Problem 45 - Essential Calculus Early Transcendentals

Question

Find the exact value of each expression (without a calculator.) $$ \begin{array}{l}{\text { (a) } \log _{2} 6-\log _{2} 15+\log _{2} 20} \\\ {\text { (b) } \log _{3} 100-\log _{3} 18-\log _{3} 50}\end{array} $$

Step-by-Step Solution

Answer

a) 3, b) -2

Step 1: Apply the Logarithmic Property for Subtraction

For expression (a), we know that the property of logarithms, \( \log_b A - \log_b B = \log_b \left( \frac{A}{B} \right) \), allows us to combine the logs. So, \( \log_2 6 - \log_2 15 = \log_2 \left( \frac{6}{15} \right) = \log_2 \left( \frac{2}{5} \right) \).

Step 2: Apply the Logarithmic Property for Addition

Now, use the property \( \log_b A + \log_b B = \log_b (A \times B) \). Applying this to the remaining expression, \( \log_2 \left( \frac{2}{5} \right) + \log_2 20 = \log_2 \left( \frac{2}{5} \times 20 \right) = \log_2 8 \).

Step 3: Simplify the Expression

Since we know \( 8 = 2^3 \), we can find \( \log_2 8 = 3 \) because the power to which 2 must be raised to get 8 is 3. Thus, the exact value of expression (a) is 3.

Step 4: Apply the Logarithmic Property for Subtraction (b)

For expression (b), apply the subtraction rule. \( \log_3 100 - \log_3 18 = \log_3 \left( \frac{100}{18} \right) = \log_3 \left( \frac{50}{9} \right) \).

Step 5: Further Subtraction for (b)

Use the logarithm property again to handle the second subtraction: \( \log_3 \left( \frac{50}{9} \right) - \log_3 50 = \log_3 \left( \frac{\frac{50}{9}}{50} \right) = \log_3 \left( \frac{1}{9} \right) \).

Step 6: Final Simplification for (b)

Recognize that \( \log_3 \left( \frac{1}{9} \right) \) is the same as \( \log_3 (3^{-2}) = -2 \), because \( 3^{-2} = \frac{1}{9} \). Therefore, the exact value of expression (b) is \(-2\).

Key Concepts

Logarithmic Properties

Logarithmic properties are essential tools that help us manipulate logarithmic expressions easily. These properties include the product rule, quotient rule, and power rule. They allow us to simplify complex logarithmic expressions into more manageable forms.

• **Product Rule:** This rule states that the logarithm of a product is equal to the sum of the logarithms of its factors, expressed as \( \log_b (A \times B) = \log_b A + \log_b B \). It helps in breaking down or combining logarithmic expressions into simpler or combined forms.
• **Quotient Rule:** According to this rule, the logarithm of a quotient is the difference of the logarithms, which is \( \log_b \left( \frac{A}{B} \right) = \log_b A - \log_b B \). This is particularly useful in simplifying expressions where divisions inside the logarithm can be split apart.
• **Power Rule:** This rule is about exponents within a logarithmic expression: \( \log_b (A^n) = n \cdot \log_b A \).

Understanding these properties facilitates simplifying logarithmic expressions and is critical in solving equations without a calculator. By recognizing that expressions can be rewritten using these rules, we open the door to finding exact values.

Simplifying Logarithms

Simplifying logarithms involves applying the logarithmic properties to condense or break down expressions into simpler forms. This process helps us transform an unwieldy expression into a form that's much easier to interpret and solve.For example, let's look at the expression \( \log_2 6 - \log_2 15 + \log_2 20 \). Using the **quotient rule**, we first combine the subtraction part: \( \log_2 6 - \log_2 15 = \log_2 \left( \frac{6}{15} \right) = \log_2 \left( \frac{2}{5} \right) \). Then, using the **product rule**, \( \log_2 \left( \frac{2}{5} \right) + \log_2 20 = \log_2 \left( \frac{2}{5} \times 20 \right) = \log_2 8 \).This simplification now leads to an exact expression that can be evaluated easily. Simplification highlights the power of understanding and applying the properties correctly to manipulate expressions into basic and recognizable forms.

Exact Values Without Calculator

Finding the exact values of logarithmic expressions without a calculator relies heavily on recognizing and simplifying expressions to a form that can be easily evaluated.
In our exercise examples, after simplifying an expression to forms \( \log_2 8 \) and \( \log_3 \left( \frac{1}{9} \right) \), we leverage basic powers and roots:

• The expression \( \log_2 8 \) asks for the exponent to which 2 must be raised to yield 8. Since 8 is \( 2^3 \), the answer is simply 3.
• For \( \log_3 \left( \frac{1}{9} \right) \), it's about expressing \( \frac{1}{9} \) as \( 3^{-2} \), which tells us directly that the value is -2.

Recognizing powers and decomposition of numbers into base powers are key skills in finding exact values quickly and accurately. This not only improves problem-solving efficiency but also aids deeper comprehension of logarithmic concepts.