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RS Aggarwal Summary: logarithms
Introduction
Logarithms form a critical component of quantitative aptitude sections in competitive examinations. This chapter is frequently tested through direct formula-based questions, simplification problems, and complex numerical reasoning. Understanding logarithmic properties and their applications is essential for solving problems quickly and accurately. Questions from this chapter typically carry significant weightage and test both conceptual clarity and computational speed.
1. Definition of Logarithm
If \(a\) is a positive real number, other than 1, and \(a^m = x\), then we write:
\[m = \log_a x\]
This reads as "the logarithm of \(x\) to the base \(a\) is \(m\)."
Examples:
- \(10^3 = 1000 \Rightarrow \log_{10} 1000 = 3\)
- \(3^4 = 81 \Rightarrow \log_3 81 = 4\)
- \(2^{-3} = \frac{1}{8} \Rightarrow \log_2\left(\frac{1}{8}\right) = -3\)
- \((0.1)^2 = 0.01 \Rightarrow \log_{0.1}(0.01) = 2\)
Solved Example 1(i): Evaluate \(\log_3 27\)
Solution:
\[\log_3 27 = \log_3 3^3 = 3 \log_3 3 = 3 \quad [\because \log_3 3 = 1]\]
\[\log_3 27 = \log_3 3^3 = 3 \log_3 3 = 3 \quad [\because \log_3 3 = 1]\]
2. Properties of Logarithms
Basic Properties:
- Product Rule: \(\log_a(xy) = \log_a x \log_a y\)
- Quotient Rule: \(\log_a\left(\frac{x}{y}\right) = \log_a x - \log_a y\)
- Identity: \(\log_x x = 1\)
- Logarithm of 1: \(\log_a 1 = 0\)
- Power Rule: \(\log_a(x^p) = p(\log_a x)\)
- Reciprocal Rule: \(\log_a\left(\frac{1}{x}\right) = -\log_a x\)
- Change of Base Formula: \(\log_a x = \frac{\log_b x}{\log_b a}\)
- Exponential Identity: \(a^{\log_a x} = x\)
- Inverse Property: \(\log_a(x^y) = y \log_a x\)
- Fractional Power: \(\log_{a^{p/q}} x = \frac{p}{q} \log_a x\)
Note: When base is not mentioned, it is taken as 10.
Example 2(iii): Evaluate \(36^{\log_6 4}\)
Solution:
We know that \(a^{\log_a x} = x\)
\[36^{\log_6 4} = 6^{2 \log_6 4} = 6^{\log_6 4^2} = 6^{\log_6 16} = 16\]
We know that \(a^{\log_a x} = x\)
\[36^{\log_6 4} = 6^{2 \log_6 4} = 6^{\log_6 4^2} = 6^{\log_6 16} = 16\]
Example 4(i): Evaluate \(\log_5 3 \times \log_{27} 25\)
Solution:
$ \log_5 3 \times \log_{27} 25 = \frac{\log 3}{\log 5} \times \frac{\log 25}{\log 27}= \frac{\log 3}{\log 5} \times \frac{\log 5^2}{\log 3^3} = \frac{\log 3}{\log 5} \times \frac{2 \log 5}{3 \log 3} = \frac{2}{3} \ $
$ \log_5 3 \times \log_{27} 25 = \frac{\log 3}{\log 5} \times \frac{\log 25}{\log 27}= \frac{\log 3}{\log 5} \times \frac{\log 5^2}{\log 3^3} = \frac{\log 3}{\log 5} \times \frac{2 \log 5}{3 \log 3} = \frac{2}{3} \ $
3. Common Logarithms
Logarithms to the base 10 are known as common logarithms. When no base is mentioned, base 10 is assumed.
Example 11: If \(\log_{10} 2 = 0.30103\), find the value of \(\log_{10} 50\).
Solution:
$ \log_{10} 50 = \log_{10}\left(\frac{100}{2}\right) = \log_{10} 100 - \log_{10} 2\\ = 2 - 0.30103 = 1.69897 $
$ \log_{10} 50 = \log_{10}\left(\frac{100}{2}\right) = \log_{10} 100 - \log_{10} 2\\ = 2 - 0.30103 = 1.69897 $
4. Characteristic and Mantissa
The logarithm of a number contains two parts:
- Characteristic: The integral part
- Mantissa: The decimal part
Characteristic Rules:
Case I: When the number is greater than 1
Characteristic = (Number of digits to the left of decimal point) - 1
Case II: When the number is less than 1
Characteristic = -(Number of zeros between decimal point and first significant digit) - 1
Written as \(\bar{1}, \bar{2}, \bar{3}\) (bar notation) instead of -1, -2, -3
Examples Table:
Note: Mantissa is found using logarithm tables and is always positive.
Example 13: If \(\log 2 = 0.30103\), find the number of digits in \(2^{56}\).
Solution:
\[\log(2^{56}) = 56 \log 2 = 56 \times 0.30103 = 16.85768\] Its characteristic is 16. Hence, the number of digits in \(2^{56}\) is 17.
\[\log(2^{56}) = 56 \log 2 = 56 \times 0.30103 = 16.85768\] Its characteristic is 16. Hence, the number of digits in \(2^{56}\) is 17.
5. Solving Logarithmic Equations
Example 6: If \(\log_2[\log_3(\log_2 x)] = 1\), find the value of \(x\).
Solution:
$ \log_2[\log_3(\log_2 x)] = 1 \Rightarrow \log_3(\log_2 x) = 2^1 = 2 \Rightarrow \log_2 x =3^2 = 9 \Rightarrow x = 2^9 = 512 $
$ \log_2[\log_3(\log_2 x)] = 1 \Rightarrow \log_3(\log_2 x) = 2^1 = 2 \Rightarrow \log_2 x =3^2 = 9 \Rightarrow x = 2^9 = 512 $
Example 7: If \(\log_{10}(x^2 - 6x 45) = 2\), find the value of \(x\).
Solution:
$ \log_{10}(x^2 - 6x 45) = 2\Rightarrow x^2 - 6x 45 = 10^2 = 100 \Rightarrow x^2 - 6x - 55 = 0 \Rightarrow x^2 - 11x 5x - 55 = 0 \Rightarrow x(x - 11) 5(x - 11)= 0 \Rightarrow (x - 11)(x 5) = 0 \Rightarrow x = 11 \text{ or } x = -5 \ $
$ \log_{10}(x^2 - 6x 45) = 2\Rightarrow x^2 - 6x 45 = 10^2 = 100 \Rightarrow x^2 - 6x - 55 = 0 \Rightarrow x^2 - 11x 5x - 55 = 0 \Rightarrow x(x - 11) 5(x - 11)= 0 \Rightarrow (x - 11)(x 5) = 0 \Rightarrow x = 11 \text{ or } x = -5 \ $
6. Simplification Using Logarithmic Properties
Example 5: Simplify: \(\log\left(\frac{75}{16}\right) - 2\log\left(\frac{5}{9}\right) \log\left(\frac{32}{243}\right)\)
Solution:
$ \log\left(\frac{75}{16}\right) - \log\left(\frac{5}{9}\right)^2 \log\left(\frac{32}{243}\right) $ = $ \log\left(\frac{75}{16}\right) -\log\left(\frac{25}{81}\right) \log\left(\frac{32}{243}\right) $ = $ \log\left[\frac{75}{16} \times \frac{32}{243} \times \frac{81}{25}\right]$= $ \log 2 $
$ \log\left(\frac{75}{16}\right) - \log\left(\frac{5}{9}\right)^2 \log\left(\frac{32}{243}\right) $ = $ \log\left(\frac{75}{16}\right) -\log\left(\frac{25}{81}\right) \log\left(\frac{32}{243}\right) $ = $ \log\left[\frac{75}{16} \times \frac{32}{243} \times \frac{81}{25}\right]$= $ \log 2 $
Example 8: Find the value of \(x\) which satisfies \(\log_{10} 3 \log_{10}(4x 1) = \log_{10}(x 1) 1\)
Solution:
$ \log_{10} 3 \log_{10}(4x 1) = \log_{10}(x 1) 1 \Leftrightarrow \log_{10} 3 \log_{10}(4x 1) = \log_{10}(x 1) \log_{10} 10 \Leftrightarrow \log_{10}[3(4x 1)]= \log_{10}[10(x 1)] \Leftrightarrow 3(4x 1) = 10(x 1) \Leftrightarrow 12x 3 = 10x 10 \Leftrightarrow 2x = 7 \Leftrightarrow x = \frac{7}{2} $
$ \log_{10} 3 \log_{10}(4x 1) = \log_{10}(x 1) 1 \Leftrightarrow \log_{10} 3 \log_{10}(4x 1) = \log_{10}(x 1) \log_{10} 10 \Leftrightarrow \log_{10}[3(4x 1)]= \log_{10}[10(x 1)] \Leftrightarrow 3(4x 1) = 10(x 1) \Leftrightarrow 12x 3 = 10x 10 \Leftrightarrow 2x = 7 \Leftrightarrow x = \frac{7}{2} $
7. Advanced Applications
Example 9: Simplify: \(\frac{1}{\log_{xy}(xyz)} \frac{1}{\log_{yz}(xyz)} \frac{1}{\log_{zx}(xyz)}\)
Solution:
$ \log_{xyz}(xy) \log_{xyz}(yz) \log_{xyz}(zx) = \log_{xyz}(xy \times yz \times zx) = \log_{xyz}(xyz)^2 = 2 \log_{xyz}(xyz) = 2 \times 1 = 2 $
$ \log_{xyz}(xy) \log_{xyz}(yz) \log_{xyz}(zx) = \log_{xyz}(xy \times yz \times zx) = \log_{xyz}(xyz)^2 = 2 \log_{xyz}(xyz) = 2 \times 1 = 2 $
Example 10: If \(\log_a b = \frac{1}{2}\), \(\log_b c = \frac{1}{3}\), and \(\log_c a = \frac{1}{k}\), find the value of \(k\).
Solution:
$ \log_a b = \frac{1}{2}, \quad \log_b c = \frac{1}{3}, \quad \log_c a = \frac{1}{k} \Rightarrow \frac{\log b}{\log a} = \frac{1}{2}, \quad \frac{\log c}{\log b} = \frac{1}{3}, \quad \frac{\log a}{\log c} = \frac{1}{k} \Rightarrow \frac{\log b}{\log a} \times \frac{\log c}{\log b} \times \frac{\log a}{\log c} = \frac{1}{2} \times \frac{1}{3} \times \frac{1}{k} \Rightarrow 1 = \frac{1}{k} \Rightarrow k = 30 $
$ \log_a b = \frac{1}{2}, \quad \log_b c = \frac{1}{3}, \quad \log_c a = \frac{1}{k} \Rightarrow \frac{\log b}{\log a} = \frac{1}{2}, \quad \frac{\log c}{\log b} = \frac{1}{3}, \quad \frac{\log a}{\log c} = \frac{1}{k} \Rightarrow \frac{\log b}{\log a} \times \frac{\log c}{\log b} \times \frac{\log a}{\log c} = \frac{1}{2} \times \frac{1}{3} \times \frac{1}{k} \Rightarrow 1 = \frac{1}{k} \Rightarrow k = 30 $
Complete Chapter Practice Examples
1. Example 1(ii): Evaluate \(\log_7\left(\frac{1}{343}\right)\)
Solution: \(\log_7\left(\frac{1}{343}\right) = \log_7(7^{-3}) = -3 \log_7 7 = -3\)
2. Example 1(iii): Evaluate \(\log_{100}(0.01)\)
Solution: \(\log_{100}(0.01) = \log_{100}\left(\frac{1}{100}\right) = \log_{100}(100)^{-1} = -1 \log_{100} 100 = -1\)
3. Example 1(iv): Evaluate \(\log_8 128\)
Solution: \(\log_8 128 = \log_{2^3}(2^7) = \frac{\log 2^7}{\log 2^3} = \frac{7}{3}\)
4. Example 2(i): Evaluate \(\log_7 1\)
Solution: We know that \(\log_a 1 = 0\), so \(\log_7 1 = 0\)
5. Example 2(ii): Evaluate \(\log_{34} 34\)
Solution: We know that \(\log_a a = 1\), so \(\log_{34} 34 = 1\)
6. Example 3: If \(\log_{10}(8^x) = \frac{1}{3}\), find the value of \(x\).
Solution:
$ \log_{10}(8^x) = \frac{1}{3} \Leftrightarrow 8^x = 10^{1/3}\Leftrightarrow (2^3)^x = 10^{1/3} \Leftrightarrow 2^{3x} = 10^{1/3} \ $
$ \log_{10}(8^x) = \frac{1}{3} \Leftrightarrow 8^x = 10^{1/3}\Leftrightarrow (2^3)^x = 10^{1/3} \Leftrightarrow 2^{3x} = 10^{1/3} \ $
7. Example 4(ii): Evaluate \(\log_9 27 - \log_{27} 9\)
Solution:
Let \(\log_9 27 = n\). Then \(9^n = 27 \Leftrightarrow 3^{2n} = 3^3 \Leftrightarrow 2n = 3 \Leftrightarrow n = \frac{3}{2}\)
Let \(\log_{27} 9 = m\). Then \(27^m = 9 \Leftrightarrow 3^{3m} = 3^2 \Leftrightarrow 3m = 2 \Leftrightarrow m = \frac{2}{3}\)
\[\therefore \log_9 27 - \log_{27} 9 = (n - m) = \left(\frac{3}{2} - \frac{2}{3}\right) = \frac{5}{6}\]
Let \(\log_9 27 = n\). Then \(9^n = 27 \Leftrightarrow 3^{2n} = 3^3 \Leftrightarrow 2n = 3 \Leftrightarrow n = \frac{3}{2}\)
Let \(\log_{27} 9 = m\). Then \(27^m = 9 \Leftrightarrow 3^{3m} = 3^2 \Leftrightarrow 3m = 2 \Leftrightarrow m = \frac{2}{3}\)
\[\therefore \log_9 27 - \log_{27} 9 = (n - m) = \left(\frac{3}{2} - \frac{2}{3}\right) = \frac{5}{6}\]
8. Example 12(i): If \(\log 2 = 0.3010\) and \(\log 3 = 0.4771\), find \(\log 25\).
Solution:
$ \log 25 = \log\left(\frac{100}{4}\right) = \log 100 - \log 4 = 2 - 2 \log 2 = 2 - 2 \times 0.3010 = 1.398 $
$ \log 25 = \log\left(\frac{100}{4}\right) = \log 100 - \log 4 = 2 - 2 \log 2 = 2 - 2 \times 0.3010 = 1.398 $
9. Example 12(ii): If \(\log 2 = 0.3010\) and \(\log 3 = 0.4771\), find \(\log 4.5\).
Solution:
$ \log 4.5 = \log\left(\frac{9}{2}\right) = \log 9 - \log 2 = 2 \log 3 - \log 2 = 2 \times 0.4771 - 0.3010 = 0.6532 \ $
$ \log 4.5 = \log\left(\frac{9}{2}\right) = \log 9 - \log 2 = 2 \log 3 - \log 2 = 2 \times 0.4771 - 0.3010 = 0.6532 \ $
10. Example 3 (Modified): If \(\log_{10}\left(\frac{8^x}{3}\right) = 1\), find \(x\).
Solution:
$ \log_{10}\left(\frac{8^x}{3}\right) = 1\Rightarrow \frac{8^x}{3} = 10^1 = 10 \Rightarrow 8^x = 30 \Rightarrow (2^3)^x = 30 \Rightarrow 2^{3x} = 30 \Rightarrow x = \frac{\log_2 30}{3} \ $
$ \log_{10}\left(\frac{8^x}{3}\right) = 1\Rightarrow \frac{8^x}{3} = 10^1 = 10 \Rightarrow 8^x = 30 \Rightarrow (2^3)^x = 30 \Rightarrow 2^{3x} = 30 \Rightarrow x = \frac{\log_2 30}{3} \ $
Key Points to Remember
- ✓ Always check if base is mentioned; assume base 10 if not specified
- ✓ \(\log_a 1 = 0\) and \(\log_a a = 1\) are fundamental identities
- ✓ Use change of base formula when bases differ in multiplication/division
- ✓ Characteristic determines the number of digits in a number
- ✓ Practice converting between exponential and logarithmic forms
- ✓ Master product, quotient, and power rules for quick simplification
The document RS Aggarwal Summary: logarithms is a part of the SSC CGL Course Famous Books Course for Competitive Exams.
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FAQs on RS Aggarwal Summary: logarithms
| 1. What is the definition of a logarithm? |  |
Ans. A logarithm is the exponent or power to which a base must be raised to produce a given number. In mathematical terms, if bᵡ = a, then log₍b₎ a = x, where b is the base, a is the number, and x is the logarithm.
| 2. What are the key properties of logarithms? | |
Ans. The key properties of logarithms include: 1. log₍b₎ (xy) = log₍b₎ x + log₍b₎ y (Product Property) 2. log₍b₎ (x/y) = log₍b₎ x - log₍b₎ y (Quotient Property) 3. log₍b₎ (x^n) = n * log₍b₎ x (Power Property) 4. log₍b₎ b = 1 (Logarithm of the base) 5. log₍b₎ 1 = 0 (Logarithm of one)
| 3. What are common logarithms and how are they used? | |
Ans. Common logarithms are logarithms with a base of 10, denoted as log₁₀. They are widely used in scientific calculations and various fields such as engineering and finance, particularly for simplifying calculations involving exponential growth or decay.
| 4. What is the difference between characteristic and mantissa in logarithms? | |
Ans. In logarithms, the characteristic is the integer part of the logarithm, indicating the order of magnitude, while the mantissa is the fractional part, which provides more precise information about the value. For example, in log₁₀ 250, the characteristic is 2 and the mantissa is approximately 0.39794.
| 5. How can one simplify logarithmic expressions using properties of logarithms? | |
Ans. Logarithmic expressions can be simplified using properties such as the product, quotient, and power properties. For instance, to simplify log₁₀ (1000/10), one can apply the quotient property: log₁₀ 1000 - log₁₀ 10 = 3 - 1 = 2. This approach can make complex expressions easier to evaluate or solve.
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