Class 9: Logarithms – Exercise 8.1 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Write the following in the from of logarithms:

(i) $\displaystyle 2^6 = 64 \Longleftrightarrow \log_2 64 = 6$

(ii) $\displaystyle 10^4 = 10000 \Longleftrightarrow \log_{10} 10000 = 4$

(iii) $\displaystyle 3^5 = 243 \Longleftrightarrow \log_3 243 = 5$

(iv) $\displaystyle 3^{-3} = \frac{1}{27} \Longleftrightarrow \log_3 {\frac{1}{27}} = -3$

(v) $\displaystyle 10^{-3} = 0.001 \Longleftrightarrow \log_{10} 0.001 = -3$

(vi) $\displaystyle 7^2 = 49 \Longleftrightarrow \log_7 49 = 2$

(vii) $\displaystyle 2^{-6} = \frac{1}{64} \Longleftrightarrow \log_2 {\frac{1}{64}} = -6$

(viii) $\displaystyle 2^{\frac{3}{2}} = 8 \Longleftrightarrow \log_4 8 = \frac{3}{2}$

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Question 2: Find the value of $\displaystyle x$ :

(i) $\displaystyle \log_3 x = 4 \hspace{0.5cm} \Rightarrow x = 3^4 \hspace{0.5cm} \Rightarrow x = 81$

(ii) $\displaystyle \log_4 x = 3 \hspace{0.5cm} \Rightarrow x = 4^3 \hspace{0.5cm} \Rightarrow x = 64$

(iii) $\displaystyle \log_{\sqrt{3}} x = 4 \hspace{0.5cm} \Rightarrow x = (\sqrt{3})^4 \hspace{0.5cm} \Rightarrow x = 9$

(iv) $\displaystyle \log_{10} x = -3 \hspace{0.5cm} \Rightarrow x = 10^{-3} \hspace{0.5cm} \Rightarrow x = 0.001$

(v) $\displaystyle \log_4 x = 1.5 \hspace{0.5cm} \Rightarrow x = 4^{1.5} \hspace{0.5cm} \Rightarrow x = 8$

(vi) $\displaystyle \log_8 x = \frac{2}{3} \hspace{0.5cm} \Rightarrow x = 8^{\frac{2}{3}} \hspace{0.5cm} \Rightarrow x = 4$

(vii) $\displaystyle \log_{125} x = \frac{1}{6} \hspace{0.5cm} \Rightarrow x = 125^{\frac{1}{6}} \hspace{0.5cm} \Rightarrow x = \sqrt{5}$

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Question 3: Solve for $\displaystyle x$ :

(i) $\displaystyle \log_{\sqrt{5}} x = 4 \hspace{0.5cm} \Rightarrow (\sqrt{5})^4 = x \hspace{0.5cm} \Rightarrow x = 25$

(ii) $\displaystyle \log_{x} 0.0001 = -4 \hspace{0.5cm} \Rightarrow x^{-4} = 0.0001 \hspace{0.5cm} \Rightarrow x = 10$

(iii) $\displaystyle \log_{\sqrt{3}} (x-1) = 2 \hspace{0.5cm} \Rightarrow (\sqrt{3})^2 = (x-1) \hspace{0.5cm} \Rightarrow x = 4$

(iv) $\displaystyle \log_{3} (x^2+5) = 2 \hspace{0.5cm} \Rightarrow 3^{2} = x^2+5 \hspace{0.5cm} \Rightarrow x = \pm 2$

(v) $\displaystyle \log_{10} (3x-2) = 1 \hspace{0.5cm} \Rightarrow 10^{1} = 3x-2 \hspace{0.5cm} \Rightarrow x = 4$

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Question 4: Express the following in exponential form:

(i) $\displaystyle \log_5 25 = 2 \hspace{0.5cm} \Rightarrow 5^2 = 25$

(ii) $\displaystyle \log_4 64 = 3 \hspace{0.5cm} \Rightarrow 4^3 = 64$

(iii) $\displaystyle \log_{10} 0.001 = -3 \hspace{0.5cm} \Rightarrow 10^{-3} = 0.001$

(iv) $\displaystyle \log_{10} 1000 = 3 \hspace{0.5cm} \Rightarrow 10^3 = 1000$

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Question 5: Find the value:

(i) $\displaystyle \log_{\sqrt{3}} 3\sqrt{3} - \log_5 (0.25) = \log_{\sqrt{3}} 3\sqrt{3} - \log_5 (0.5)^2 = 3 + 2 \log_5 5 = 5$

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Question 6: If $\displaystyle \log_2 y = x$ , find the value of $\displaystyle 8^x$ in terms of $\displaystyle y$ .

Answer:

$\displaystyle \log_2 y = x \hspace{0.5cm} \Rightarrow 2^x = y$

$\displaystyle \text{Therefore } 2^{3x} = y^3 \hspace{0.5cm} \Rightarrow 8^x = y^3$

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Question 7: If $\displaystyle \log_{10} x = a$ , find the value of $\displaystyle 10^{2a-1}$ in terms of $\displaystyle x$ .

Answer:

$\displaystyle \log_{10} x = a \hspace{0.5cm} \Rightarrow 10^a = x$

$\displaystyle \text{Therefore } 10^{2a-1} = \frac{10^{2a}}{10} = \frac{x^2}{10} = 0.1x^2$

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Question 8: Given $\displaystyle \log_{10} x = a, \ \log_{10} y = b$ and $\displaystyle \log_{10} z = c$ , write down $\displaystyle 10^{2a-1}$ in terms of $\displaystyle x$ and $\displaystyle 10^{3b-1}$ in terms of $\displaystyle y$ . If $\displaystyle \log_{10} u = 2a + \frac{b}{2} -3c$ , express $\displaystyle u$ in terms of $\displaystyle x, \ y$ and $\displaystyle z$ .

Answer:

$\displaystyle \log_{10} x = a \hspace{0.5cm} \Rightarrow 10^a = x$

$\displaystyle \log_{10} y = b \hspace{0.5cm} \Rightarrow 10^b = y$

$\displaystyle \log_{10} z = c \hspace{0.5cm} \Rightarrow 10^c = z$

$\displaystyle \text{Therefore } 10^{2a-3} = \frac{1}{1000} . 10^{2a} = \frac{x^2}{1000}$

$\displaystyle 10^{3b-1} = \frac{1}{10} . 10^{3b} = \frac{y^3}{10}$

$\displaystyle \log_{10} u = 2a + \frac{b}{2} -3c$

$\displaystyle = 2 \log_{10} x + \frac{1}{2} \log_{10} y - 3 \log_{10} z$

$\displaystyle =\log_{10} x^2 + \log_{10}y^{\frac{1}{2}} - \log_{10} z^3$

$\displaystyle = \log_{10} \frac{x^2 \sqrt{y}}{z^3}$

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Question 9: If $\displaystyle \log_{10} x = 2a, \ 2\log_{10} y = b$ , then write $\displaystyle 10^a$ in terms of $\displaystyle x, 10^{2b-1}$ in terms of $\displaystyle y$ . Also, if $\displaystyle \log_{10} z = 3a - 2b$ , express $\displaystyle z$ in terms of $\displaystyle x$ and $\displaystyle y$ .

Answer:

$\displaystyle \log_{10} x = 2a \hspace{0.5cm} \Rightarrow 10^{2a} = x$

$\displaystyle 2 \log_{10} y = b \hspace{0.5cm} \Rightarrow \log_{10} y^2 = b \hspace{0.5cm} \Rightarrow 10^b = y^2$

$\displaystyle 10^a = (10^{2a})^{\frac{1}{2}} = x^{\frac{1}{2}}$

$\displaystyle 10^{2b+1} = 10. 10^{2b} = 10. y^4$

$\displaystyle \log_{10} z = 3a - 2b = \frac{3}{2} \log_{10} x - 4 \log_{10} y$

$\displaystyle = \log_{10} x^{\frac{3}{2}} - \log_{10} y^4$

$\displaystyle = \log_{10} \frac{x^{\frac{3}{2}}}{y^4}$

$\displaystyle \text{Therefore } z = \frac{x^{\frac{3}{4}}}{y^4}$

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Question 10: If $\displaystyle x = \log_{10} a$ and $\displaystyle y = \log_{10} b$ , express $\displaystyle \frac{a^3}{b^2}$ in terms of $\displaystyle x$ and $\displaystyle y$ .

Answer:

$\displaystyle \log_{10} a = x \hspace{0.5cm} \Rightarrow 10^x = a$

$\displaystyle \log_{10} b = y \hspace{0.5cm} \Rightarrow 10^y = b$

$\displaystyle \text{Therefore } \frac{a^3}{b^2} = \frac{10^{3x}}{10^{2y}} = 10^{3x-2y}$

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Question 11: If $\displaystyle \log_{10} x = a$ and $\displaystyle \log_{10} y = b$ , find the value of $\displaystyle xy$ .

Answer:

$\displaystyle \log_{10} x = a \hspace{0.5cm} \Rightarrow 10^a = x$

$\displaystyle \log_{10} y = b \hspace{0.5cm} \Rightarrow 10^b = y$

$\displaystyle \text{Therefore } xy = 10^a. 10^b = 10^{a+b}$

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Question 12: If $\displaystyle \log_{10} x = y$ , express $\displaystyle 10^{2y - 3}$ in terms of $\displaystyle x$ .

Answer:

$\displaystyle \log_{10} x = y \hspace{0.5cm} \Rightarrow 10^y = x$

$\displaystyle \text{Therefore } 10^{2y-3} = \frac{10^{2y}}{1000} = \frac{x^2}{1000}$

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Question 13: If $\displaystyle \log_2 x = a$ and $\displaystyle \log_3 y = a$ , find $\displaystyle 12^{2a-1}$ in terms of $\displaystyle x$ and $\displaystyle y$ .

Answer:

$\displaystyle \log_2 x = a \hspace{0.5cm} \Rightarrow 2^a = x$

$\displaystyle \log_3 y = a \hspace{0.5cm} \Rightarrow 3^a = y$

$\displaystyle \text{Therefore } 12^{2a-1} = \frac{1}{12} (12^{2a}) = \frac{1}{12} (3 \times 4)^{2a} = \frac{1}{12} 3^{2a}. 4^{2a} = \frac{1}{12} y^2x^4$

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ICSE / ISC / CBSE Mathematics Portal for K12 Students

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Class 9: Logarithms – Exercise 8.1

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