Problem 18 - Precalculus : Building Concepts and Connections

Question

In Exercises \(15-20,\) use the properties of logarithms to simplify each expression by eliminating all exponents and radicals. Assume that \(x, y > 0\). $$\log \sqrt[5]{x^{2}} \sqrt{y^{5}}$$

Step-by-Step Solution

Answer

\(\frac{2}{5} \log(x) + \log(y)\)

Step 1: Breaking down the complex roots

We can first simplify the complex root expression outside the logarithm by recognizing it as the product of two square roots, so it can be rewritten: \(\sqrt[5]{x^2}y.\)

Step 2: Applying the logarithm identity

The given expression is now in a form that a logarithm identity can be applied to - that identity is the product rule which states that \(\log(AB) = \log(A) + \log(B)\). Applying this rule we get the following expression: \(\log(\sqrt[5]{x^2}) + \log(y).\)

Step 3: Simplifying the Logarithm

There is another logarithm identity that could be used to further simplify this expression - the power rule, which states that \(\log(A^n) = n \cdot \log(A)\). Using this rule will give us \(\frac{2}{5} \log(x) + \log(y)\). Notice that we represented the fifth root as a power of 2/5, which is equivalent.

Key Concepts

Understanding the Properties of Logarithms

Logarithms have certain properties that make it easier to simplify complex expressions. These properties include the product rule, quotient rule, and power rule. Each of these rules allows us to transform and simplify logarithmic expressions in a structured way.

When simplifying logarithms, these properties act like tools. They help break down more complicated expressions into simpler parts, making them easier to work with. Let's dive into some of these essential properties to see how they work.

• Product Rule: Allows you to split logs of products into sums.
• Quotient Rule: Transforms logs of quotients into differences.
• Power Rule: Moves exponents in logs to become coefficients.

Applying the Product Rule in Simplifying Logs

The product rule for logarithms states that \(\log(AB) = \log(A) + \log(B)\).

This rule comes in handy when you need to break down a logarithm of a product. By applying the product rule, the expression \(\log(\sqrt[5]{x^2} \sqrt{y^5})\) can be rewritten as \(\log(\sqrt[5]{x^2}) + \log(\sqrt{y^5})\).

This is because the expression inside the logarithm is a product of two terms. Breaking it into two separate logs makes further steps of simplification more manageable.
Remember: The objective of using the product rule is to transform a complex multiplicative expression into an easier-to-handle additive one.

Using the Power Rule for Simplification

The power rule for logarithms states that \(\log(A^n) = n \cdot \log(A)\).

This rule is particularly useful when dealing with roots and exponents within a logarithmic expression. For example, since \(\sqrt[5]{x^2}\) is equivalent to \(x^{\frac{2}{5}}\), you can apply the power rule to simplify \(\log(\sqrt[5]{x^2})\) to \(\frac{2}{5} \log(x)\).

The same applies for \(\sqrt{y^5}\), which could be simplified to \(\frac{5}{2} \log(y)\). Using the power rule transforms complex exponents and roots into coefficients, greatly simplifying the expression.
This transformation makes it straightforward to compress complex computations into simpler operations.

Simplifying Expressions Effectively

The ultimate goal of using logarithmic properties is to achieve a simplified expression that’s manageable. By employing the product and power rules effectively, you simplify the task of dealing with complicated logarithmic expressions.

In the given exercise, by applying these rules, the expression \(\log(\sqrt[5]{x^2} \sqrt{y^5})\) was successfully simplified to \(\frac{2}{5} \log(x) + \frac{5}{2} \log(y)\).

Consider these steps when simplifying:

• Break down components: Use the product rule to split into simpler parts.
• Apply power adjustments: Use the power rule to manage exponents or roots.
• Simplify incrementally: Work through the expression step by step for clarity.

By following these methods, handling logarithmic expressions becomes much more approachable.