Problem 35 - College Algebra

Question

In Exercises \(1-40,\) use properties of logarithms to expand each logarithmic expression as much as possible. Where possible, evaluate logarithmic expressions without using a calculator. $$ \log _{5} \sqrt[3]{\frac{x^{2} y}{25}} $$

Step-by-Step Solution

Answer

The expanded logarithmic expression is \( \frac{2}{3}*\log _{5} {x} + \frac{1}{3}*\log _{5} {y} - \frac{2}{3}\)

Step 1: Identify the root, base, and the expression inside the root

Here, the logarithmic base is 5, the index of the root is 3, and the expression inside the root is \(\frac{x^{2} y}{25}\)

Step 2: Apply the property of the root inside the logarithm

We can apply our third mentioned property of logarithm. So, \( \log _{5} \sqrt[3]{\frac{x^{2} y}{25}} = \frac{1}{3} * \log _{5} {(\frac{x^{2} y}{25})}\)

Step 3: Apply the property of the logarithm of a quotient

Now let's apply the first mentioned property of logarithm. So, \( \frac{1}{3} * \log _{5} {(\frac{x^{2} y}{25})} = \frac{1}{3} * (\log _{5} {x^{2} y} - \log _{5} {25}) \)

Step 4: Apply the property of the logarithm of a product

We apply the property that says that the logarithm of a product is the sum of the logarithms. So, \( \frac{1}{3} * (\log _{5} {x^{2} y} - \log _{5} {25}) = \frac{1}{3} * (\log _{5} {x^{2}} + \log _{5} {y} - \log _{5} {25}) \)

Step 5: Apply the property of the logarithm of a power

Applying the second mentioned property of logarithms gives \( \frac{1}{3} * (\log _{5} {x^{2}} + \log _{5} {y} - \log _{5} {25}) = \frac{1}{3} * (2*\log _{5} {x} + \log _{5} {y} - \log _{5} {25}) \)

Step 6: Evaluate the constant logarithm

\(\log _{5} {25}\) is a constant logarithm, because both the number and the base are known. Since \(5^2 = 25\), \(\log _{5} {25} = 2\). Substituting this into our expression gives \( \frac{1}{3} * (2*\log _{5} {x} + \log _{5} {y} - 2) = \frac{2}{3}*\log _{5} {x} + \frac{1}{3}*\log _{5} {y} - \frac{2}{3}\)

Key Concepts

Logarithmic Expressions

Logarithmic expressions represent the exponents or powers to which a base must be raised to produce a certain number. A simple example is the expression \( \text{log}_b(x) \), which asks for the power that the base \( b \) is raised to get \( x \). Understanding these expressions is fundamental when working with exponential and logarithmic functions.

For instance, in the given exercise, the expression \( \text{log}_5 \ text{log}_5 (x^2) \), the base is \( 5 \), and we want to know the exponent that \( 5 \) is raised to get \( x^2 \). This is a foundational concept that allows for further manipulation of logarithmic expressions using properties of logarithms.

Expanding Logarithms

Expanding logarithms involves using logarithmic identities to rewrite logarithmic expressions in a form that involves sums, differences, and multiples of simpler logarithms. It’s a crucial skill for simplifying complex logs and solving logarithmic equations.

In our exercise, expanding the logarithm \( \text{log}_5 \ text{log}_5 (y) \). These identities are pivotal to breaking down complicated logarithms into more manageable parts.

Logarithm Properties

The properties of logarithms are derived from the properties of exponents and include the product, quotient, and power rules. They allow us to manipulate logarithmic expressions and solve logarithmic equations easily.

• Product Rule: \( \text{log}_b(M \ text{log}_b(M) + \text{log}_b(N) \).
• Quotient Rule: \( \text{log}_b\text{log}_b\text{log}_b(M) - \text{log}_b(N) \).
• Power Rule: \( \text{log}_b(M^k) = k \text{log}_b(M) \).

These properties were sequentially applied to expand the initial logarithmic expression in the textbook exercise, demonstrating how to systematically simplify and evaluate the expression.

Evaluating Logarithms

Evaluating logarithms means finding the numerical value of logarithmic expressions, often without a calculator. This is possible when the argument of the log is a power of the log's base or can be transformed into such a form through algebraic manipulation.

For example, in the final step of our exercise, knowing that \( 5^2 = 25 \) allows us to evaluate \( \text{log}_5 25 \) directly as \( 2 \). This step is critical as it taps into prior knowledge of exponents to simplify the logarithm into a recognizable form. When logarithms include known values, they become much easier to work with, and evaluating them becomes a straightforward task.