Problem 24 - Algebra and Trigonometry

Question

Evaluate the expression. $$ \begin{array}{llll}{\text { (a) }\log _{3} 125} & {\text { (b) } \log _{49} 7} & {\text { (c) } \log _{8} \sqrt{3}}\end{array} $$

Step-by-Step Solution

Answer

(a) \( \approx 4.3944 \), (b) \( = \frac{1}{2} \), (c) \( \approx 0.2644 \).

Step 1: Understand the expression (a)

The expression \( \log_{3} 125 \) asks for the power to which the base 3 must be raised to yield 125. Since 125 is not a simple power of 3, it's more practical to calculate the value using a calculator or logarithm conversion formulas.

Step 2: Evaluate expression (a) using change of base formula

Apply the change of base formula: \( \log_{3} 125 = \frac{\log_{10} 125}{\log_{10} 3} \). Calculate the values: \( \log_{10} 125 \approx 2.0969 \) and \( \log_{10} 3 \approx 0.4771 \). Then, \( \frac{2.0969}{0.4771} \approx 4.3944 \).

Step 3: Understand the expression (b)

The expression \( \log_{49} 7 \) asks for the power to which 49 must be raised to equal 7. Notice that 49 is the square of 7, i.e., \( 49 = 7^2 \).

Step 4: Solve expression (b) using properties of logarithms

Since 49 can be written as \( 7^2 \), use the property \( \log_{a^b} c = \frac{1}{b} \log_{a} c \). Therefore, \( \log_{49} 7 = \frac{1}{2} \log_{7} 7 \). Since \( \log_{7} 7 = 1 \), it follows that \( \log_{49} 7 = \frac{1}{2} \cdot 1 = \frac{1}{2} \).

Step 5: Understand the expression (c)

The expression \( \log_{8} \sqrt{3} \) asks for the power to which 8 needs to be raised to produce \( \sqrt{3} \). Note that \( \sqrt{3} = 3^{1/2} \).

Step 6: Solve expression (c)

Using the property \( \log_{a} b^c = c \cdot \log_{a} b \), rewrite the expression as \( \log_{8} 3^{1/2} = \frac{1}{2} \cdot \log_{8} 3 \). Using a calculator, find \( \log_{8} 3 \approx 0.5288 \). Therefore, \( \frac{1}{2} \cdot 0.5288 \approx 0.2644 \).

Step 7: Final Statement

The solutions are: (a) \( \log_{3} 125 \approx 4.3944 \), (b) \( \log_{49} 7 = \frac{1}{2} \), and (c) \( \log_{8} \sqrt{3} \approx 0.2644 \).

Key Concepts

Change of Base Formula

The change of base formula is a handy tool when dealing with logarithms of arbitrary bases. It's especially useful when your calculator does not have a button for that base. The formula works by expressing a logarithm as the ratio of common or natural logarithms. It is defined as:

• For any positive numbers a, b, and c where a and c are not equal to 1, the formula is: \( \log_{a} b = \frac{\log_{c} b}{\log_{c} a} \).

To use this formula, you choose a base c that your calculator can handle, typically 10 or e.
For example, to evaluate \( \log_{3} 125 \), you express it as \( \frac{\log_{10} 125}{\log_{10} 3} \). This allows you to calculate it using logs to base 10.
You calculate both the numerator and the denominator separately using your calculator and then divide them to find the result. This approach greatly simplifies the process of evaluating complex logarithmic expressions with unfamiliar bases.

Properties of Logarithms

Logarithms have several properties that make them extremely useful in simplifying and solving expressions. Some key properties include:

• Product Property: \( \log_{a}(xy) = \log_{a}x + \log_{a}y \)
• Quotient Property: \( \log_{a}\left(\frac{x}{y}\right) = \log_{a}x - \log_{a}y \)
• Power Property: \( \log_{a}(x^{n}) = n \cdot \log_{a}x \)
• Change of Base Formula: \( \log_{a}b = \frac{\log_{c}b}{\log_{c}a} \)

In the case of \( \log_{49} 7 \), we notice that 49 is a power of 7, specifically \( 49 = 7^2 \). This allows us to use the power property: \( \log_{49} 7 = \frac{1}{2} \log_{7} 7 \).
Since \( \log_{7} 7 = 1 \), it simplifies to \( \frac{1}{2} \cdot 1 = \frac{1}{2} \). This makes evaluating logarithms in such a form straightforward and helps find solutions quickly.

Logarithm Evaluation

Evaluating logarithms means finding the value of the exponent that turns the base into the number. This often requires using a combination of known log values, properties of logarithms, or approximations via calculator.

• For example, to evaluate \( \log_{8} \sqrt{3} \), consider recognizing \( \sqrt{3} \) as \( 3^{1/2} \) before applying the properties.
• By using the power property, you rewrite it as \( \log_{8} 3^{1/2} = \frac{1}{2} \cdot \log_{8} 3 \).

Calculate \( \log_{8} 3 \) using your calculator. You might need to apply the change of base formula \( \frac{\log_{10} 3}{\log_{10} 8} \) to convert it into a form your calculator can handle, giving \( \log_{8} 3 \approx 0.5288 \).
Multiply this by \( \frac{1}{2} \) to get the evaluated log value of \( \log_{8} \sqrt{3} \approx 0.2644 \). By understanding these processes, evaluating more complex logarithmic expressions becomes manageable and less tedious.