Problem 27 - Intermediate Algebra

Question

For Problems \(21-40\), evaluate each expression. $$ \log _{7} \sqrt{7} $$

Step-by-Step Solution

Answer

The evaluated expression is \( \frac{1}{2} \).

Step 1: Understand the Expression

The exercise requires evaluating the logarithm \( \log_{7} \sqrt{7} \). Here, \( \log_{7} \) is a logarithm with base 7 and \( \sqrt{7} \) is the square root of 7.

Step 2: Rewrite the Square Root

Recognize that \( \sqrt{7} \) can be rewritten as \( 7^{1/2} \). This is because the square root of a number is the same as raising that number to the power of \( 1/2 \).

Step 3: Apply Logarithmic Identity

Use the logarithmic identity \( \log_{b} (a^{c}) = c \log_{b} a \). In our expression, substitute \( a = 7 \), \( b = 7 \), and \( c = 1/2 \). Thus, \( \log_{7} (7^{1/2}) = \frac{1}{2} \log_{7} 7 \).

Step 4: Evaluate the Simple Logarithm

Recognize that \( \log_{7} 7 = 1 \) since the logarithm of a number to its own base is always 1. Substitute this back into the identity from Step 3 to get \( \frac{1}{2} \times 1 = \frac{1}{2} \).

Step 5: Verify Your Solution

Reassure the correctness by acknowledging that the simplification and application of logarithmic identities were done correctly, leading you to the result. The evaluated expression is \( \frac{1}{2} \).

Key Concepts

Logarithmic Identity

A logarithmic identity is a mathematical rule that helps us simplify logarithmic expressions. One of the most useful identities is:

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

This identity means if you have a number raised to a power inside a logarithm, you can pull that power out front. This can make the calculation simpler.

In our exercise, we used this identity to evaluate \( \log_{7} \sqrt{7} \). We noticed that the square root \( \sqrt{7} \) is the same as raising 7 to the power of 1/2, or \( 7^{1/2} \).

By applying the identity, we moved the exponent 1/2 in front of the logarithm, resulting in the expression \( \frac{1}{2} \log_{7} 7 \). This step is crucial because it simplifies the expression considerably and allows us to evaluate the logarithm easily.

Logarithmic Evaluation

Evaluating a logarithm means finding out what it equals. Once we simplified \( \log_{7} \sqrt{7} \) to \( \frac{1}{2} \log_{7} 7 \), the evaluation became straightforward.

We know another important rule of logarithms:

• A logarithm of a number to its own base equals 1.

That's because the base raised to the power of 1 gives you the same number back. In this exercise, \( \log_{7} 7 = 1 \) because 7 to the power of 1 is 7 itself.
Substituting this back into our simplified expression gives \( \frac{1}{2} \times 1 \). Calculating this product results in 0.5, which is the final evaluation of the original logarithmic expression.

Mathematical Expressions

In mathematics, expressions are combinations of numbers, variables, and operation symbols representing a certain quantity.

In the context of the exercise, the original problem \( \log_{7} \sqrt{7} \) is a mathematical expression that involves both a logarithm and a radical.

To effectively solve such an expression, it is often helpful to rewrite components of the expression, so they are easier to work with, as demonstrated.

• We transformed \( \sqrt{7} \) into \( 7^{1/2} \).
• Then, we used logarithmic identities to simplify and solve the expression.

This kind of manipulation is an essential skill in mathematics, as it helps to tackle complex expressions by breaking them down into more manageable parts. Understanding these manipulations allows students to evaluate and simplify a range of different mathematical problems.