Problem 55 - A Graphical Approach to Precalculus with Limits

Question

Suppose \(f(x)=\log _{a} x\) and \(f(3)=2 .\) Determine each function value. $$f\left(\frac{1}{9}\right)$$

Step-by-Step Solution

Answer

\(f\left(\frac{1}{9}\right) = -4\).

Step 1: Understanding the Given Information

We are given a function \( f(x) = \log_a x \) and a specific value \( f(3) = 2 \). This means that \( \log_a 3 = 2 \). According to the logarithm properties, this implies \( a^2 = 3 \).

Step 2: Writing the Function Value to Determine

The goal is to determine \( f \left( \frac{1}{9} \right) \), which means finding \( \log_a \left( \frac{1}{9} \right) \).

Step 3: Utilizing Logarithm Properties

To determine \( f \left( \frac{1}{9} \right) \), we use the property \( \log_a \left( \frac{1}{b} \right) = - \log_a b \). Therefore, \( \log_a \left( \frac{1}{9} \right) = - \log_a 9 \).

Step 4: Connecting to Known Values

Since \( 9 = 3^2 \), we have \( \log_a 9 = \log_a (3^2) = 2 \cdot \log_a 3 = 2 \cdot 2 = 4 \).

Step 5: Finding the Function Value

Substituting back, we find \( \log_a \left( \frac{1}{9} \right) = -4 \). So, \( f \left( \frac{1}{9} \right) = -4 \).

Key Concepts

Understanding Logarithm Properties

Logarithm properties are rules that help us to manipulate and simplify logarithmic expressions. These properties are vital when solving logarithmic functions or equations. One key property used frequently is the change of base rule:

• \( \log_a\left(\frac{1}{b}\right) = -\log_a b \)

This means the logarithm of a reciprocal is the negative logarithm of the number itself. Also, if we raise a number to a power, the logarithm of that power is the product of the power and the logarithm of the base:

• \( \log_a\left(b^c\right) = c \cdot \log_a b \)

In the original solution, these rules are applied to determine the logarithm of \( \frac{1}{9} \). By recognizing that \( 9 = 3^2 \), we convert \( \log_a 9 \) into a form that uses known values. This makes complex logarithmic calculations much more manageable and logical.

Solving Exponential Equations

Exponential equations can sometimes be solved with the help of logarithms. If an equation involves an exponential form like \( a^2 = 3 \), logarithms can help to express it in a logarithmic function:

• For \( f(x) = \log_a x \), if \( f(3) = 2 \), then \( \log_a 3 = 2 \) means \( a^2 = 3 \).

Logarithms essentially reverse the process of exponentiation. If you turn "raising a" into "log based a", you step back into a realm where multiplication and division look like addition and subtraction.Using known logarithmic values, you can reconstruct exponential properties of unknown bases in problems that require such a step. This method was used to determine the original base for the function \( f(x) = \log_a x \), thereby allowing further calculations without manually working out the base's calculations.

Function Evaluation in Logarithms

Function evaluation often means substituting a value into a function and finding the result. In the context of logarithmic functions, it involves determining the logarithmic value of various expressions. Initially, the problem was about finding \( f\left(\frac{1}{9}\right) \) based on the known \( f(3) = 2 \). After determining the equivalent base using properties of logarithms and exponential calculations, you substitute it back into the function.

• For example, substituting \( \frac{1}{9} \) as \( \log_a\left(\frac{1}{9}\right) = -\log_a 9 \).

Using the pre-determined value \( \log_a 9 = 4 \), the function value becomes \( -4 \), indicating that \( f\left(\frac{1}{9}\right) = -4 \).This systematic method of function evaluation using known values and properties aids in solving complex problems without straightforward inputs.