Understanding Logarithms
A logarithm answers the question: To what power must a base be raised, to produce a given number? Formally, for a positive real number \(a \neq 1\), and \(x > 0\), the logarithm of \(x\) to the base \(a\) is defined as the exponent \(y\) such that
\( a^y = x \quad \Longleftrightarrow \quad y = \log_a x \)
Here, \(a\) is the base, \(x\) is the argument, and \(y\) is the logarithm. For example, since \(2^3 = 8\), we have \( \log_2 8 = 3 \).
Common and Natural Logarithms
Two special logarithms are widely used:
- Common logarithm: Base 10, written as \( \log x \) or \( \log_{10} x \). For example, \( \log 100 = 2 \) because \(10^2 = 100\).
- Natural logarithm: Base \(e \approx 2.718\), written as \( \ln x \). For example, \( \ln e = 1 \) because \( e^1 = e \).
Relationship Between Logarithms and Exponents
Logarithms are the inverse operation of exponentiation. This means:
\( \log_a (a^x) = x \quad \text{and} \quad a^{\log_a x} = x \)
This inverse relationship allows us to solve equations involving exponents by converting them into logarithmic form.
Laws of Logarithms
Logarithms follow specific laws that simplify expressions and solve equations efficiently. For \(a > 0\), \(a \neq 1\), and positive \(x, y\), the key laws are:
- Product Rule: \[ \log_a (xy) = \log_a x + \log_a y \] This means the logarithm of a product is the sum of the logarithms.
- Quotient Rule: \[ \log_a \left(\frac{x}{y}\right) = \log_a x - \log_a y \] The logarithm of a quotient is the difference of the logarithms.
- Power Rule: \[ \log_a (x^k) = k \log_a x \] The logarithm of a power is the exponent times the logarithm of the base.
Why These Laws Hold
These laws come directly from the properties of exponents. For example, the product rule:
If \( \log_a x = m \) and \( \log_a y = n \), then \( x = a^m \) and \( y = a^n \). So,
\[ xy = a^m \times a^n = a^{m+n} \quad \Rightarrow \quad \log_a (xy) = m + n = \log_a x + \log_a y \]
Change of Base Formula
Sometimes, you need to evaluate logarithms with bases that your calculator does not support. The change of base formula allows you to rewrite logarithms in terms of any other base, usually 10 or \(e\):
\[ \log_a x = \frac{\log_b x}{\log_b a} \]
Here, \(b\) is any positive number different from 1. For example, to calculate \( \log_2 7 \) using common logarithms:
\[ \log_2 7 = \frac{\log 7}{\log 2} \]
Solving Logarithmic Equations
To solve equations involving logarithms, use the laws of logarithms and the definition of logarithms to rewrite the equation in exponential form or simplify it.
Example: Solve for \(x\):
\[ \log_3 (x+1) = 2 \]
Using the definition, rewrite as:
\[ x + 1 = 3^2 = 9 \quad \Rightarrow \quad x = 8 \]
Applications in Entrance Exams
Logarithms are frequently tested in entrance exams like NDA for their properties, simplifications, and solving equations. Understanding the laws and being able to apply them quickly is essential.
Common Mistakes to Avoid
- Forgetting that the argument of a logarithm must be positive.
- Mixing up the base and the argument.
- Incorrectly applying the laws, for example, \( \log (x + y) \neq \log x + \log y \).
- Ignoring the domain restrictions when solving equations.
Exam Tips
- Always check the domain of the logarithmic expressions.
- Use the change of base formula when needed.
- Practice simplifying expressions using the laws of logarithms.
- Be comfortable converting between exponential and logarithmic forms.
Worked Examples
Example 1 (Easy): Find \( \log_5 125 \).
Solution:
Since \(125 = 5^3\), by definition, \[ \log_5 125 = \log_5 (5^3) = 3 \]
Example 2 (Medium): Simplify \( \log_2 8 + \log_2 4 \).
Solution:
Using the product rule, \[ \log_2 8 + \log_2 4 = \log_2 (8 \times 4) = \log_2 32 \] Since \(32 = 2^5\), \[ \log_2 32 = 5 \]
Example 3 (Medium): Solve for \(x\): \[ \log_3 (x) + \log_3 (x-2) = 3 \]
Solution:
Use the product rule: \[ \log_3 [x(x-2)] = 3 \] Rewrite in exponential form: \[ x(x-2) = 3^3 = 27 \] Expand: \[ x^2 - 2x = 27 \] Bring all terms to one side: \[ x^2 - 2x - 27 = 0 \] Solve quadratic: \[ x = \frac{2 \pm \sqrt{4 + 108}}{2} = \frac{2 \pm \sqrt{112}}{2} = \frac{2 \pm 4\sqrt{7}}{2} = 1 \pm 2\sqrt{7} \] Since \(x > 0\) and \(x-2 > 0\), check values: - \(x = 1 + 2\sqrt{7} \approx 6.29\) (valid) - \(x = 1 - 2\sqrt{7} < 0\) (invalid)
Thus, \[ \boxed{x = 1 + 2\sqrt{7}} \]
Example 4 (Hard): Express \( \log_2 10 \) in terms of common logarithms.
Solution:
Using the change of base formula, \[ \log_2 10 = \frac{\log 10}{\log 2} \] Since \( \log 10 = 1 \), \[ \log_2 10 = \frac{1}{\log 2} \approx \frac{1}{0.3010} \approx 3.3219 \]
Example 5 (Hard): Solve for \(x\): \[ 2 \log_5 (x) - \log_5 (4) = 1 \]
Solution:
Use the power rule: \[ \log_5 (x^2) - \log_5 (4) = 1 \] Use quotient rule: \[ \log_5 \left(\frac{x^2}{4}\right) = 1 \] Convert to exponential form: \[ \frac{x^2}{4} = 5^1 = 5 \] Multiply both sides by 4: \[ x^2 = 20 \] Take square root: \[ x = \pm \sqrt{20} = \pm 2\sqrt{5} \] Since the argument of the logarithm must be positive, \(x > 0\), so \[ \boxed{x = 2\sqrt{5}} \]
Example 6 (Medium): If \( \log_a 3 = 0.5 \) and \( \log_a 5 = 1.2 \), find \( \log_a \frac{45}{2} \).
Solution:
Express \( \frac{45}{2} \) as \( \frac{3^2 \times 5}{2} \): \[ \log_a \frac{45}{2} = \log_a (3^2) + \log_a 5 - \log_a 2 \] Using power rule: \[ = 2 \log_a 3 + \log_a 5 - \log_a 2 = 2(0.5) + 1.2 - \log_a 2 = 1 + 1.2 - \log_a 2 = 2.2 - \log_a 2 \] Without \( \log_a 2 \), this is the simplified form. If \( \log_a 2 \) is known, substitute to find the numerical value.
