Problem 106

Question

The equations are identities because they are true for all real numbers. Use properties of logarithms to simplify the expression on the left side of the equation so that it equals the expression on the right side, where \(x\) is any real number. $$\frac{1}{2} \ln \left(\frac{x^{2}}{7}\right)-\frac{1}{2} \ln \left(\frac{x^{4}+x^{2}}{7}\right)=\ln \sqrt{\frac{1}{x^{2}+1}}$$

Step-by-Step Solution

Verified
Answer
The expressions on both sides are equivalent after simplification.
1Step 1: Apply Logarithm Quotient Rule
The expression on the left is \( \frac{1}{2} \ln \left( \frac{x^2}{7} \right) - \frac{1}{2} \ln \left( \frac{x^4 + x^2}{7} \right) \). We can apply the quotient rule for logarithms, \( \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) \), to combine these logs: \( \frac{1}{2} \ln \left( \frac{x^2}{7} \left/ \frac{x^4 + x^2}{7} \right. \right) \).
2Step 2: Simplify Fraction Inside Logarithm
Simplifying the fraction inside the logarithm: \( \frac{\frac{x^2}{7}}{\frac{x^4 + x^2}{7}} = \frac{x^2}{x^4 + x^2} \) because the \(7\)s cancel out. So the expression becomes \( \frac{1}{2} \ln \left( \frac{x^2}{x^4 + x^2} \right) \).
3Step 3: Factor and Further Simplify
Factor \( x^2 \) from the denominator: \( \frac{x^2}{x^2(x^2 + 1)} = \frac{1}{x^2 + 1} \). Then, the expression becomes \( \frac{1}{2} \ln \left( \frac{1}{x^2 + 1} \right) \).
4Step 4: Expand Exponent by Property of Logarithms
Using the property of logarithms, \( a \ln(b) = \ln(b^a) \), we rewrite \( \frac{1}{2} \ln \left( \frac{1}{x^2 + 1} \right) \) as \( \ln \left( \left( \frac{1}{x^2 + 1} \right)^{1/2} \right) \).
5Step 5: Express in Final Form
Since \( \left( \frac{1}{x^2 + 1} \right)^{1/2} = \sqrt{\frac{1}{x^2 + 1}} \), the expression becomes \( \ln \sqrt{\frac{1}{x^2 + 1}} \). This matches the expression on the right side of the original equation.

Key Concepts

Properties of LogarithmsLogarithm Quotient RuleSimplifying ExpressionsReal Numbers
Properties of Logarithms
Logarithms are a powerful mathematical tool that help us simplify expressions, especially when dealing with exponential equations. They have several important properties that we use in various calculations:
  • Product Rule: The logarithm of a product is the sum of the logarithms of the factors, i.e., \( \ln(a \cdot b) = \ln(a) + \ln(b) \).

  • Quotient Rule: The logarithm of a quotient is the difference of the logarithms, i.e., \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \).

  • Power Rule: The logarithm of a power is the exponent times the logarithm of the base, i.e., \( \ln(a^b) = b \ln(a) \).

By using these properties, we can manipulate and simplify logarithmic expressions effortlessly. These rules are fundamental when performing algebraic manipulations or solving logarithmic equations.
Logarithm Quotient Rule
The logarithm quotient rule is one of the key tools used to combine logarithms into a simplified form. When you encounter a difference of logs, like \( \ln(a) - \ln(b) \), the quotient rule allows you to rewrite it as a single logarithm: \( \ln\left(\frac{a}{b}\right) \).

This transformation can be especially useful in simplifying expressions where multiple logarithms are present. By reducing the expression to a single logarithm, further simplification becomes more straightforward.This rule is applied in Step 1 of the solution where the expression \( \frac{1}{2} \ln\left(\frac{x^2}{7}\right) - \frac{1}{2} \ln\left(\frac{x^4 + x^2}{7}\right) \) is transformed into a combined logarithmic expression. This is the foundation for simplifying expressions involving logarithms.
Simplifying Expressions
Simplifying expressions involves making them easier to read or work with without changing their value. This can involve factoring, canceling common terms, or using algebraic identities to rewrite the expressions in a more concise way.

In the context of the given solution, simplifying the fraction \( \frac{\frac{x^2}{7}}{\frac{x^4 + x^2}{7}} \) by canceling the common term 7 results in \( \frac{x^2}{x^4 + x^2} \). Further simplification through factoring reveals \( \frac{x^2}{x^2(x^2 + 1)} = \frac{1}{x^2 + 1} \).

These simplifications help to reduce complexity and pave the way for finer manipulations, such as applying the logarithm properties to reach the final form of a given expression.
Real Numbers
Real numbers are all numbers that can be found on the number line. This includes all the rational numbers like fractions and integers, as well as all the irrational numbers which cannot be expressed as simple fractions, such as \( \pi \) or \( \sqrt{2} \).

In the context of logarithmic identities, it's crucial to understand that logarithms apply to positive real numbers only. The variable \( x \) in the original exercise represents any real number that satisfies this condition. This means \( x \) can be any non-negative real number because logarithms of non-positive numbers are undefined in the context of real numbers.

Whenever working with logarithms and real numbers, remember that ensuring the base and argument within a logarithmic function are positive is fundamental. This ensures that the operations and identities applied are valid and accurate.