Problem 82 - Intermediate Algebra

Question

Write each logarithmic expression as one logarithm. See Example 7. $$ \frac{1}{4}\left[\log _{r}\left(n^{2}-16\right)-\log _{r}(n-4)\right] $$

Step-by-Step Solution

Answer

\( \log_{r}((n+4)^{1/4}) \)

Step 1: Apply the logarithm subtraction rule

The expression given is \( \frac{1}{4}\left[\log_{r}(n^{2}-16)-\log_{r}(n-4)\right] \). We start by using the property \( \log_{b}(A) - \log_{b}(B) = \log_{b}(\frac{A}{B}) \). Therefore, \( \log_{r}(n^{2}-16) - \log_{r}(n-4) = \log_{r}\left(\frac{n^{2}-16}{n-4}\right) \).

Step 2: Simplify the inside expression

Inside the logarithm, we have \( \frac{n^{2}-16}{n-4} \). Notice that \( n^{2}-16 \) is a difference of squares and can be factored into \( (n-4)(n+4) \). Thus, \( \frac{n^{2}-16}{n-4} = \frac{(n-4)(n+4)}{n-4} = n+4 \), assuming \( n eq 4 \).

Step 3: Apply the constant multiplier

After simplification, the expression becomes \( \frac{1}{4}\log_{r}(n+4) \). The constant \( \frac{1}{4} \) in front of the logarithm can be applied using the power rule for logarithms, \( c \cdot \log_{b}(A) = \log_{b}(A^c) \). Therefore, \( \frac{1}{4} \log_{r}(n+4) = \log_{r}((n+4)^{1/4}) \).

Key Concepts

Understanding the Logarithm Subtraction Rule

In order to simplify the given expression, we start with the concept of the *logarithm subtraction rule*. This rule is very useful when you have the subtraction of two logarithms with the same base. The rule states that \( \log_{b}(A) - \log_{b}(B) = \log_{b}\left( \frac{A}{B} \right) \).This allows us to condense two logarithm expressions into a single one. When applying this rule, the numerators and denominators in the arguments of the logarithms switch place and become a division in the new single logarithm argument.
Using this rule simplifies expressions and eases further manipulation. Always make sure that the bases of the logarithms you are dealing with are the same, and the expressions inside the logarithms are positive. This rule is pivotal for computing logarithms efficiently and is used frequently in algebraic calculations.

Importance of Factorization

Factorization is a mathematical process used to break down an expression into a product of simpler factors. In our exercise, we encounter the expression \( n^2-16 \). Recognizing this as a difference of squares is crucial.
The difference of squares formula is written as \( a^2-b^2 = (a-b)(a+b) \). Here, replace \( a \) with \( n \) and \( b \) with 4, to get \( n^2-16 = (n-4)(n+4) \). Factorization simplifies expressions and is an essential skill for reducing complex algebraic expressions effectively.
Once factored, these expressions allow simplification of the overall problem by canceling out terms, as long as they don't result in a zero denominator or any undefined terms. In our solution, canceling \( n-4 \) results in a much simpler expression to manage in subsequent steps.

Utilizing the Power Rule for Logarithms

The power rule of logarithms is another powerful tool that helps in manipulating expressions where a constant is multiplied with a logarithm. This rule states that \( c \cdot \log_{b}(A) = \log_{b}\left(A^c\right) \). This can be especially useful in cases like our exercise where you multiply a logarithm by a fraction or other constants.
In the solution, after simplifying inside the logarithm, we have \( \frac{1}{4} \log_{r}(n+4) \). By applying the power rule, this expression simplifies to a single logarithm: \( \log_{r}((n+4)^{1/4}) \).Using the power rule saves computation time and reduces complexity, which is especially handy in solving more challenging and lengthy logarithmic expressions. Understanding when and how to use this rule, allows for converting multiplication outside a logarithm into an exponent inside the logarithm efficiently.