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Chapter 9: Problem 54

If \(\log _{b} 3=0.5\) and \(\log _{b} 5=0.7,\) evaluate each expression. See Example 6 $$ \log _{b} 25 $$

Short Answer

Expert verified
The value of \( \log_b 25 \) is 1.4.

Step by step solution

01

Understand the Question

The question asks us to evaluate \( \log_b 25 \) given \( \log_b 3 = 0.5 \) and \( \log_b 5 = 0.7 \). Our goal is to express \( 25 \) in terms of \( 3 \) and \( 5 \), if possible.
02

Identify the Base-Number Relationship

Notice that \( 25 \) can be expressed as \( 5^2 \). This relationship allows us to use the properties of logarithms to simplify \( \log_b 25 \).
03

Apply Logarithm Power Rule

Apply the power rule of logarithms, which states that \( \log_b (a^n) = n \cdot \log_b a \). Hence, \( \log_b (5^2) = 2 \cdot \log_b 5 \).
04

Substitute Given Values

Substitute the given value for \( \log_b 5 \). We have \( 2 \cdot \log_b 5 = 2 \cdot 0.7 = 1.4 \). This simplifies the expression.
05

Conclusion

Thus, the value of \( \log_b 25 \) is 1.4 based on the properties of logarithms and the given values.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Logarithm Properties
Logarithms have a set of essential properties that make them incredibly useful in simplifying and solving mathematical problems. Let's highlight a few key properties:
  • Product Property: The logarithm of a product is the sum of the logarithms of the factors, i.e., \( \log_b (xy) = \log_b x + \log_b y \)
  • Quotient Property: The logarithm of a quotient is the difference between the logarithms, i.e., \( \log_b \left( \frac{x}{y} \right) = \log_b x - \log_b y \)
  • Power Property: The logarithm of a power can be simplified by multiplying the exponent by the logarithm of the base, i.e., \( \log_b (x^n) = n \cdot \log_b x \)
These properties make it easier to break down complex logarithmic expressions into simpler components. They are crucial in mathematical analysis, allowing for better manipulation and evaluation of expressions.
Logarithm Power Rule
The power rule of logarithms is a powerful tool that simplifies the logarithm of a number raised to a power. If you have an expression of the form \( \log_b (a^n) \), this rule lets you take the exponent, \(n\), and multiply it by the logarithm of the base, \(a\). Therefore, \( \log_b (a^n) = n \cdot \log_b a \). This transformation is especially handy when dealing with exponential bases in logarithms.

As seen in the exercise, we can simplify \( \log_b (5^2) \) using the power rule. When applied, this becomes \( 2 \cdot \log_b 5 \). By using substitution with known logarithmic values, it allows complex calculations to become simple arithmetic calculations.
Logarithmic Evaluation
Evaluating a logarithm is about finding the exponent to which the base must be raised to get a particular number. For instance, when you evaluate \( \log_b 25 = 1.4 \), you are essentially saying that \( b^{1.4} = 25 \).

The step-by-step solution shows how we use known values (\( \log_b 3 \) and \( \log_b 5 \)) to compute \( \log_b 25 \). By expressing 25 as \( 5^2 \), we apply the power rule to simplify the expression. Finally, substitution with the given \( \log_b 5 = 0.7 \) leads to easy calculation of \(16\). This evaluates \( \log_b 25 \) accurately, showcasing the effectiveness of combining logarithmic properties with specific values.
Base Conversion in Logarithms
Converting bases in logarithms is sometimes needed to evaluate logs with non-standard bases. While not required in our specific problem, understanding this technique is valuable for other logarithmic computations. The change of base formula is:
  • \( \log_b x = \frac{\log_k x}{\log_k b} \)
This formula allows transformation of any logarithm base into another, typically to one that a calculator can handle like base 10 or base \(e\) (the natural base). This conversion can simplify computational efforts and make logarithmic evaluations easier for numbers not readily solvable in a given base.

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