Problem 16 - Intermediate Algebra

Question

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

Step-by-Step Solution

Answer

For \( x = 25 \), \( f(x) = 2 \); for \( x = \frac{1}{25} \), \( f(x) = -2 \).

Step 1: Understand the Function

The function given is \( f(x) = \log_{5}x \), which means we're finding the power to which the base 5 must be raised to obtain \( x \).

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

To find \( f(25) \), we solve \( \log_{5}25 \). Since \( 25 = 5^2 \), it follows that \( \log_{5}25 = 2 \). Thus, \( f(25) = 2 \).

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

For \( f(\frac{1}{25}) \), we solve \( \log_{5}(\frac{1}{25}) \). Since \( \frac{1}{25} = 5^{-2} \), it follows that \( \log_{5}(\frac{1}{25}) = -2 \). Thus, \( f(\frac{1}{25}) = -2 \).

Step 4: Fill in the Table

Using the values calculated, complete the table: \[\begin{array}{|c|c|}\hline x & f(x) \\hline 25 & 2\\hline \frac{1}{25} & -2 \\hline\end{array}\]

Key Concepts

Understanding Logarithmic Equations

Logarithmic equations involve expressions where the unknown variable appears inside a logarithm. In such equations, we solve for the unknown by utilizing our understanding of logarithmic functions. For instance, in the function \[ f(x) = \log_{5}x \]we ask what power we must raise the base (5) to get the number (x). If we take an example from our exercise, where we were asked to determine \( f(25) \),we need to determine the power of 5 that results in 25. Since \( 25 = 5^2 \),we know that the solution is 2; therefore, \( f(25) = 2 \).This process shows the fundamental nature of solving logarithmic equations:

• Identify the base of the logarithm.
• Convert the number inside the logarithm into a power of the base.
• The exponent is the solution to the equation.

Base Change in Logarithms

When working with logarithms, the base can be changed to simplify calculations or when converting between different logarithmic scales. The base change formula is particularly useful. It states:\[ \log_{b}x = \frac{\log_{c}x}{\log_{c}b} \]Here, \( b \)is the original base and\( c \)is the new base. This formula can transform any logarithm into one with any base, including the natural logarithm \( \ln \)or the common logarithm \( \log \),which is base 10.Imagine you're stuck without a calculator that supports base 5, and you need to evaluate something like \( \log_{5}25 \).Using the base change formula, you could transform this into a common logarithm:\[ \log_{5}25 = \frac{\log_{10}25}{\log_{10}5} \]By doing so, you can utilize any standard scientific calculator that supports base 10 to find the values for \( \log_{10}25 \)and \( \log_{10}5 \).This technique simplifies calculations and can aid in solving more complex logarithmic equations.

Properties of Logarithms

Logarithms follow specific rules known as the properties of logarithms, which simplify complex expressions and equations. Understanding these properties aids immensely in solving logarithmic problems:

• Product Property: \( \log_{b}(xy) = \log_{b}x + \log_{b}y \)
• Quotient Property: \( \log_{b}\frac{x}{y} = \log_{b}x - \log_{b}y \)
• Power Property: \( \log_{b}(x^n) = n\log_{b}x \)
• Change of Base Formula: As mentioned, \( \log_{b}x = \frac{\log_{c}x}{\log_{c}b} \)

Let's consider the example \( \log_{5}(\frac{1}{25}) \).We used the properties of exponents and logarithms to rewrite \( \frac{1}{25} \)as \( 5^{-2} \).Applying the power property, we found:\[ \log_{5}(\frac{1}{25}) = \log_{5}(5^{-2}) = -2 \]Understanding these basics is key. They help in rearranging, simplifying, and calculating logarithmic expressions efficiently.