Problem 16 - Intermediate Algebra

Question

Fill in the blanks. $$ \text { Does } \frac{\log 7}{\log 3}=\log 7-\log 3 ? $$

Step-by-Step Solution

Answer

No, \( \frac{\log 7}{\log 3} \neq \log 7 - \log 3 \).

Step 1: Understanding the Expression

The expression compares two quantities: \( \frac{\log 7}{\log 3} \) and \( \log 7 - \log 3 \). We need to understand whether these two expressions are equivalent.

Step 2: Simplifying the Left Side

The left side of the equation is \( \frac{\log 7}{\log 3} \). This expression represents the change of base formula, meaning that \( \frac{\log 7}{\log 3} = \log_3 7 \).

Step 3: Simplifying the Right Side

The right side of the equation is \( \log 7 - \log 3 \). This expression can be simplified using the property of logarithms: \( \log a - \log b = \log \left(\frac{a}{b}\right) \). Therefore, \( \log 7 - \log 3 = \log \left( \frac{7}{3} \right) \).

Step 4: Comparing the Results

Now, compare the two simplified forms: \( \log_3 7 \) from the left side and \( \log \left( \frac{7}{3} \right) \) from the right side. These values represent different mathematical expressions unless they are numerically proven equivalent, which requires specific values, and generally, they are not equal.

Key Concepts

Change of Base Formula

In mathematics, logarithms allow us to solve exponential problems in a more manageable way by translating multiplication into addition. One of the handy tools when dealing with logarithms is the change of base formula. It is especially useful when you want to evaluate logarithms but only have a calculator that computes logarithms to certain bases like 10 or e.

The formula is given by:

• \( \log_b a = \frac{\log_c a}{\log_c b} \)

This formula allows you to compute a logarithm with an arbitrary base by using any other base you prefer. It essentially shows that the logarithm of a number with a specific base can be expressed as a ratio of logarithms with a different base.

For instance, the expression \( \frac{\log 7}{\log 3} \) uses the base 10 common logarithm to transform it into \( \log_3 7 \), establishing a practical way to handle calculations when the direct base is not available.

Properties of Logarithms

Logarithmic functions have properties that help us simplify complex logarithmic expressions, making them easier to understand and solve. One such useful property is the subtraction property:

• \( \log_b a - \log_b c = \log_b \left( \frac{a}{c} \right) \)

This property implies that the difference between two logarithms with the same base can be rewritten as a single logarithm of a fraction. It's an incredibly helpful tool for combining separate logarithmic terms into one.

Using this property allows us to simplify expressions like \( \log 7 - \log 3 \) into \( \log \left( \frac{7}{3} \right) \). Thus, instead of evaluating two separate logarithmic values, we can deal with a single, often simpler, term.

Understanding these properties empowers you to accurately manipulate logarithmic equations and enhances your algebraic problem-solving skills.

Logarithmic Expressions

Logarithmic expressions can often seem tricky at first glance. However, by breaking them down and understanding their fundamental properties, these expressions become much more approachable.

• Logarithms are, fundamentally, the inverse of exponentiation. If \( b^x = a \), then \( \log_b a = x \).
• They help in reducing the complexity of multiplication and division into addition and subtraction.

For example, the expression \( \frac{\log 7}{\log 3} = \log_3 7 \) translates a fraction into a simple form using a different base that might make evaluation easier or more meaningful in context.

Moreover, when you encounter subtraction in logarithmic forms such as \( \log 7 - \log 3 \), you can use properties of logarithms to rewrite it. Understanding how to manipulate these expressions is crucial, especially when solving real-world problems or in scenarios involving growth, decay, and data analysis.

By mastering these basic concepts, you obtain the tools needed to tackle advanced logarithmic challenges with confidence.