Problem 52 - Algebra 2

Question

Assume that \(\log 4 \approx 0.6021, \log 5 \approx 0.6990,\) and \(\log 6 \approx 0.7782 .\) Use the properties of logarithms to evaluate each expression. Do not use your calculator. \(\log 25\)

Step-by-Step Solution

Answer

Therefore, \( \log 25 \approx 1.3980 \)

Step 1: Expression Rewrite

Rewrite \( 25 \) as \( 5^2 \) and apply the property of logarithms, \( \log a^2 = 2 \log a \), to the expression \( \log 25 \). This gives \( \log 25 = \log 5^2 = 2 \log 5 \)

Step 2: Substitute Logarithm Value

Now substitute the given value for \( \log 5 \) into the rewritten expression: \( 2 * \log 5 = 2 * 0.6990 \)

Step 3: Calculate Result

Perform the multiplication to obtain the value of \( \log 25 \): \( 2 * 0.6990 = 1.3980 \)

Key Concepts

Logarithmic Expressions

Logarithmic expressions are mathematical notations that can seem intimidating at first. However, they are simply another way to express exponential relationships. A logarithm answers the question: "To what power must the base be raised, to yield a specific number?" For instance, if we say \( \log_b a = c \), it means the base \( b \) raised to the power \( c \) equals \( a \).

When working with logarithmic expressions, it's crucial to understand their properties and how they relate to exponentiation. Just like addition and multiplication, logarithms have specific properties that simplify computations. These include:

• The product property: \( \log_b (mn) = \log_b m + \log_b n \)
• The quotient property: \( \log_b \left( \frac{m}{n} \right) = \log_b m - \log_b n \)
• The power property: \( \log_b (a^n) = n \log_b a \)

Using these properties can transform complex logarithmic expressions into manageable calculations.

Logarithm Calculations

Calculating logarithms often involves leveraging their properties to simplify the expression. This is particularly useful when a calculator is not accessible or when only specific logarithmic values are provided. In such cases, knowing these core properties becomes vital.

In the given exercise, we aim to calculate \( \log 25 \) using the properties of logarithms. Recognizing that \( 25 = 5^2 \), we can apply the power property of logarithms: \( \log (a^n) = n \log a \). This simplifies to \( \log 25 = 2 \log 5 \).

Once rewritten, you can substitute the known value of \( \log 5 \). Calculating becomes a matter of simple arithmetic by multiplying: \( 2 \times 0.6990 = 1.3980 \). This demonstrates how powerful understanding logarithm calculations can be in solving seemingly complex expressions without extensive tools.

Exponent Rules

Exponent rules are the backbone of logarithmic reasoning, as both concepts are two sides of the same mathematical coin. Understanding these rules is essential for manipulating and simplifying expressions involving powers and roots.

The key exponent rules include:

• Product of same bases: \( a^m \times a^n = a^{m+n} \)
• Quotient of same bases: \( \frac{a^m}{a^n} = a^{m-n} \)
• Power to a power: \( (a^m)^n = a^{m\cdot n} \)
• Zero exponent: \( a^0 = 1 \)

To solve the original exercise, knowing \( 25 = 5^2 \) is essential, revealing a deeper connection between exponential expressions and their logarithmic equivalents. By translating the exponential form to logarithmic form, calculations can become straightforward. For instance, the power to a power rule directly informs us how to handle logs of powers, as demonstrated in the solution for \( \log 25 \), where you see \( 2 \log 5 \). Grasping these rules means unlocking the full potential of working with logs and exponents.