Problem 19 - Algebra and Trigonometry

Question

\(15-24\) Evaluate the expression. $$ \begin{array}{llll}{\text { (a) } \log _{3}\left(\frac{1}{27}\right)} & {\text { (b) } \log _{10} \sqrt{10}} & {} & {\text { (c) } \log _{5} 0.2}\end{array} $$

Step-by-Step Solution

Answer

(a) -3, (b) 0.5, (c) -1

Step 1: Understanding Logarithms

Recall that if \( \log_b(x) = y \), then \( b^y = x \). This will guide us in evaluating each log expression.

Step 2: Calculating \( \log_3\left(\frac{1}{27}\right) \)

Notice that \( \frac{1}{27} = 27^{-1} \) and since \( 27 = 3^3 \), we have \( \frac{1}{27} = (3^3)^{-1} = 3^{-3} \). So \( \log_3\left(\frac{1}{27}\right) = \log_3(3^{-3}) \). Using the property \( \log_b(b^x) = x \), we find \( \log_3(3^{-3}) = -3 \).

Step 3: Calculating \( \log_{10} \sqrt{10} \)

Recognize that \( \sqrt{10} = 10^{1/2} \). Therefore, \( \log_{10} \sqrt{10} = \log_{10}(10^{1/2}) \). Using the same property, \( \log_b(b^x) = x \), we get \( \log_{10} \sqrt{10} = 1/2 \).

Step 4: Calculating \( \log_5(0.2) \)

Rewrite 0.2 as \( \frac{1}{5} \). Hence, \( \log_5(0.2) = \log_5\left(\frac{1}{5}\right) \). Recognize that \( \frac{1}{5} = 5^{-1} \). Therefore, \( \log_5(5^{-1}) = -1 \).

Step 5: Summarize the Results

Compile the results from each part: (a) \( \log_3\left(\frac{1}{27}\right) = -3 \); (b) \( \log_{10} \sqrt{10} = \frac{1}{2} \); (c) \( \log_5(0.2) = -1 \).

Key Concepts

Properties of Logarithms

Logarithms are mathematical functions that help us express powers and exponents in a different form. They follow specific rules, known as properties, that simplify the process of working with them.

• Product Rule: The logarithm of a product is the sum of the logarithms. Formula: \( \log_b(MN) = \log_b(M) + \log_b(N) \).
• Quotient Rule: The logarithm of a quotient is the difference of the logarithms. Formula: \( \log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N) \).
• Power Rule: The logarithm of a number raised to a power is the power times the logarithm of the number. Formula: \( \log_b(M^n) = n \cdot \log_b(M) \).
• Base Rule: \( \log_b(b) = 1 \) because any number to the power of 1 is itself.

These properties allow for easier manipulation and evaluation of logarithmic expressions, which you'll see in the examples above.

Evaluating Logarithmic Expressions

When evaluating logarithmic expressions, you often need to transform or simplify the expression using the properties of logarithms.
For example, consider \( \log_3\left(\frac{1}{27}\right) \). By rewriting \( \frac{1}{27} \) as \( 3^{-3} \), you use the Power Rule to find that \( \log_3\left(3^{-3}\right) = -3 \).
Similarly, for \( \log_{10} \sqrt{10} \), rewrite \( \sqrt{10} \) as \( 10^{1/2} \), making it simple to apply the Power Rule and evaluate it as \( \frac{1}{2} \).
Understanding how to rewrite numbers and apply these properties helps in simplifying and evaluating logarithmic expressions effectively.

Change of Base Formula

While the exercise doesn’t explicitly require the change of base formula, it's a useful tool for many logarithmic evaluations.
This formula allows you to evaluate logs with bases that aren’t immediately usable with a calculator.
The formula is \( \log_b(a) = \frac{\log_c(a)}{\log_c(b)} \), where \( c \) is a new base that you can choose, often 10 or \( e \), for convenience.
For instance, if you wanted to evaluate \( \log_2(8) \) on a calculator, you could use the change of base formula: \( \log_2(8) = \frac{\log_{10}(8)}{\log_{10}(2)} \).
Using the formula gives access to a broader range of calculations when practical tools are limited.

Exponential Expressions

Exponential expressions and functions involve numbers written in the form \( b^x \), where \( b \) is the base and \( x \) is the exponent.
Understanding exponential expressions is pivotal for grasping logs, as logarithms are essentially the inverse operations of exponentials.
When solving logs, you often rewrite numbers in their exponential form to utilize properties of logs, such as converting \( \frac{1}{27} \) to \( 3^{-3} \).
Exponential expressions are evident in natural growth processes, compound interest, and scientific phenomena, making them relevant and practical in real-world scenarios.
Mastering exponentials and their connection to logs will lead to a better understanding of both mathematical concepts.