n$ .

Solution

Apply the quotient law of logarithms.

According to the Quotient Law $\log_{a} (\frac{m}{n}) = \log_{a} m - \log_{a} n$ :

\begin{matrix} \log (\frac{11}{5}) & = \log 11 - \log 5 \end{matrix}

(Note: If evaluated numerically using log tables: $\log 11 - \log 5 \approx 1.0414 - 0.6990 = 0.3424$ ).

Answer

$\log 11 - \log 5 (\approx 0.3424)$

Q3Part (ii)

MediumApproved by Miss Aniya • Aug 24, 2026

$\log_{5} \sqrt{8 a^{6}}$

Hint

Convert radical to exponent $\frac{1}{2}$ , write $8 = 2^{3}$ , and apply power and product laws: $\frac{3}{2} \log_{5} 2 + 3 \log_{5} a$ .

Solution

Convert radical form to exponential form.
$\begin{matrix} \log_{5} \sqrt{8 a^{6}} & = \log_{5} (8 a^{6})^{\frac{1}{2}} \end{matrix}$
Apply the power law of logarithms.
$\begin{matrix} = \frac{1}{2} \log_{5} (8 a^{6}) \end{matrix}$
Apply product law and express $8 = 2^{3}$ .
$\begin{matrix} = \frac{1}{2} [\log_{5} (2^{3}) + \log_{5} (a^{6})] \\ = \frac{1}{2} [3 \log_{5} 2 + 6 \log_{5} a] \\ = \frac{3}{2} \log_{5} 2 + 3 \log_{5} a \end{matrix}$

Answer

$\frac{3}{2} \log_{5} 2 + 3 \log_{5} a$

Q3Part (iii)

EasyApproved by Miss Alisha • Aug 29, 2026

$\ln (\frac{a^{2} b}{c})$

Hint

Apply quotient law, product law, and power law: $2 \ln a + \ln b - \ln c$ .

Solution

Apply the quotient law of logarithms.
$\begin{matrix} \ln (\frac{a^{2} b}{c}) & = \ln (a^{2} b) - \ln c \end{matrix}$
Apply product and power laws.
$\begin{matrix} = \ln (a^{2}) + \ln b - \ln c \\ = 2 \ln a + \ln b - \ln c \end{matrix}$

Answer

$2 \ln a + \ln b - \ln c$

Q3Part (iv)

MediumApproved by Miss Alisha • Aug 29, 2026

$\log {(\frac{x y}{z})}^{\frac{1}{9}}$

Hint

Apply power law with exponent $\frac{1}{9}$ , then expand: $\frac{1}{9} \log x + \frac{1}{9} \log y - \frac{1}{9} \log z$ .

Solution

Apply the power law of logarithms.
$\begin{matrix} \log {(\frac{x y}{z})}^{\frac{1}{9}} & = \frac{1}{9} \log (\frac{x y}{z}) \end{matrix}$
Apply quotient and product laws.
$\begin{matrix} = \frac{1}{9} [\log (x y) - \log z] \\ = \frac{1}{9} [\log x + \log y - \log z] \\ = \frac{1}{9} \log x + \frac{1}{9} \log y - \frac{1}{9} \log z \end{matrix}$

Answer

$\frac{1}{9} \log x + \frac{1}{9} \log y - \frac{1}{9} \log z$

Q3Part (v)

MediumApproved by Wasif • Aug 30, 2026

$\ln \sqrt[3]{16 x^{3}}$

Hint

Convert cube root to exponent $\frac{1}{3}$ , write $16 = 2^{4}$ , and expand to $\frac{4}{3} \ln 2 + \ln x$ .

Solution

Convert cube root to exponential form.
$\begin{matrix} \ln \sqrt[3]{16 x^{3}} & = \ln (16 x^{3})^{\frac{1}{3}} \end{matrix}$
Apply the power law of logarithms.
$\begin{matrix} = \frac{1}{3} \ln (16 x^{3}) \end{matrix}$
Express $16 = 2^{4}$ and expand.
$\begin{matrix} = \frac{1}{3} [\ln (2^{4}) + \ln (x^{3})] \\ = \frac{1}{3} [4 \ln 2 + 3 \ln x] \\ = \frac{4}{3} \ln 2 + \frac{1}{3} (3 \ln x) \\ = \frac{4}{3} \ln 2 + \ln x \end{matrix}$

Answer

$\frac{4}{3} \ln 2 + \ln x$

Q3Part (vi)

MediumApproved by Miss Alisha • Aug 29, 2026

$\log_{2} {(\frac{1 - a}{b})}^{5}$

Hint

Apply power law with exponent $5$ , then expand: $5 \log_{2} (1 - a) - 5 \log_{2} b$ .

Solution

Apply the power law of logarithms.
$\begin{matrix} \log_{2} {(\frac{1 - a}{b})}^{5} & = 5 \log_{2} (\frac{1 - a}{b}) \end{matrix}$
Apply the quotient law of logarithms.
$\begin{matrix} = 5 [\log_{2} (1 - a) - \log_{2} b] \\ = 5 \log_{2} (1 - a) - 5 \log_{2} b \end{matrix}$

Answer

$5 \log_{2} (1 - a) - 5 \log_{2} b$

Q4Question 4

Hard

Find the value of $x$ in the following equations:

Q4Part (i)

EasyApproved by Miss Alisha • Aug 29, 2026

$\log 2 + \log x = 1$

Hint

Combine using product law: $\log (2 x) = 1 ⟹ 2 x = 10^{1} ⟹ x = 5$ .

Solution

Apply the product law of logarithms.
$\begin{matrix} \log 2 + \log x & = 1 \\ \log_{10} (2 x) & = 1 \end{matrix}$
Convert to exponential form.
$\begin{matrix} 2 x & = 10^{1} \\ 2 x & = 10 \end{matrix}$
Solve for $x$ .
$\begin{matrix} x & = \frac{10}{2} \\ x & = 5 \end{matrix}$

Answer

$x = 5$

Q4Part (ii)

EasyApproved by Miss Alisha • Aug 29, 2026

$\log_{2} x + \log_{2} 8 = 5$

Hint

Combine using product law: $\log_{2} (8 x) = 5 ⟹ 8 x = 2^{5} = 32 ⟹ x = 4$ .

Solution

Apply the product law of logarithms.
$\begin{matrix} \log_{2} x + \log_{2} 8 & = 5 \\ \log_{2} (8 x) & = 5 \end{matrix}$
Convert to exponential form.
$\begin{matrix} 8 x & = 2^{5} \\ 8 x & = 32 \end{matrix}$
Solve for $x$ .
$\begin{matrix} x & = \frac{32}{8} \\ x & = 4 \end{matrix}$

Answer

$x = 4$

Q4Part (iii)

MediumApproved by Miss Alisha • Aug 30, 2026

$(81)^{x} = (243)^{x + 2}$

Hint

Express $81 = 3^{4}$ and $243 = 3^{5}$ , then equate exponents: $4 x = 5 (x + 2)$ .

Solution

Express both bases as powers of $3$ .

Since $81 = 3^{4}$ and $243 = 3^{5}$ :
$\begin{matrix} (81)^{x} & = (243)^{x + 2} \\ (3^{4})^{x} & = (3^{5})^{x + 2} \end{matrix}$
Multiply exponents.
$\begin{matrix} 3^{4 x} & = 3^{5 (x + 2)} \\ 3^{4 x} & = 3^{5 x + 10} \end{matrix}$
Equate exponents and solve for $x$ .
$\begin{matrix} 4 x & = 5 x + 10 \\ 4 x - 5 x & = 10 \\ - x & = 10 \\ x & = - 10 \end{matrix}$

Answer

$x = - 10$

Q4Part (iv)

MediumApproved by Miss Alisha • Aug 30, 2026

${(\frac{1}{27})}^{x - 6} = 27$

Hint

Write $\frac{1}{27} = 27^{- 1}$ , then equate exponents: $- (x - 6) = 1 ⟹ - x + 6 = 1 ⟹ x = 5$ .

Solution

Express left-hand side with base $27$ .

Since $\frac{1}{27} = 27^{- 1}$ and $27 = 27^{1}$ :
$\begin{matrix} {(\frac{1}{27})}^{x - 6} & = 27 \\ (27^{- 1})^{x - 6} & = 27^{1} \end{matrix}$
Multiply exponents.
$\begin{matrix} 27^{- (x - 6)} & = 27^{1} \\ 27^{- x + 6} & = 27^{1} \end{matrix}$
Equate exponents and solve for $x$ .
$\begin{matrix} - x + 6 & = 1 \\ - x & = 1 - 6 \\ - x & = - 5 \\ x & = 5 \end{matrix}$

Answer

$x = 5$

Q4Part (v)

MediumApproved by Miss Alisha • Aug 29, 2026

$\log (5 x - 10) = 2$

Hint

Convert to exponential form: $5 x - 10 = 10^{2} = 100 ⟹ 5 x = 110 ⟹ x = 22$ .

Solution

Convert to exponential form (base 10).
$\begin{matrix} \log_{10} (5 x - 10) & = 2 \\ 5 x - 10 & = 10^{2} \\ 5 x - 10 & = 100 \end{matrix}$
Solve for $x$ .
$\begin{matrix} 5 x & = 100 + 10 \\ 5 x & = 110 \\ x & = \frac{110}{5} \\ x & = 22 \end{matrix}$

Answer

$x = 22$

Q4Part (vi)

HardApproved by Miss Alisha • Aug 30, 2026

$\log_{2} (x + 1) - \log_{2} (x - 4) = 2$

Hint

Combine using quotient law: $\log_{2} (\frac{x + 1}{x - 4}) = 2 ⟹ \frac{x + 1}{x - 4} = 2^{2} = 4$ .

Solution

Apply the quotient law of logarithms.
$\begin{matrix} \log_{2} (x + 1) - \log_{2} (x - 4) & = 2 \\ \log_{2} (\frac{x + 1}{x - 4}) & = 2 \end{matrix}$
Convert to exponential form.
$\begin{matrix} \frac{x + 1}{x - 4} & = 2^{2} \\ \frac{x + 1}{x - 4} & = 4 \end{matrix}$
Cross-multiply and solve for $x$ .
$\begin{matrix} x + 1 & = 4 (x - 4) \\ x + 1 & = 4 x - 16 \\ 1 + 16 & = 4 x - x \\ 17 & = 3 x \\ x & = \frac{17}{3} = 5 \frac{2}{3} \end{matrix}$

Answer

$x = \frac{17}{3} = 5 \frac{2}{3}$

Q5Question 5

Hard

Find the values of the following with the help of logarithm table:

Q5Part (i)

MediumApproved by Miss Alisha • Aug 29, 2026

$\frac{3.68 \times 4.21}{5.234}$

Hint

Take log on both sides: $\log x = \log (3.68) + \log (4.21) - \log (5.234)$ . Then find $x = antilog (\log x)$ .

Solution

Let $x = \frac{3.68 \times 4.21}{5.234}$ and take log on both sides.
$\begin{matrix} \log x & = \log (\frac{3.68 \times 4.21}{5.234}) \\ \log x & = \log (3.68) + \log (4.21) - \log (5.234) \end{matrix}$
Find logarithms using log table.
- $\log (3.68) = 0.5658$
- $\log (4.21) = 0.6243$
- $\log (5.234) = 0.7188$
Calculate $\log x$ .
$\begin{matrix} \log x & = 0.5658 + 0.6243 - 0.7188 \\ = 1.1901 - 0.7188 \\ = 0.4713 \end{matrix}$
Take antilogarithm on both sides.
$\begin{matrix} x & = antilog (0.4713) \\ x & = 2.960 \end{matrix}$

Answer

$x = 2.960$

Q5Part (ii)

MediumApproved by Miss Alisha • Aug 29, 2026

$4.67 \times 2.11 \times 2.397$

Hint

Take log on both sides: $\log x = \log (4.67) + \log (2.11) + \log (2.397)$ . Then find $x = antilog (\log x)$ .

Solution

Let $x = 4.67 \times 2.11 \times 2.397$ and apply log.
$\begin{matrix} \log x & = \log (4.67) + \log (2.11) + \log (2.397) \end{matrix}$
Find logarithms from table.
- $\log (4.67) = 0.6693$
- $\log (2.11) = 0.3243$
- $\log (2.397) = 0.3797$
Sum the logarithms.
$\begin{matrix} \log x & = 0.6693 + 0.3243 + 0.3797 \\ = 1.3733 \end{matrix}$
Take antilogarithm.
$\begin{matrix} x & = antilog (1.3733) \\ x & = 23.62 \end{matrix}$

Answer

$x = 23.62$

Q5Part (iii)

HardApproved by Miss Alisha • Aug 29, 2026

$\frac{(20.46)^{2} \times (2.4122)}{754.3}$

Hint

Apply log laws: $\log x = 2 \log (20.46) + \log (2.4122) - \log (754.3)$ . Then find antilog.

Solution

Let $x = \frac{(20.46)^{2} \times 2.4122}{754.3}$ and take log on both sides.
$\begin{matrix} \log x & = 2 \log (20.46) + \log (2.4122) - \log (754.3) \end{matrix}$
Find logarithms from table.
- $\log (20.46) = 1.3109 ⟹ 2 \log (20.46) = 2.6218$
- $\log (2.4122) = 0.3824$
- $\log (754.3) = 2.8776$
Compute $\log x$ .
$\begin{matrix} \log x & = 2.6218 + 0.3824 - 2.8776 \\ = 3.0042 - 2.8776 \\ = 0.1266 \end{matrix}$
Take antilogarithm.
$\begin{matrix} x & = antilog (0.1266) \\ x & = 1.339 \end{matrix}$

Answer

$x = 1.339$

Q5Part (iv)

HardApproved by Miss Alisha • Aug 29, 2026

$\frac{\sqrt[3]{9.364} \times 21.64}{3.21}$

Hint

Apply log laws: $\log x = \frac{1}{3} \log (9.364) + \log (21.64) - \log (3.21)$ . Then find antilog.

Solution

Let $x = \frac{(9.364)^{1 / 3} \times 21.64}{3.21}$ and take log on both sides.
$\begin{matrix} \log x & = \frac{1}{3} \log (9.364) + \log (21.64) - \log (3.21) \end{matrix}$
Find logarithms from table.
- $\log (9.364) = 0.9715 ⟹ \frac{1}{3} (0.9715) = 0.3238$
- $\log (21.64) = 1.3353$
- $\log (3.21) = 0.5065$
Calculate $\log x$ .
$\begin{matrix} \log x & = 0.3238 + 1.3353 - 0.5065 \\ = 1.6591 - 0.5065 \\ = 1.1526 \end{matrix}$
Take antilogarithm.
$\begin{matrix} x & = antilog (1.1526) \\ x & = 14.21 \end{matrix}$

Answer

$x = 14.21$

Question 6

MediumApproved by Miss Alisha • Aug 29, 2026

The formula to measure the magnitude of earthquakes is given by $M = \log_{10} (\frac{A}{A_{0}})$ . If amplitude ( $A$ ) is $10,000$ and reference amplitude ( $A_{0}$ ) is $10$ . What is the magnitude of the earthquake?

Hint

Substitute $A = 10,000$ and $A_{0} = 10$ . Then $M = \log_{10} (1000) = \log_{10} (10^{3}) = 3$ .

Solution

Identify given data.
- Wave amplitude ( $A$ ) $= 10,000$
- Reference amplitude ( $A_{0}$ ) $= 10$
Substitute into the magnitude formula.
$\begin{matrix} M & = \log_{10} (\frac{A}{A_{0}}) \\ = \log_{10} (\frac{10,000}{10}) \\ = \log_{10} (1000) \end{matrix}$
Evaluate using power law $1000 = 10^{3}$ .
$\begin{matrix} M & = \log_{10} (10^{3}) \\ = 3 \log_{10} 10 \\ = 3 (1) \\ = 3 \end{matrix}$

Answer

$M = 3$

Question 7

HardApproved by Miss Alisha • Aug 29, 2026

Abdullah invested Rs. 100,000 in a saving scheme and gains interest at the rate of 5% per annum so that the total value of this investment after $t$ years is Rs $y$ . This is modelled by an equation $y = 100,000 (1.05)^{t}, t \geq 0$ . Find after how many years the investment will be double.

Hint

Set $y = 200,000$ (doubled) $⟹ (1.05)^{t} = 2$ . Take log: $t = \frac{\log 2}{\log 1.05}$ .

Solution

Set up the doubling condition.

Initial investment $= Rs. 100,000$ . Doubled value ( $y$ ) $= 2 \times 100,000 = Rs. 200,000$ .
$\begin{matrix} 200,000 & = 100,000 (1.05)^{t} \\ \frac{200,000}{100,000} & = (1.05)^{t} \\ (1.05)^{t} & = 2 \end{matrix}$
Take common logarithm on both sides.
$\begin{matrix} \log_{10} ({1.05}^{t}) & = \log_{10} (2) \\ t \cdot \log_{10} (1.05) & = \log_{10} (2) \\ t & = \frac{\log_{10} (2)}{\log_{10} (1.05)} \end{matrix}$
Substitute logarithm values from tables.
- $\log_{10} (2) \approx 0.3010$
- $\log_{10} (1.05) \approx 0.0212$
t=0.30100.0212t≈14.2 years

Answer

$t \approx 14.2 years$

Question 8

MediumApproved by Miss Alisha • Aug 29, 2026

Huria is hiking up a mountain where the temperature (T) decreases by 3% (or a factor of 0.97) for every 100 metres gained in altitude. The initial temperature ( $T_{i}$ ) at sea level is $20^{\circ} C$ . Using the formula $T = T_{i} \times {0.97}^{\frac{h}{100}}$ , calculate the temperature at an altitude ( $h$ ) of 500 metres.

Hint

Substitute $h = 500 ⟹ \frac{h}{100} = 5$ . Calculate $T = 20 \times (0.97)^{5}$ using logarithms: $\log T = \log 20 + 5 \log (0.97)$ .

Solution

Substitute given parameters into formula.
- $T_{i} = 20^{\circ} C$
- $h = 500 m ⟹ \frac{h}{100} = \frac{500}{100} = 5$
T=20×(0.97)5
Take logarithm on both sides.
$\begin{matrix} \log T & = \log (20) + 5 \log (0.97) \end{matrix}$
Substitute table values.
- $\log (20) = 1.3010$
- $\log (0.97) = \overline{1} .9868 = - 0.0132$
log⁡T=1.3010+5(−0.0132)=1.3010−0.0660=1.2350
Take antilogarithm.
$\begin{matrix} T & = antilog (1.2350) \\ T & \approx {17.18}^{\circ} C \end{matrix}$

Answer

$T \approx {17.18}^{\circ} C (or {17.2}^{\circ} C)$

The exercises and question numbering reproduced on these pages are from Mathematics for Class 9 (National Curriculum of Pakistan 2023), published by the Punjab Curriculum and Textbook Board (PCTB), Lahore, authored by Muhammad Akhtar Shirani, Madiha Mahmood, and Ghulam Murtaza. PCTB holds the copyright in the original textbook. PrepSure is not affiliated with, endorsed by, or sponsored by PCTB. The worked solutions, explanations, hints and method notes are PrepSure's own original work, written and reviewed by our team. If you hold rights in this material and believe anything here exceeds fair use, write to us and we will take it down.

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