Problem 16 - Elementary and Intermediate Algebra

Question

Complete the table of values. $$ \begin{aligned} &f(x)=\log _{5} x\\\ &\begin{array}{|c|c|} \hline x & f(x) \\ \hline 25 & \\ \hline \frac{1}{25} & \\ \hline \end{array} \end{aligned} $$

Step-by-Step Solution

Answer

\( f(25) = 2 \) and \( f\left(\frac{1}{25}\right) = -2 \).

Step 1: Understanding logarithm function

The function given is \( f(x) = \log_{5}(x) \). This function asks for the exponent to which the base 5 must be raised to produce \( x \). For example, if \( x = 25 \), then \( f(x) \) is the exponent to which 5 must be raised to result in 25.

Step 2: Calculate \( f(25) \)

For \( x = 25 \), we need to find \( f(25) = \log_{5}(25) \). Since \( 25 = 5^2 \), it follows that \( \log_{5}(25) = 2 \). Therefore, \( f(25) = 2 \).

Step 3: Calculate \( f\left(\frac{1}{25}\right) \)

For \( x = \frac{1}{25} \), we find \( f\left(\frac{1}{25}\right) = \log_{5}\left(\frac{1}{25}\right) \). We know that \( \frac{1}{25} = 5^{-2} \), therefore, \( \log_{5}\left(\frac{1}{25}\right) = -2 \). Thus, \( f\left(\frac{1}{25}\right) = -2 \).

Step 4: Complete the Table

Now we have calculated the values for each specified \( x \). The table should be filled in as follows: \[ \begin{array}{|c|c|} \hline x & f(x) \ \hline 25 & 2 \ \hline \frac{1}{25} & -2 \ \hline \end{array} \]

Key Concepts

Exponential Functions

Exponential functions are a type of mathematical function where a constant base is raised to a variable exponent. These functions are expressed in the form of \( f(x) = a^x \) where \( a \) is the base (a constant) and \( x \) is the exponent (a variable). This concept is crucial when dealing with logarithmic functions as they are, in fact, the inverse of exponential functions. Logarithms ask the question, "to what power must the base be raised, in order to get a certain value?"Understanding this relationship is essential as it allows you to convert between exponential and logarithmic forms easily. For example:

• \( a^b = c \) is the same as \( b = \log_a(c) \).
• If \( 5^2 = 25 \), then \( \log_5(25) = 2 \).

Exponential functions grow rapidly, and each unit increase to the exponent results in multiplication by the base value itself. Recognizing how exponential growth works helps in evaluating and understanding logarithmic expressions.

Base of a Logarithm

The base of a logarithm is an essential component that determines the rate at which the logarithmic function scales. When you see a logarithmic expression such as \( \log_5(x) \), the number 5 is the base of the logarithm. This means we are considering powers of 5. The base tells us what number we repeatedly multiply to reach a given value.The concept of a base is crucial because it directly influences how these calculations are performed:

• The equation \( \log_b(x) = y \) means that \( b^y = x \).
• This is why \( \log_5(25) = 2 \) because \( 5^2 = 25 \).

Choosing a different base changes the calculation. Logarithms with different bases translate problems into different scales. Common bases, such as 10 and \( e \) (approximately 2.718), often appear in scientific calculations and natural phenomena, but any positive number can serve as a base for logarithms.

Properties of Logarithms

Logarithms have several important properties that make them useful in simplifying expressions and solving equations. Understanding these properties allows you to manipulate logarithms and make complex calculations more manageable.One fundamental property is the power rule, which states that \( \log_b(x^y) = y \cdot \log_b(x) \). This allows you to "bring down" exponents, which helps in simplifying logarithmic expressions. For example:

• If \( x = 25 = 5^2 \), then \( \log_5(25) = 2 \cdot \log_5(5) = 2 \cdot 1 = 2 \).

Another useful property is the change of base formula, which enables you to convert logarithms from one base to another: \( \log_b(x) = \frac{\log_k(x)}{\log_k(b)} \), where \( k \) is any positive number (commonly 10 or \( e \)).Moreover, understanding the fact that \( \log_b(\frac{1}{x}) = -\log_b(x) \) helps to simplify calculations involving fractions. For example, since \( \frac{1}{25} = 5^{-2} \), then \( \log_5\left(\frac{1}{25}\right) = -2 \) because \( \log_5(5^{-2}) = -2 \cdot \log_5(5) = -2 \cdot 1 = -2 \).These properties make logarithms a versatile tool in mathematical problem solving, providing a powerful way to simplify and solve equations involving powers and roots.