Problem 32 - Calculus of a Single Variable

Question

In Exercises 29–34, write the expression as a logarithm of a single quantity. $$ 2[\ln x-\ln (x+1)-\ln (x-1)] $$

Step-by-Step Solution

Answer

Therefore, the expression \(2[\ln x-\ln (x+1)-\ln (x-1)]\) simplifies to \(\ln(\frac{x^2}{(x+1)^2*(x-1)^2})\).

Step 1: Distribute the constant

Multiply the expression inside the brackets by the constant 2. This gives \(2\ln x - 2\ln(x+1) - 2\ln(x-1)\).

Step 2: Use the power rule of logarithms

The power rule of logarithms states that \(a\ln b = \ln(b^a)\). Applying this rule, the expression becomes \(\ln(x^2) - \ln((x+1)^2) - \ln((x-1)^2)\).

Step 3: Use the quotient rule of logarithms

The quotient rule for logarithms states that \(\ln(a) - \ln(b) = \ln(a/b)\). Applying this rule, the final expression is \(\ln(\frac{x^2}{(x+1)^2*(x-1)^2})\).

Key Concepts

Power Rule of Logarithms

Logarithms often come with exponents, and the power rule of logarithms is a handy tool in simplifying such expressions. This rule simply states that the logarithm of a number raised to a power is equal to the power multiplied by the logarithm of the number, expressed as \( a \log_b{c} = \log_b{c^a} \).
For example, if we have an expression like \(2\ln x\), we can use the power rule to rewrite it as \(\ln x^2\). This transformation greatly simplifies the task of combining logarithmic terms by allowing us to work with the exponents directly. It's particularly powerful in calculus exercises where exponents frequently appear and must be manipulated for differentiating or integrating logarithmic expressions.

Quotient Rule of Logarithms

When faced with logarithms of ratios, the quotient rule of logarithms comes to the rescue. This rule helps us combine two logarithms that are being subtracted. It states that the difference between the logarithm of two numbers is the logarithm of the quotient of those numbers: \( \log_b{a} - \log_b{c} = \log_b{\frac{a}{c}} \).

It's incredibly useful because it turns what would be a complicated expression into a much simpler one. For instance, if we apply it to \( \ln(x^2) - \ln((x+1)^2) \), we get \( \ln(\frac{x^2}{(x+1)^2}) \), consolidating the terms into a single logarithmic expression. The quotient rule shines in calculus where the combination and simplification of logarithmic expressions often herald the next steps of a problem.

Logarithmic Expressions

Logarithmic expressions, such as \( \ln(x) \) or \( \log_{10}(x) \), represent the power to which a base must be raised to produce a given number. They're reverse operations of exponentiation and are essential in solving equations involving exponents. When these expressions involve terms that can be combined or simplified, like in the case of the exercise, understanding how to apply the properties of logarithms becomes crucial.

With a strong grasp of rules like the power and quotient rule, students can break down complex logarithmic expressions into more manageable pieces. These skills are not only foundational for algebra but are also applied in higher-level math like calculus, economics, and the sciences where logarithmic functions model phenomena such as sound intensity, earthquakes, and acidity.

Calculus Exercises

Calculus exercises involving logarithms test a student's ability to apply various properties and rules to differentiate and integrate. These exercises often require manipulating expressions to a form that is more suitable for the calculus operations. In our example, simplifying \(2[\ln x-\ln (x+1)-\ln (x-1)]\) paves the way for further application in calculus, such as finding derivatives or integrals.

As students progress through calculus, they'll encounter logarithmic differentiation, a method where taking the natural log of both sides of an equation simplifies the differentiation process. Understanding how to simplify logarithmic expressions beforehand makes such advanced techniques more approachable and highlights the interconnectedness of these mathematical concepts.