Problem 2 - Precalculus Mathematics for Calculus

Question

The logarithm of a quotient of two numbers is the same as the _______ of the logarithms of these numbers. So \(\log _{5}\left(\frac{25}{125}\right)=\) ________ - ________.

Step-by-Step Solution

Answer

Difference; 2; 3

Step 1: Understanding the Logarithm Quotient Rule

The logarithm of a quotient is understood by the identity that the logarithm of a quotient is equal to the difference of the logarithms: \[ \log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N) \] This means for the statement given, \( \log_5\left(\frac{25}{125}\right) = \log_5(25) - \log_5(125) \).

Step 2: Calculating Individual Logarithms

Let's calculate \( \log_5(25) \) and \( \log_5(125) \). We need to express these numbers in terms of base 5. Since \( 25 = 5^2 \), \( \log_5(25) = 2 \). Since \( 125 = 5^3 \), \( \log_5(125) = 3 \).

Step 3: Applying the Logarithmic Identity

Substitute the calculated logarithms into the identity: \[ \log_5\left(\frac{25}{125}\right) = 2 - 3 \]

Step 4: Solving the Expression

Calculate the difference: \[ 2 - 3 = -1 \] So, \( \log_5\left(\frac{25}{125}\right) = -1 \).

Key Concepts

Logarithm Properties

Logarithms have several useful properties that make them powerful in mathematics. Let's explore a key property used frequently: the logarithm quotient rule. This rule states that the logarithm of a quotient is equal to the difference of the logarithms. Specifically, for any positive numbers \(M\) and \(N\) and a base \(b > 0\), \(b eq 1\), the rule is expressed as:

• \(\log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N)\)

This rule is extremely helpful because it allows complex division calculations to be broken down into simpler subtraction operations involving logarithms.
Another key property is the fact that logs convert multiplication into addition, another simplification tool, but here we are focusing on the quotient rule. Understanding this concept helps to ease calculations in algebra, calculus, and beyond.

Exponentiation

Exponentiation is at the heart of understanding logarithms. It involves raising a number (the base) to a power. For instance, given the number \(5\) raised to the power of \(2\), it is \(5^2 = 25\). This connection is crucial when solving logarithmic problems where you often reverse the exponential function.
The essence of logarithms is to find the power that a base must be raised to obtain a certain number. For example, \(\log_5(25)\) asks, "What power must 5 be raised to get 25?" The answer is \(2\) because \(5^2 = 25\). The notion works symmetrically for division, i.e., the quotient, as seen in our exercise, where we find that \(\log_5(125) = 3\) because \(5^3 = 125\).
Understanding this exponentiation relationship is fundamental to mastering logarithms as it forms the basis of how we interpret and solve logarithmic expressions.

Problem Solving in Logarithms

Problem-solving with logarithms often involves identifying the right properties and applying them correctly. In our example, we solve \(\log_5\left(\frac{25}{125}\right)\) by employing the logarithm quotient rule. First, we need to identify each component: the numbers for which we calculate individual logarithms.
We expressed \(25\) and \(125\) as powers of \(5\), computing \(\log_5(25) = 2\) and \(\log_5(125) = 3\).

• Step 1: Use the quotient rule: \(\log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N)\)
• Step 2: Express each number in terms of base \(5\).
• Step 3: Substitute the values to find: \(\log_5\left(\frac{25}{125}\right) = 2 - 3\).
• Step 4: Solve by simple subtraction yielding \( -1 \).

This systematic approach underscores the importance of breaking problems into smaller parts, simplifying expressions with rules, and using exponential relationships.