Problem 50 - College Algebra and Calculus: An Applied Approach

Question

In Exercises, write the expression as the logarithm of a single quantity. $$ \frac{1}{2} \ln (x-2)+\frac{3}{2} \ln (x+2) $$

Step-by-Step Solution

Answer

The given expression, when written as the logarithm of a single quantity, becomes \( \ln [(x^2 - 4)^{1/2}(x^2 + 4x + 4)^{3/2}] \)

Step 1: Simplify terms

The given expression can be expressed as: \( \frac{1}{2} \ln (x-2)+ \frac{3}{2} \ln (x+2) \). The coefficient of the logarithm can be written as the exponent of its argument using logarithmic property, so simplify the expression into: \( \ln [(x-2)^{1/2}]+ \ln [(x+2)^{3/2}] \).

Step 2: Apply logarithmic product rule

The sum of two logarithms equals the logarithm of the product of their arguments. Applying this property: \( \ln [(x-2)^{1/2}(x+2)^{3/2}] \).

Step 3: Consolidate into a single log expression

Lastly, the expression inside the log function can be simplified and written as a single quantity. This results in the following simplified expression: \( \ln [(x^2 - 4)^{1/2}(x^2 + 4x + 4)^{3/2}] \)

Key Concepts

Logarithm Properties

Whenever you're dealing with logarithmic expressions, understanding the basic properties is essential. One crucial property is that logarithms can help transform multiplication into addition and vice versa. Another is how coefficients in front of logarithms can be rewritten as exponents. This is called the power rule and is given by:

• \( a \ln(b) = \ln(b^a) \)

In the provided exercise, coefficients like \( \frac{1}{2} \) and \( \frac{3}{2} \) are transformed into exponents for their respective terms. This means \( \frac{1}{2} \ln (x-2) \) becomes \( \ln [(x-2)^{1/2}] \). Similarly, \( \frac{3}{2} \ln (x+2) \) becomes \( \ln [(x+2)^{3/2}] \). By applying this property, expressions get much simpler and easier to work with.
This concept is fundamental for transforming and simplifying complex logarithmic expressions.

Logarithmic Product Rule

The logarithmic product rule states that the sum of two logarithms with the same base is equal to the logarithm of the product of their arguments. Specifically,

• \( \ln(a) + \ln(b) = \ln(ab) \)

This is handy for condensing logarithmic expressions. In the exercise, once the coefficients were turned into exponents, you could apply this rule. Taking \( \ln [(x-2)^{1/2}] + \ln [(x+2)^{3/2}] \), you use the product rule to combine it into a single logarithm:

• \( \ln [(x-2)^{1/2}(x+2)^{3/2}] \)

This simplifies multiple terms into a single expression, demonstrating the power of logarithmic properties in breaking down and combining expressions.

Exponents in Logarithms

Exponents in logarithms allow us to take seemingly complex expressions and simplify them substantially. By turning coefficients into exponents on their respective base numbers, complex products and sums within logarithms can be managed effectively. For example, when addressing expressions like \( \frac{1}{2} \ln (x-2) \) in the task, it is simplified by transforming this into \( \ln [(x-2)^{1/2}] \).

• This utilizes the power property: \( a \ln(b) = \ln(b^a) \).

Once these transformations are made, the expression becomes more compact and ready for further simplification using properties like the product rule.
This manipulation of exponents is a core skill when solving logarithmic expressions, making it easier to deal with equations and perform operations such as integration or differentiation in calculus.