Problem 16 - Intermediate Algebra

Question

Complete each solution. $$\log \frac{r}{s t}=\log r-\log (\square)$$ $$\quad\quad\quad\quad\quad\quad=\log r-(\log\square +\log t)$$ $$\quad\quad\quad\quad=\log r-\log s\square\square$$

Step-by-Step Solution

Answer

Fill the blanks with \( st \), \( s \), and \( t \) respectively.

Step 1: Recall Properties of Logarithms

Recall that the property \( \log \frac{a}{b} = \log a - \log b \) allows us to separate the logarithm of a fraction into a difference of two logarithms.

Step 2: Identify the Denominator of the Fraction

The given fraction is \( \frac{r}{s t} \). Thus, according to the property, it is split as follows: \( \log \frac{r}{st} = \log r - \log(st) \). The denominator \( st \) is what is subtracted in the first part of the solution.

Step 3: Apply the Product Rule for Logs

The expression \( \log(st) \) can be expanded using the product rule for logarithms: \( \log(st) = \log s + \log t \). Thus, in the context of the problem, \( \log r - \log(st) = \log r - (\log s + \log t) \).

Step 4: Identify Complete Solution for Blanks

The question provides blanks for where expressions should be added. In the first equation, \( \log r - \log(\square) = \log r - \log(st) \), thus \( \square = st \). In the second equation, \( \log r - (\log \square + \log t) \), thus \( \square = s \). For the third equation, \( \log r - \log s\square \square \), noting that \( \log(st) = \log s + \log t \), the operation implies \( \square \square = t \), identifying the blank entries in each case.

Key Concepts

Logarithmic Identities

Logarithmic identities are essential mathematical tools that simplify complex logarithmic expressions. They help us to break down and understand logarithms by expressing them in simpler terms. One of the primary logarithmic identities is the quotient rule, which states:

• \( \log \frac{a}{b} = \log a - \log b \)

This identity is particularly useful when dealing with a logarithm of a fraction, allowing us to split it into the difference of two individual logarithms. The quotient rule simplifies expressions by breaking them into manageable parts, making it easier to perform operations like addition and subtraction on logarithms. Understanding and applying logarithmic identities is crucial in solving equations where logarithms are present, as seen in the exercise where \( \log \frac{r}{st} \) is split into \( \log r - \log(st) \). This foundational identity is pivotal to further manipulations such as logarithmic expansions and application of other properties.

Logarithmic Expansion

Logarithmic expansion is a method used to break down complex logarithmic expressions into simpler components. This is done by applying various properties and identities of logarithms. A common tool used in logarithmic expansion is the product rule, which is an integral part of the detailed solution we explored:

• \( \log(ab) = \log a + \log b \)

When you encounter expressions like \( \log(st) \), they can be expanded using this product rule into \( \log s + \log t \). This not only simplifies the expression but also makes it easier to identify each term and manipulate them when solving logarithmic equations. Logarithmic expansion facilitates calculations by linearizing the otherwise multiplicative relationships within the logarithms. Through expansion, we convert products and quotients inside a logarithm into sums and differences outside, respectively. This step is vital for simplifying complex logarithmic equations as shown when the product \( st \) was transformed into the sum of \( \log s \) and \( \log t \).

Product Rule for Logarithms

The product rule for logarithms is a powerful technique used in the simplification and calculation of logarithmic expressions. It states:

• \( \log(ab) = \log a + \log b \)

This rule allows us to decompose a logarithm of a product into the sum of two separate logarithms, each representing one of the factors in the product. In the context of solving logarithmic equations, the product rule is invaluable because it lets you distribute the logarithm across a multiplication, turning a potentially complex problem into a simpler one by transforming a product into an addition. When applying this to the problem statement given in the original solution steps, we used the product rule to break down \( \log(st) \) in the expression \( \log \frac{r}{st} \). By expanding it to \( \log s + \log t \), we could effectively insert these terms into the entire expression as \( \log r - (\log s + \log t) \), making it easier to see what components needed to be added or subtracted to achieve the correct solution.