To thoroughly understand the properties of logarithms, we will start from their definition. From here, we will demonstrate, step by step, the main rules that allow us to simplify and manipulate logarithmic expressions. Each property will be accompanied by a worked-out example to practice what has been learned.
Definition. Given a positive real number \( x > 0 \) and a base \( b > 0 \) with \( b \neq 1 \), the logarithm of \( x \) to the base \( b \), denoted as \( \log_b(x) \), is the exponent \( y \) such that \( b^y = x \). Formally:
\[ \log_b(x) = y \iff b^y = x \]
Table of Contents
Fundamental Identity
The fundamental identity of logarithms states that if we calculate \( b^{\log_b(a)} \), the result is \( a \). This property is essential for solving logarithmic and exponential equations.
\[ b^{\log_b(a)} = a \quad \text{where} \quad a > 0 \]
This is a direct consequence of the definition of logarithm. In fact, \( \log_b(a) \) is the exponent to which \( b \) must be raised to produce \( a \).
Exercise. Compute \( 3^{\log_3(81)} \).
Solution. Using the property \( b^{\log_b(a)} = a \), we can write:
\[ 3^{\log_3(81)} = 81 \]
Answer: \( 81 \).
Exponent Rule
The exponent rule allows us to compute the logarithm of a power. This rule transforms the logarithm of a power into the product of the exponent and the logarithm of the base of the power.
\[ \log_b(x^n) = n \cdot \log_b(x) \quad \text{where} \quad x > 0 \]
Proof. Let \( k = \log_b(x) \). By the definition of logarithm, this means \( b^k = x \). Therefore:
\[ \log_b(x^n) = \log_b((b^k)^n) = \log_b(b^{kn}) = kn = n \cdot \log_b(x) \]
Exercise. Simplify \( \log_2(32^3) \).
Solution. Using the rule \( \log_b(x^n) = n \cdot \log_b(x) \):
\[ \log_2(32^3) = 3 \cdot \log_2(32) \]
Since \( 32 = 2^5 \), we have:
\[ \log_2(32) = 5 \]
Hence:
\[ \log_2(32^3) = 3 \cdot 5 = 15 \]
Answer: \( 15 \).
Product Rule
The product rule states that the logarithm of a product is equal to the sum of the logarithms of its factors. This rule is fundamental for simplifying expressions involving products.
\[ \log_b(x \cdot y) = \log_b(x) + \log_b(y) \quad \text{where} \quad x, y > 0 \]
Proof. Let \( k = \log_b(x) \) and \( h = \log_b(y) \). By the definition of logarithm: \( b^k = x \) and \( b^h = y \). Thus:
\[ x \cdot y = b^k \cdot b^h = b^{k+h} \]
\[ \log_b(x \cdot y) = \log_b(b^{k+h}) = k + h = \log_b(x) + \log_b(y) \]
Exercise. Compute \( \log_5(25) + \log_5(4) \) and compare it to \( \log_5(100) \).
Solution. Using the product rule:
\[ \log_5(25) + \log_5(4) = \log_5(25 \cdot 4) \]
Since \( 25 \cdot 4 = 100 \), we have:
\[ \log_5(25) + \log_5(4) = \log_5(100) \]
As \( 100 = 5^2 \), it follows that:
\[ \log_5(100) = 2 \]
Answer: \( 2 \).
Quotient Rule
The quotient rule states that the logarithm of a quotient is equal to the difference of the logarithms. This rule is useful for simplifying expressions involving fractions.
\[ \log_b \left( \frac{x}{y} \right) = \log_b(x) - \log_b(y) \quad \text{where} \quad x, y > 0 \]
Proof. Let \( k = \log_b(x) \) and \( h = \log_b(y) \). By the definition of logarithm: \( b^k = x \) and \( b^h = y \). Thus:
\[ \frac{x}{y} = \frac{b^k}{b^h} = b^{k-h} \]
\[ \log_b \left( \frac{x}{y} \right) = \log_b(b^{k-h}) = k - h = \log_b(x) - \log_b(y) \]
Exercise. Simplify \( \log_3(81) - \log_3(9) \).
Solution. Using the quotient rule:
\[ \log_3(81) - \log_3(9) = \log_3\left(\frac{81}{9}\right) \]
Since \( \frac{81}{9} = 9 \), we have:
\[ \log_3(81) - \log_3(9) = \log_3(9) \]
As \( 9 = 3^2 \), it follows that:
\[ \log_3(9) = 2 \]
Answer: \( 2 \).
Change of Base Formula
The change of base formula allows us to rewrite a logarithm in one base in terms of logarithms in another base. This is especially useful when the base of the logarithm is not directly supported by calculators.
The formula is:
\[ \log_b(x) = \frac{\log_k(x)}{\log_k(b)} \quad \text{with} \quad x, b, k > 0 \quad \text{and} \quad b, k \neq 1 \]
Proof. By the definition of the logarithm, we know that \( b^{\log_b(x)} = x \). Now, consider the logarithm of both sides in base \( k \):
\[ \log_k(b^{\log_b(x)}) = \log_k(x) \]
Using the power rule \( \log_k(a^n) = n \cdot \log_k(a) \), this becomes:
\[ \log_b(x) \cdot \log_k(b) = \log_k(x) \]
Dividing through by \( \log_k(b) \), we obtain:
\[ \log_b(x) = \frac{\log_k(x)}{\log_k(b)} \]
Exercise. Rewrite \( \log_5(125) \) in terms of natural logarithms (\( \ln \)).
Solution. Using the change of base formula with \( k = e \):
\[ \log_5(125) = \frac{\ln(125)}{\ln(5)} \]
Since \( 125 = 5^3 \), we can simplify:
\[ \ln(125) = \ln(5^3) = 3 \cdot \ln(5) \]
Substituting this back, we have:
\[ \log_5(125) = \frac{3 \cdot \ln(5)}{\ln(5)} = 3 \]
Result: \( 3 \).