Problem 82 - Precalculus with Calculus Previews

Question

Show that \(\left(\log _{10} e\right)\left(\log _{e} 10\right)=1 .\) Can you generalize this result?

Step-by-Step Solution

Answer

The expression is equal to 1, and this result can be generalized to \( (\log_a b)(\log_b a) = 1 \) for any positive numbers \(a\) and \(b\).

Step 1: Understand the Problem

We are asked to prove that \(\left(\log_{10} e\right)(\log_{e} 10) = 1\). This requires using properties of logarithms to show the equality.

Step 2: Apply Change of Base Formula

Recall the change of base formula: \(\log_{b} a = \frac{\log_{c} a}{\log_{c} b}\). Using this, we can express \(\log_{10} e\) as \(\frac{\log_{e} e}{\log_{e} 10}\), knowing that \(\log_{e} e = 1\).

Step 3: Simplify the Expression

Since \(\log_{e} e = 1\), \(\log_{10} e = \frac{1}{\log_{e} 10}\). Multiply by \(\log_{e} 10\), giving \(\frac{1}{\log_{e} 10} \cdot \log_{e} 10\).

Step 4: Solve the Expression

Notice that \(\frac{1}{\log_{e} 10} \cdot \log_{e} 10 = 1\) since the terms are multiplicative inverses. This confirms \(\left(\log_{10} e\right)(\log_{e} 10) = 1\).

Step 5: Generalization

The result can be generalized: for any positive numbers \(a\) and \(b\), \(\left(\log_{a} b\right)(\log_{b} a) = 1\), since they are multiplicative inverses due to the change of base property.

Key Concepts

Change of Base Formula

The Change of Base Formula is a powerful tool in logarithms that allows us to switch from one base to another. This is especially useful when you are working with bases that aren't easy to calculate directly. The formula is given by:

• \( \log_{b} a = \frac{\log_{c} a}{\log_{c} b} \)

Here’s how it works: You can express \( \log_{b} a \) in terms of another base \( c \). For instance, when calculating \( \log_{10} e \), it’s more convenient to convert it into a natural logarithm \( \log_{e} \) because we commonly have access to this in calculators. In mathematical form: \( \log_{10} e = \frac{\log_{e} e}{\log_{e} 10} \). Since \( \log_{e} e \) equals 1, this simplifies the expression to \( \frac{1}{\log_{e} 10} \). This formula is critical as it allows flexibility in dealing with different logarithmic bases.

Multiplicative Inverses

Multiplicative inverses play a crucial role in solving logarithmic expressions, especially when related to the identity property—that is, any number multiplied by its multiplicative inverse equals 1. In the context of logarithms, consider two expressions such as \( \log_{10} e \) and \( \log_{e} 10 \). These are multiplicative inverses of each other, meaning:

• \( \log_{10} e \cdot \log_{e} 10 = 1 \)

This relationship holds because by applying the change of base formula, \( \log_{10} e \) becomes \( \frac{1}{\log_{e} 10} \). When multiplied by \( \log_{e} 10 \), these values respectively cancel each other out, beautifully simplifying to 1. Understanding this concept helps you maneuver through complex logarithmic scenarios by recognizing these reciprocal properties.

Logarithmic Identities

Logarithmic identities simplify complex problems by reducing them to manageable components. They are fundamental in proving various expressions, including the multiplicative inverses we discussed. One essential identity is:

• \( \log_{b} a \times \log_{a} b = 1 \)

This demonstrates that switching the base and argument in logarithms not only changes the perspective of the expression but also yields consistent identities. Another important identity to remember is:

• \( \log_{a} a = 1 \)

This means that any number raised to its logarithmic base equals 1, a key concept that simplifies many calculations. In real-world applications, these identities support solving equations efficiently by transforming and decomposing complex logarithmic relationships. Their versatility underpins the computational foundation of logarithmic calculations.