Problem 1 - A Graphical Approach to College Algebra

Question

In Exercises 1 and \(2,\) match the logarithm in Column I with its value in Column II. Remember that \(\log _{a} x\) is the exponent to which a must be raised in order to obtain \(x\). (a) \(\log _{2} 16\) (b) \(\log _{3} 1\) (c) \(\log _{10} 0.1\) (d) \(\log _{2} \sqrt{2}\) (e) \(\log _{e} \frac{1}{e^{2}}\) (f) \(\log _{1 / 2} 8\) A. 0 B. \(\frac{1}{2}\) C. 4 D. \(-3\) E. \(-1\) F. \(-2\)

Step-by-Step Solution

Answer

(a) C, (b) A, (c) E, (d) B, (e) F, (f) D.

Step 1: Understanding Logarithms

Logarithms are the inverse of exponential functions. The expression \( \log_{a} x \) asks the question: "To what power must \( a \) be raised to obtain \( x \)?" This understanding will help us match the logarithms to their correct values.

Step 2: Calculate \( \log_{2} 16 \)

The equation asks \( 2^y = 16 \). Since \( 2^4 = 16 \), we have \( y = 4 \). Therefore, \( \log_{2} 16 = 4 \), which matches with C.

Step 3: Calculate \( \log_{3} 1 \)

Any number raised to the power of 0 equals 1, i.e., \( 3^0 = 1 \). Therefore, \( \log_{3} 1 = 0 \), which matches with A.

Step 4: Calculate \( \log_{10} 0.1 \)

The task is to find \( y \) such that \( 10^y = 0.1 \). Since \( 0.1 = 10^{-1} \), it follows that \( \log_{10} 0.1 = -1 \), which matches with E.

Step 5: Calculate \( \log_{2} \sqrt{2} \)

We know \( \sqrt{2} = 2^{1/2} \). So, \( \log_{2} \sqrt{2} = 1/2 \), which matches with B.

Step 6: Calculate \( \log_{e} \frac{1}{e^{2}} \)

Find \( y \) such that \( e^y = \frac{1}{e^2} \). This translates to \( y = -2 \) because \( e^{-2} = \frac{1}{e^2} \). Therefore, \( \log_{e} \frac{1}{e^{2}} = -2 \), which matches with F.

Step 7: Calculate \( \log_{1/2} 8 \)

We look for \( y \) such that \( (\frac{1}{2})^y = 8 \). Noting that \( 8 = (\frac{1}{2})^{-3} \), it follows that \( y = -3 \). Therefore, \( \log_{1/2} 8 = -3 \), which matches with D.

Key Concepts

Logarithmic Functions

Logarithmic functions are a powerful mathematical tool that helps us understand how often a certain base needs to be multiplied by itself to achieve a certain value. A fundamental property of logarithms is that they are the inverse of exponential functions, which means they "undo" what exponentiation "does". This relationship is critical because it allows us to solve problems involving exponential growth or decay by using logarithms.

When we write a logarithmic expression like \( \log_{a} x \), we're asking: "To what power must the base \( a \) be raised to get \( x \)?" This question forms the backbone of many mathematical and scientific applications of logarithms.

• For example, in the case of \( \log_{2} 16 \), we're essentially asking: What exponent \( y \) satisfies \( 2^y = 16 \)? The answer is \( y = 4 \), as \( 2^4 = 16 \).
• In another instance, \( \log_{10} 0.1 \) represents the power that 10 must be raised to result in 0.1. Since \( 0.1 = 10^{-1} \), the answer is \( -1 \).

Understanding logarithms thus involves recognizing these expressions and using them to find the unknown exponent.

Exponential Functions

Exponential functions are characterized by their constant base and a variable exponent, expressed in the form \( a^x \). They play a crucial role in various fields, such as finance, biology, and physics, due to their ability to model rapid growth or decay over time.

In these functions, the base \( a \) is a positive real number, and the exponent \( x \) can be any real number. A key characteristic of exponential functions is their tendency to grow increasingly faster or decrease rapidly depending on the base. When the base is greater than one, as in \( 2^x \), the function exhibits exponential growth. Conversely, a base between zero and one, like \( (\frac{1}{2})^x \), leads to exponential decay.

• For instance, in \( 2^x \), as \( x \) increases, \( 2^x \) grows dramatically; in our problem, we needed \( 2^4 \) to achieve 16.
• On the other hand, with a base like \( \frac{1}{2} \), we use negative exponents to signify multiplying its reciprocal, as seen in \( (\frac{1}{2})^{-3} = 8 \).

By recognizing these patterns, you can tackle various challenges involving exponential relationships.

Inverse Operations

Inverse operations are mathematical processes that essentially "undo" each other. Understanding inverse operations is critical when dealing with functions, especially logarithmic and exponential functions, since each is the inverse of the other.

The inverse relationship between logarithms and exponentials can be described succinctly:

• Taking the logarithm of a number undoes its exponentiation. In other words, if you exponentiate a number using a base \( a \), using \( \log_{a} \) will reverse that process.
• Conversely, if you take a logarithm and then exponentiate it with the corresponding base, you will end up with the original number.

For example, consider \( \log_{2} 16 \). If we expand it using exponentiation, \( 2^4 = 16 \). Logarithmic calculations reveal this inverse pattern: \( \log_{a} x = y \) translates to \( a^y = x \).

Recognizing that these operations are inverses helps simplify complex problems and ensures accuracy in mathematical computations. Whether facing growth patterns through exponentials or breaking them down using logarithms, knowing this interplay is invaluable.