Exercise 2.4 — Class 9 Mathematics Solutions

Solution

1. Apply the quotient law of logarithms.

   $$\begin{aligned} \log_2 18 - \log_2 9 &= \log_2 \left(\frac{18}{9}\right) \end{aligned}$$

2. Simplify the fraction.

   $$\begin{aligned} &= \log_2 (2) \end{aligned}$$

3. Evaluate using base identity $\log_a a = 1$.

   $$\begin{aligned} &= 1 \end{aligned}$$

Solution

1. Express arguments as powers of base $2$.

   Since $64 = 2^6$:

   $$\begin{aligned} \log_2 64 + \log_2 2 &= \log_2 (2^6) + \log_2 2 \end{aligned}$$

2. Apply the power law of logarithms.

   $$\begin{aligned} &= 6\log_2 2 + \log_2 2 \end{aligned}$$

3. Substitute the base identity $\log_2 2 = 1$.

   $$\begin{aligned} &= 6(1) + 1 \\ &= 6 + 1 \\ &= 7 \end{aligned}$$

Solution

1. Apply the power law to the first term.

   $$\begin{aligned} \frac{1}{3}\log_3 8 - \log_3 18 &= \log_3 \left(8^{1/3}\right) - \log_3 18 \end{aligned}$$

2. Simplify $8^{1/3} = (2^3)^{1/3} = 2$.

   $$\begin{aligned} &= \log_3 2 - \log_3 18 \end{aligned}$$

3. Apply the quotient law of logarithms.

   $$\begin{aligned} &= \log_3 \left(\frac{2}{18}\right) \\ &= \log_3 \left(\frac{1}{9}\right) \end{aligned}$$

4. Express as a power of base $3$ and evaluate.

   $$\begin{aligned} &= \log_3 (3^{-2}) \\ &= -2\log_3 3 \\ &= -2(1) \\ &= -2 \end{aligned}$$

Solution

1. Apply the power law to the first term.

   $$\begin{aligned} 2\log 2 + \log 25 &= \log(2^2) + \log 25 \\ &= \log 4 + \log 25 \end{aligned}$$

2. Apply the product law of logarithms.

   $$\begin{aligned} &= \log(4 \times 25) \\ &= \log 100 \end{aligned}$$

3. Evaluate base 10 logarithm.

   $$\begin{aligned} &= \log_{10}(10^2) \\ &= 2\log_{10} 10 \\ &= 2(1) \\ &= 2 \end{aligned}$$

Solution

1. Express arguments as powers of their bases.

   Since $64 = 4^3$ and $25 = 5^2$:

   $$\begin{aligned} \frac{1}{3}\log_4 64 + 2\log_5 25 &= \frac{1}{3}\log_4 (4^3) + 2\log_5 (5^2) \end{aligned}$$

2. Apply the power law to each term.

   $$\begin{aligned} &= \frac{1}{3}(3\log_4 4) + 2(2\log_5 5) \\ &= 1\log_4 4 + 4\log_5 5 \end{aligned}$$

3. Substitute $\log_4 4 = 1$ and $\log_5 5 = 1$.

   $$\begin{aligned} &= 1(1) + 4(1) \\ &= 1 + 4 \\ &= 5 \end{aligned}$$

Solution

1. Convert decimal to fraction.

   Since $0.25 = \frac{1}{4}$:

   $$\begin{aligned} \log_3 12 + \log_3 0.25 &= \log_3 12 + \log_3 \left(\frac{1}{4}\right) \end{aligned}$$

2. Apply the product law of logarithms.

   $$\begin{aligned} &= \log_3 \left(12 \times \frac{1}{4}\right) \\ &= \log_3 (3) \end{aligned}$$

3. Apply the base identity.

   $$\begin{aligned} &= 1 \end{aligned}$$

Solution

1. Apply the power law of logarithms.

   $$\begin{aligned} \frac{1}{2}\log 25 + 2\log 3 &= \log\left(25^{\frac{1}{2}}\right) + \log(3^2) \end{aligned}$$

2. Simplify powers.

   $$\begin{aligned} &= \log 5 + \log 9 \end{aligned}$$

3. Apply the product law of logarithms.

   $$\begin{aligned} &= \log(5 \times 9) \\ &= \log 45 \end{aligned}$$

Solution

1. Apply the quotient law of logarithms.

   $$\begin{aligned} \log 9 - \log\left(\frac{1}{3}\right) &= \log\left(\frac{9}{\frac{1}{3}}\right) \end{aligned}$$

2. Simplify the fraction.

   $$\begin{aligned} &= \log(9 \times 3) \\ &= \log 27 \end{aligned}$$

Solution

1. Apply the power law of logarithms.

   $$\begin{aligned} \log_5 b^2 \cdot \log_a 5^3 &= (2\log_5 b) \cdot (3\log_a 5) \\ &= 6 \cdot (\log_5 b \cdot \log_a 5) \end{aligned}$$

2. Apply change of base rule $\log_x y = \frac{\log y}{\log x}$.

   $$\begin{aligned} &= 6 \cdot \left(\frac{\log b}{\log 5} \cdot \frac{\log 5}{\log a}\right) \\ &= 6 \cdot \frac{\log b}{\log a} \\ &= 6\log_a b \end{aligned}$$

3. Express as a single logarithm.

   $$\begin{aligned} &= \log_a(b^6) \end{aligned}$$

Solution

1. Apply the power law of logarithms.

   $$\begin{aligned} 2\log_3 x + \log_3 y &= \log_3(x^2) + \log_3 y \end{aligned}$$

2. Apply the product law of logarithms.

   $$\begin{aligned} &= \log_3(x^2 y) \end{aligned}$$

Solution

1. Apply the power law of logarithms.

   $$\begin{aligned} 4\log_5 x - \log_5 y + \log_5 z &= \log_5(x^4) - \log_5 y + \log_5 z \end{aligned}$$

2. Group positive terms and apply product law.

   $$\begin{aligned} &= [\log_5(x^4) + \log_5 z] - \log_5 y \\ &= \log_5(x^4 z) - \log_5 y \end{aligned}$$

3. Apply the quotient law of logarithms.

   $$\begin{aligned} &= \log_5\left(\frac{x^4 z}{y}\right) \end{aligned}$$

Solution

1. Apply the power law of logarithms.

   $$\begin{aligned} 2\ln a + 3\ln b - 4\ln c &= \ln(a^2) + \ln(b^3) - \ln(c^4) \end{aligned}$$

2. Combine terms using product and quotient laws.

   $$\begin{aligned} &= \ln(a^2 b^3) - \ln(c^4) \\ &= \ln\left(\frac{a^2 b^3}{c^4}\right) \end{aligned}$$

Solution

1. Apply the quotient law of logarithms.

   According to the Quotient Law $\log_a\left(\frac{m}{n}\right) = \log_a m - \log_a n$:

   $$\begin{aligned} \log\left(\frac{11}{5}\right) &= \log 11 - \log 5 \end{aligned}$$

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

Solution

1. Convert radical form to exponential form.

   $$\begin{aligned} \log_5 \sqrt{8a^6} &= \log_5 (8a^6)^{\frac{1}{2}} \end{aligned}$$

2. Apply the power law of logarithms.

   $$\begin{aligned} &= \frac{1}{2}\log_5 (8a^6) \end{aligned}$$

3. Apply product law and express $8 = 2^3$.

   $$\begin{aligned} &= \frac{1}{2}[\log_5(2^3) + \log_5(a^6)] \\ &= \frac{1}{2}[3\log_5 2 + 6\log_5 a] \\ &= \frac{3}{2}\log_5 2 + 3\log_5 a \end{aligned}$$

Solution

1. Apply the quotient law of logarithms.

   $$\begin{aligned} \ln\left(\frac{a^2 b}{c}\right) &= \ln(a^2 b) - \ln c \end{aligned}$$

2. Apply product and power laws.

   $$\begin{aligned} &= \ln(a^2) + \ln b - \ln c \\ &= 2\ln a + \ln b - \ln c \end{aligned}$$

Solution

1. Apply the power law of logarithms.

   $$\begin{aligned} \log\left(\frac{xy}{z}\right)^{\frac{1}{9}} &= \frac{1}{9}\log\left(\frac{xy}{z}\right) \end{aligned}$$

2. Apply quotient and product laws.

   $$\begin{aligned} &= \frac{1}{9}[\log(xy) - \log z] \\ &= \frac{1}{9}[\log x + \log y - \log z] \\ &= \frac{1}{9}\log x + \frac{1}{9}\log y - \frac{1}{9}\log z \end{aligned}$$

Solution

1. Convert cube root to exponential form.

   $$\begin{aligned} \ln\sqrt[3]{16x^3} &= \ln(16x^3)^{\frac{1}{3}} \end{aligned}$$

2. Apply the power law of logarithms.

   $$\begin{aligned} &= \frac{1}{3}\ln(16x^3) \end{aligned}$$

3. Express $16 = 2^4$ and expand.

   $$\begin{aligned} &= \frac{1}{3}[\ln(2^4) + \ln(x^3)] \\ &= \frac{1}{3}[4\ln 2 + 3\ln x] \\ &= \frac{4}{3}\ln 2 + \frac{1}{3}(3\ln x) \\ &= \frac{4}{3}\ln 2 + \ln x \end{aligned}$$

Solution

1. Apply the power law of logarithms.

   $$\begin{aligned} \log_2\left(\frac{1-a}{b}\right)^5 &= 5\log_2\left(\frac{1-a}{b}\right) \end{aligned}$$

2. Apply the quotient law of logarithms.

   $$\begin{aligned} &= 5[\log_2(1-a) - \log_2 b] \\ &= 5\log_2(1-a) - 5\log_2 b \end{aligned}$$

Solution

1. Apply the product law of logarithms.

   $$\begin{aligned} \log 2 + \log x &= 1 \\ \log_{10}(2x) &= 1 \end{aligned}$$

2. Convert to exponential form.

   $$\begin{aligned} 2x &= 10^1 \\ 2x &= 10 \end{aligned}$$

3. Solve for $x$.

   $$\begin{aligned} x &= \frac{10}{2} \\ x &= 5 \end{aligned}$$

Solution

1. Apply the product law of logarithms.

   $$\begin{aligned} \log_2 x + \log_2 8 &= 5 \\ \log_2(8x) &= 5 \end{aligned}$$

2. Convert to exponential form.

   $$\begin{aligned} 8x &= 2^5 \\ 8x &= 32 \end{aligned}$$

3. Solve for $x$.

   $$\begin{aligned} x &= \frac{32}{8} \\ x &= 4 \end{aligned}$$

Solution

1. Express both bases as powers of $3$.

   Since $81 = 3^4$ and $243 = 3^5$:

   $$\begin{aligned} (81)^x &= (243)^{x+2} \\ (3^4)^x &= (3^5)^{x+2} \end{aligned}$$

2. Multiply exponents.

   $$\begin{aligned} 3^{4x} &= 3^{5(x+2)} \\ 3^{4x} &= 3^{5x + 10} \end{aligned}$$

3. Equate exponents and solve for $x$.

   $$\begin{aligned} 4x &= 5x + 10 \\ 4x - 5x &= 10 \\ -x &= 10 \\ x &= -10 \end{aligned}$$

Solution

1. Express left-hand side with base $27$.

   Since $\frac{1}{27} = 27^{-1}$ and $27 = 27^1$:

   $$\begin{aligned} \left(\frac{1}{27}\right)^{x-6} &= 27 \\ (27^{-1})^{x-6} &= 27^1 \end{aligned}$$

2. Multiply exponents.

   $$\begin{aligned} 27^{-(x-6)} &= 27^1 \\ 27^{-x + 6} &= 27^1 \end{aligned}$$

3. Equate exponents and solve for $x$.

   $$\begin{aligned} -x + 6 &= 1 \\ -x &= 1 - 6 \\ -x &= -5 \\ x &= 5 \end{aligned}$$

Solution

1. Convert to exponential form (base 10).

   $$\begin{aligned} \log_{10}(5x - 10) &= 2 \\ 5x - 10 &= 10^2 \\ 5x - 10 &= 100 \end{aligned}$$

2. Solve for $x$.

   $$\begin{aligned} 5x &= 100 + 10 \\ 5x &= 110 \\ x &= \frac{110}{5} \\ x &= 22 \end{aligned}$$

Solution

1. Apply the quotient law of logarithms.

   $$\begin{aligned} \log_2(x+1) - \log_2(x-4) &= 2 \\ \log_2\left(\frac{x+1}{x-4}\right) &= 2 \end{aligned}$$

2. Convert to exponential form.

   $$\begin{aligned} \frac{x+1}{x-4} &= 2^2 \\ \frac{x+1}{x-4} &= 4 \end{aligned}$$

3. Cross-multiply and solve for $x$.

   $$\begin{aligned} x + 1 &= 4(x - 4) \\ x + 1 &= 4x - 16 \\ 1 + 16 &= 4x - x \\ 17 &= 3x \\ x &= \frac{17}{3} = 5\frac{2}{3} \end{aligned}$$

Solution

1. Let $x = \frac{3.68 \times 4.21}{5.234}$ and take log on both sides.

   $$\begin{aligned} \log x &= \log\left(\frac{3.68 \times 4.21}{5.234}\right) \\ \log x &= \log(3.68) + \log(4.21) - \log(5.234) \end{aligned}$$

2. Find logarithms using log table.

   • $\log(3.68) = 0.5658$
   • $\log(4.21) = 0.6243$
   • $\log(5.234) = 0.7188$

3. Calculate $\log x$.

   $$\begin{aligned} \log x &= 0.5658 + 0.6243 - 0.7188 \\ &= 1.1901 - 0.7188 \\ &= 0.4713 \end{aligned}$$

4. Take antilogarithm on both sides.

   $$\begin{aligned} x &= \text{antilog}(0.4713) \\ x &= 2.960 \end{aligned}$$

Solution

1. Let $x = 4.67 \times 2.11 \times 2.397$ and apply log.

   $$\begin{aligned} \log x &= \log(4.67) + \log(2.11) + \log(2.397) \end{aligned}$$

2. Find logarithms from table.

   • $\log(4.67) = 0.6693$
   • $\log(2.11) = 0.3243$
   • $\log(2.397) = 0.3797$

3. Sum the logarithms.

   $$\begin{aligned} \log x &= 0.6693 + 0.3243 + 0.3797 \\ &= 1.3733 \end{aligned}$$

4. Take antilogarithm.

   $$\begin{aligned} x &= \text{antilog}(1.3733) \\ x &= 23.62 \end{aligned}$$

Solution

1. Let $x = \frac{(20.46)^2 \times 2.4122}{754.3}$ and take log on both sides.

   $$\begin{aligned} \log x &= 2\log(20.46) + \log(2.4122) - \log(754.3) \end{aligned}$$

2. Find logarithms from table.

   • $\log(20.46) = 1.3109 \implies 2\log(20.46) = 2.6218$
   • $\log(2.4122) = 0.3824$
   • $\log(754.3) = 2.8776$

3. Compute $\log x$.

   $$\begin{aligned} \log x &= 2.6218 + 0.3824 - 2.8776 \\ &= 3.0042 - 2.8776 \\ &= 0.1266 \end{aligned}$$

4. Take antilogarithm.

   $$\begin{aligned} x &= \text{antilog}(0.1266) \\ x &= 1.339 \end{aligned}$$

Solution

1. Let $x = \frac{(9.364)^{1/3} \times 21.64}{3.21}$ and take log on both sides.

   $$\begin{aligned} \log x &= \frac{1}{3}\log(9.364) + \log(21.64) - \log(3.21) \end{aligned}$$

2. Find logarithms from table.

   • $\log(9.364) = 0.9715 \implies \frac{1}{3}(0.9715) = 0.3238$
   • $\log(21.64) = 1.3353$
   • $\log(3.21) = 0.5065$

3. Calculate $\log x$.

   $$\begin{aligned} \log x &= 0.3238 + 1.3353 - 0.5065 \\ &= 1.6591 - 0.5065 \\ &= 1.1526 \end{aligned}$$

4. Take antilogarithm.

   $$\begin{aligned} x &= \text{antilog}(1.1526) \\ x &= 14.21 \end{aligned}$$

Solution

1. Identify given data.

   • Wave amplitude ($A$) $= 10{,}000$
   • Reference amplitude ($A_0$) $= 10$

2. Substitute into the magnitude formula.

   $$\begin{aligned} M &= \log_{10}\left(\frac{A}{A_0}\right) \\ &= \log_{10}\left(\frac{10{,}000}{10}\right) \\ &= \log_{10}(1000) \end{aligned}$$

3. Evaluate using power law $1000 = 10^3$.

   $$\begin{aligned} M &= \log_{10}(10^3) \\ &= 3\log_{10} 10 \\ &= 3(1) \\ &= 3 \end{aligned}$$