Problem 60 - Precalculus with Calculus Previews

Question

In Problems \(57-68\), use \(\log _{b} 4=0.6021\) and \(\log _{b} 5=\) \(0.6990\) to evaluate the given logarithm. Round your answer to four decimal places. $$ \log _{b} 625 $$

Step-by-Step Solution

Answer

\( \log_b 625 = 2.7960 \).

Step 1: Express 625 as a Power or Product

First, observe that 625 is a power of 5. Specifically, \( 625 = 5^4 \). This understanding is important to help us rewrite the logarithm using properties of logs.

Step 2: Logarithmic Identity Application

Use the property of logarithms that states \( \log_b (a^n) = n \cdot \log_b a \). Applying this identity, we get: \( \log_b 625 = \log_b (5^4) = 4 \cdot \log_b 5 \).

Step 3: Substitute Known Values

We were given that \( \log_b 5 = 0.6990 \). Substitute this into the equation: \( 4 \cdot \log_b 5 = 4 \times 0.6990 \).

Step 4: Calculate the Result

Perform the multiplication: \( 4 \times 0.6990 = 2.7960 \). This is the value of \( \log_b 625 \).

Step 5: Round to Four Decimal Places

In this problem, our result is already at four decimal places: 2.7960. Therefore, no further rounding is needed.

Key Concepts

Properties of Logarithms

Logarithms are a unique and fascinating branch of mathematics, and one of their most important aspects is their properties. These properties allow us to manipulate and understand logarithms better by simplifying complex expressions or solving equations. Here are some fundamental properties you should know:

• The Product Property: \( \log_b (xy) = \log_b x + \log_b y \). This states that the logarithm of a product is the sum of the logarithms.
• The Quotient Property: \( \log_b \left(\frac{x}{y}\right) = \log_b x - \log_b y \). This shows that the logarithm of a quotient is the difference of the logarithms.
• The Power Property: \( \log_b (x^n) = n \cdot \log_b x \). This indicates that the logarithm of a number raised to a power is the power times the logarithm of the number itself.

These properties are powerful tools for breaking down and solving logarithm-based problems, such as the given exercise where we identified 625 as \(5^4\) and then used the power property to simplify \(\log_b 625\) into a manageable form. By leveraging these properties, the task of solving logarithms becomes much more straightforward.

Logarithmic Identities

In addition to logarithmic properties, understanding logarithmic identities helps in effectively working through problems involving logarithms. These identities are essentially shortcuts or standard rules that always hold true. Some key logarithmic identities include:

• The Identity Logarithm: \( \log_b b = 1 \). This tells us that any base raised to the power one equals the base itself.
• The Zero Logarithm: \( \log_b 1 = 0 \). This notes that no matter the base (as long as it's positive), the logarithm of 1 is always 0.
• Inverse Property: \( b^{\log_b x} = x \). This identity asserts the inverse relationship between exponentials and logarithms.

These identities are helpful for simplifying equations or transforming them into a more solvable form. For example, knowing these identities can make it easier to rework an equation or check the accuracy of your solutions. They serve as a vital support structure when dealing with more complicated logarithmic expressions.

Exponents and Powers

Exponents and powers are closely related to logarithms, acting as the inverse operations. Understanding these concepts is crucial for solving logarithm problems correctly. Here's a quick rundown:

• An exponent tells us how many times a number (the base) is multiplied by itself. For instance, \(5^4\) means \(5 \times 5 \times 5 \times 5\).
• A power is the entire expression, consisting of both the base and the exponent, like \(5^4\), where 5 is the base and 4 is the exponent.
• Understanding the relationship: Logarithms can be thought of as another way of expressing exponents. If \(b^n = a\), then \(\log_b a = n\).

In the exercise provided, recognizing that 625 equals \(5^4\) was a critical step in applying the power property of logarithms. It simplified the logarithm into a multiplication problem where known values could be substituted and calculated easily. Grasping the nature of exponents and powers is fundamental to mastering related logarithmic problems.