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 solved exercise to put into 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 base \( b \), denoted by \( \log_b(x) \), 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 tells us that if we calculate \( b^{\log_b(a)} \), we obtain \( a \). This property is essential for solving logarithmic and exponential equations.
\[ b^{\log_b(a)} = a \quad \text{with} \quad a > 0 \]
This is a direct consequence of the definition of logarithm. Indeed, \( \log_b(a) \) is precisely that exponent which, when \( b \) is raised to it, gives \( a \) as a result.
Exercise. Calculate \( 3^{\log_3(81)} \).
Solution. Using the property \( b^{\log_b(a)} = a \), we can write:
\[ 3^{\log_3(81)} = 81 \]
Result: \( 81 \).
Power rule
The power rule allows us to calculate 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.
\[\log_b(x^n) = n \cdot \log_b(x) \quad \text{with} \quad x > 0 \]
Proof. Let \( k = \log_b(x) \). By definition of logarithm, this means that \( b^k = x \). 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. We use 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:
\[ \log_2(32) = 5 \]
Therefore:
\[ \log_2(32^3) = 3 \cdot 5 = 15 \]
Result: \( 15 \).
Product rule
The product rule tells us that the logarithm of a product equals the sum of the logarithms of the factors. This rule is fundamental for simplifying expressions with products.
\[\log_b(x \cdot y) = \log_b(x) + \log_b(y) \quad \text{with} \quad x, y > 0 \]
Proof. Let \( k = \log_b(x) \) and \( h = \log_b(y) \). By definition of logarithm: \( b^k = x \) and \( b^h = y \). Therefore:
\[ x \cdot y = b^k \cdot b^h = b^{k+h} \]
\[ \log_b(x \cdot y) = \log_b(b^{k+h}) = k + h = \log_b(x) + \log_b(y) \]
Exercise. Calculate \( \log_5(25) + \log_5(4) \) and compare with \( \log_5(100) \).
Solution. We use the product rule:
\[ \log_5(25) + \log_5(4) = \log_5(25 \cdot 4) \]
Since \( 25 \cdot 4 = 100 \), we have:
\[ \log_5(25) + \log_5(4) = \log_5(100) \]
Since \( 25 = 5^2 \), we have \( \log_5(25) = 2 \), and consequently:
\[ \log_5(100) = 2 + \log_5(4) \]
Result: \( \log_5(25) + \log_5(4) = \log_5(100) \).
Quotient rule
The quotient rule tells us that the logarithm of a quotient equals the difference of the logarithms. This rule is useful for simplifying expressions with fractions.
\[\log_b \left( \frac{x}{y} \right) = \log_b(x) - \log_b(y) \quad \text{with} \quad x, y > 0 \]
Proof. Let \( k = \log_b(x) \) and \( h = \log_b(y) \). By definition of logarithm: \( b^k = x \) and \( b^h = y \). Therefore:
\[ \frac{x}{y} = \frac{b^k}{b^h} = b^{k-h} \]
\[ \log_b \left( \frac{x}{y} \right) = \log_b(b^{k-h}) = k - h = \log_b(x) - \log_b(y) \]
Exercise. Simplify \( \log_3(81) - \log_3(9) \).
Solution. We use the quotient rule:
\[ \log_3(81) - \log_3(9) = \log_3\left(\frac{81}{9}\right) \]
Since \( \displaystyle\frac{81}{9} = 9 \), we have:
\[ \log_3(81) - \log_3(9) = \log_3(9) \]
Since \( 9 = 3^2 \), it follows that:
\[ \log_3(9) = 2 \]
Result: \( 2 \).
Change of base
The change of base formula allows us to express a logarithm in any base using logarithms in another base. It is particularly useful when we want to use a calculator, which often only has the \( \ln \) and \( \log_{10} \) keys.
\[\log_b(x) = \frac{\log_c(x)}{\log_c(b)} \quad \text{with} \quad x, b > 0 \quad , \quad b \neq 1 \quad , \quad c > 0 \quad ,\quad c \neq 1 \]
Exercise. Write \( \log_2(40) \) using the natural logarithm (\( \ln \)).
Solution. We use the change of base formula:
\[ \log_2(40) = \frac{\ln(40)}{\ln(2)} \]
Result: \( \displaystyle \frac{\ln(40)}{\ln(2)} \).