Problem 2 - Precalculus Mathematics for Calculus

Question

The function \(f(x)=\log _{9} x\) is the logarithm function with base ______. So \(f(9)=\) ______, \(f(1)=\) ______, \(f\left(\frac{1}{9}\right)=\) ______, \(f(81)=\) ______, and \(f(3)=\) ______.

Step-by-Step Solution

Answer

The function uses base 9. \(f(9) = 1, f(1) = 0, f(\frac{1}{9}) = -1, f(81) = 2, f(3) = 0.5\).

Step 1: Understanding the Logarithm Base

The given function is \(f(x) = \log_9 x\), meaning that the function uses the logarithm base 9.

Step 2: Calculating the Logarithm of 9

Since the function \(f(x) = \log_9 x\), \(f(9) = \log_9 9\). Any number to its base is 1, so \(\log_9 9 = 1\). Thus, \(f(9) = 1\).

Step 3: Calculating the Logarithm of 1

For any logarithmic function \(\log_b 1 = 0\), since any base raised to the power of 0 is 1. Therefore, \(f(1) = \log_9 1 = 0\).

Step 4: Calculating the Logarithm of 1/9

\(f(\frac{1}{9}) = \log_9 \frac{1}{9}\) involves finding the power to raise 9 to get \(\frac{1}{9}\). Since \(9^{-1} = \frac{1}{9}\), \(\log_9 \frac{1}{9} = -1\). Thus, \(f(\frac{1}{9}) = -1\).

Step 5: Calculating the Logarithm of 81

\(f(81) = \log_9 81\). Since 81 is \(9^2\), then \(\log_9 81 = 2\). Thus, \(f(81) = 2\).

Step 6: Calculating the Logarithm of 3

\(f(3) = \log_9 3\). We can express 3 in terms of 9 as 3 is \(9^{0.5}\). Therefore, \(\log_9 3 = 0.5\). Thus, \(f(3) = 0.5\).

Key Concepts

Logarithm Base

The concept of a logarithm base is fundamental when dealing with logarithmic functions. Essentially, the base of a logarithm is the number that is raised to a power to determine another number. For instance, in the function \(f(x) = \log_9 x\), the base is 9. This means we are asking the question, "To what power must 9 be raised to result in x?"

Understanding the base is crucial because it dictates the entire nature of the logarithmic function. Different bases will yield different results for the same value of \(x\). For example, if the base were 10 instead of 9, then the calculations of the logarithmic values would differ significantly.

It's important to note the common bases used in logarithmic functions: base 10 (known as common logarithm), base \(e\) (natural logarithm, written as \(\ln\)), and any other positive number. By comprehending the base, you can better grasp how the function behaves for different arguments and what the logarithm aims to calculate.

Evaluation of Logarithms

Evaluating logarithms involves finding the exponent that the base must be raised to obtain a certain number. Let's go through a series of evaluations using the function \(f(x) = \log_9 x\).

• \(f(9) = \log_9 9\) - Here, we look for the power to which 9 must be raised to yield 9. Since 9 to the power of 1 equals 9, \(\log_9 9\) is 1.
• \(f(1) = \log_9 1\) - Any base to the power of 0 is 1. Thus, \(\log_9 1 = 0\).
• \(f\left(\frac{1}{9}\right) = \log_9 \frac{1}{9}\) - To get \(\frac{1}{9}\) from 9, we use -1 as the exponent: \(9^{-1} = \frac{1}{9}\). Therefore, this logarithm evaluates to -1.
• \(f(81) = \log_9 81\) - Since 81 can be expressed as \(9^2\), \(\log_9 81 = 2\).
• \(f(3) = \log_9 3\) - Recognizing that 3 is \(9^{0.5}\), we find that \(\log_9 3 = 0.5\).

Evaluating logarithms requires you to identify how the number can be expressed as a power of the base, and this power is your logarithmic result.

Properties of Logarithms

Logarithms come with a set of properties that help simplify and solve expressions. These properties, when understood well, make working with logarithmic functions more intuitive.

Some primary properties include:

• Product Property: \(\log_b (xy) = \log_b x + \log_b y\), meaning the logarithm of a product is the sum of the logarithms.
• Quotient Property: \(\log_b \left(\frac{x}{y}\right) = \log_b x - \log_b y\), implying the logarithm of a quotient is the difference of the logarithms.
• Power Property: \(\log_b (x^k) = k \cdot \log_b x\), showing that the logarithm of a power is the exponent times the logarithm.
• Change of Base Formula: \(\log_b x = \frac{\log_c x}{\log_c b}\), a useful way to convert logs from one base to another.

Utilizing these properties, you can often break down complex logarithmic expressions into simpler parts, or rewrite them in a way that makes computation more straightforward. These properties provide a powerful toolkit for evaluating expressions and solving equations involving logarithms.