Introduction
Logarithms are the inverse operations of exponents, allowing us to “undo” exponentiation and solve equations involving exponential terms. We will cover the main rules of logarithms, which make it easier to manipulate logarithmic expressions.
Definition of Logarithms
A logarithm answers the question: “To what exponent must a base be raised to get a certain number?” In general, for a base \( a \), a logarithm is written as:
\[\begin{align*} \log_a(x) = y \end{align*}\]
where:
– \( a \) is the base of the logarithm, a positive number other than 1.\\
– \( x \) is the argument of the logarithm, a positive number.\\
– \( y \) is the result, or the exponent to which the base \( a \) must be raised to obtain \( x \).
For example, \( \log_2(8) = 3 \) because \( 2^3 = 8 \).
Logarithm Rules
Logarithm rules are useful for simplifying expressions and solving equations involving logarithms. Here are the key rules:
\[\begin{align*} \text{1. Product Rule:} \quad \log_a(x \cdot y) = \log_a(x) + \log_a(y) \end{align*}\]
The product rule states that the logarithm of a product is the sum of the logarithms. For example:
\[\begin{align*} \log_2(8 \cdot 4) = \log_2(8) + \log_2(4) = 3 + 2 = 5 \end{align*}\]
\[\begin{align*} \text{2. Quotient Rule:} \quad \log_a\left( \frac{x}{y} \right) = \log_a(x) – \log_a(y) \end{align*}\]
The quotient rule states that the logarithm of a quotient is the difference of the logarithms. For example:
\[\begin{align*} \log_3\left( \frac{27}{3} \right) = \log_3(27) – \log_3(3) = 3 – 1 = 2 \end{align*}\]
\[\begin{align*} \text{3. Power Rule:} \quad \log_a(x^n) = n \cdot \log_a(x) \end{align*}\]
The power rule states that the logarithm of a power is the exponent times the logarithm of the base. For example:
\[\begin{align*} \log_5(125^2) = 2 \cdot \log_5(125) = 2 \cdot 3 = 6 \end{align*}\]
\[\begin{align*} \text{4. Change of Base Rule:} \quad \log_a(x) = \frac{\log_b(x)}{\log_b(a)} \end{align*}\]
The change of base rule allows you to rewrite a logarithm in terms of a different base. This is especially useful for calculating logarithms on a calculator when the base is not 10 or \( e \). For example:
\[\begin{align*} \log_2(10) = \frac{\log_{10}(10)}{\log_{10}(2)} = \frac{1}{0.301} \approx 3.32 \end{align*}\]
\[\begin{align*} \text{5. Logarithm of 1:} \quad \log_a(1) = 0 \end{align*}\]
The logarithm of 1 is always zero, regardless of the base, because any number raised to the power of 0 is 1:
\[\begin{align*} \log_7(1) = 0 \end{align*}\]
\[\begin{align*} \text{6. Logarithm of the Base:} \quad \log_a(a) = 1 \end{align*}\]
The logarithm of a base to itself is always 1, because \( a^1 = a \):
\[\begin{align*} \log_4(4) = 1 \end{align*}\]
Practice Problems
Let’s look at some examples of how these rules can simplify expressions.
- Simplify \( \log_3(9) + \log_3(27) \):
Using the product rule:
\[\begin{align*} \log_3(9) + \log_3(27) = \log_3(9 \cdot 27) = \log_3(243) = 5 \end{align*}\] - Simplify \( \log_5\left( \dfrac{125}{5} \right) \):
Using the quotient rule:
\[\begin{align*} \log_5\left( \dfrac{125}{5} \right) = \log_5(125) – \log_5(5) = 3 – 1 = 2 \end{align*}\] - Evaluate \( \log_2(16^3) \):
Using the power rule:
\[\begin{align*} \log_2(16^3) = 3 \cdot \log_2(16) = 3 \cdot 4 = 12 \end{align*}\]
Conclusion
Logarithm rules are essential for solving logarithmic equations. By mastering these rules, you can make calculations involving exponents and logarithms much easier, setting a foundation for applications in algebra, calculus, and beyond.
