Problem 14 - Algebra and Trigonometry

Question

\(7-18\) Evaluate the expression. $$ \log _{3} 100-\log _{3} 18-\log _{3} 50 $$

Step-by-Step Solution

Answer

The expression evaluates to -2.

Step 1: Apply Logarithm Subtraction Rule

Apply the logarithm subtraction rule: \( \log_a b - \log_a c = \log_a \left( \frac{b}{c} \right) \). This transforms the expression \( \log_{3} 100 - \log_{3} 18 - \log_{3} 50 \) into \( \log_3 \left( \frac{100}{18} \right) - \log_3 50 \).

Step 2: Simplify Using the Logarithm Rule Again

Use the logarithm subtraction rule again to combine \( \log_3 \left( \frac{100}{18} \right) - \log_3 50 \) into a single logarithm: \( \log_3 \left( \frac{\left( \frac{100}{18} \right)}{50} \right) \).

Step 3: Simplify the Fraction Inside the Logarithm

Simplify the expression \( \frac{\left( \frac{100}{18} \right)}{50} \). This results in \( \frac{100}{18 \times 50} = \frac{1}{9} \).

Step 4: Evaluate the Logarithm

Evaluate \( \log_3 \left( \frac{1}{9} \right) \). Recall that \( 9 = 3^2 \), so \( \frac{1}{9} = 3^{-2} \). Thus, \( \log_3 \left( 3^{-2} \right) = -2 \).

Key Concepts

Logarithm Subtraction Rule

Logarithms are a handy mathematical tool that help us handle exponential relationships. One of the core properties of logarithms is the subtraction rule. This rule states that for logarithms with the same base, you can transform the subtraction of two logarithms into the logarithm of a quotient. Mathematically, this is expressed as \( \log_a b - \log_a c = \log_a \left( \frac{b}{c} \right) \). Breaking it down, this means:

• We are subtracting two logarithmic expressions with the same base.
• You change the subtraction into the logarithm of the fraction \( \frac{b}{c} \).

This property is useful because it simplifies the expression significantly and reduces the number of operations needed to evaluate it. For example, in the exercise, we used this rule to simplify \( \log_3 100 - \log_3 18 - \log_3 50 \) into the form \( \log_3 \left( \frac{100}{18} \right) - \log_3 50 \). By applying it step-by-step, we make complex logarithmic equations much easier to handle.

Simplifying Fractions in Logarithms

When working with logarithms, simplifying fractions inside them is an essential skill. After converting the subtraction of logarithms into a single logarithmic expression using the subtraction rule, the next step often involves simplifying the fraction within the log. Let's see how we can do this effectively.
Consider the expression \( \log_3 \frac{\left( \frac{100}{18} \right)}{50} \). Simplifying this fraction allows us to reduce its complexity. Here, follow these steps:

• First, simplify the numerator, \( \frac{100}{18} \), which is simplified to \( \frac{50}{9} \).
• Next, divide \( \frac{50}{9} \) by 50 which results in \( \frac{1}{9} \).'

By simplifying each step, you reduce the complexity of evaluating the entire logarithmic expression, making it easier to solve. These simplifications are not just number-crunching exercises—they play a critical role in evaluating logarithms efficiently.

Evaluating Logarithms

Finally, after using properties like the subtraction rule and simplifying fractions, we reach the step of evaluating the logarithm itself. Evaluating a logarithm involves finding what power the base must be raised to produce a given number.
In this exercise, we end up needing to evaluate \( \log_3 \left( \frac{1}{9} \right) \). Here’s how:

• Recognize that \( 9 \) can be written as \( 3^2 \), so \( \frac{1}{9} \) is equivalent to \( 3^{-2} \).
• Therefore, \( \log_3 \left( 3^{-2} \right) \) asks what power 3 must be raised to in order to get \( 3^{-2} \).

The answer is clearly \(-2\) because \(3^{-2}\) is indeed the power which makes \(3\) equal to \(\frac{1}{9}\). Converting expressions and evaluating logarithms like this can often simplify complex problems and should be part of anyone's toolkit when learning about logarithms.