Problem 49 - A Graphical Approach to Precalculus with Limits

Question

Evaluate each logarithm in three ways: (a) Use the definition of logarithm to find the exact value analytically. (b) Support the result of part (a) by using the change-of-base rule and common logarithms on a calculator. (c) Use a buill-in calculator finction to evaluate the logarithm. (d) Support the result of part (a) by locating the appropriate point on the graph of the finction \(y=\log _{a} x\). $$\log _{16}\left(\frac{1}{8}\right)$$

Step-by-Step Solution

Answer

The value is \(-\frac{3}{4}\).

Step 1: Understand the Definition of Logarithm

To evaluate \( \log_{16} \left( \frac{1}{8} \right) \) analytically, recall the definition: A logarithm \( \log_b (a) = c \) means that \( b^c = a \). We are looking for an exponent \( c \) such that \( 16^c = \frac{1}{8} \).

Step 2: Rewrite the Equation

Express both 16 and \( \frac{1}{8} \) as powers of 2 to make the comparison easier. \( 16 = 2^4 \) and \( \frac{1}{8} = 2^{-3} \). So the equation \( 16^c = \frac{1}{8} \) becomes \( (2^4)^c = 2^{-3} \).

Step 3: Solve for c Using Exponents

From \( (2^4)^c = 2^{-3} \), we simplify this to \( 2^{4c} = 2^{-3} \). Since the bases are equal, you can equate the exponents: \( 4c = -3 \). Solving for \( c \) gives \( c = -\frac{3}{4} \). Thus, \( \log_{16} \left( \frac{1}{8} \right) = -\frac{3}{4} \).

Step 4: Use the Change-of-Base Formula

Apply the change-of-base formula: \( \log_{16} \left( \frac{1}{8} \right) = \frac{\log \left( \frac{1}{8} \right)}{\log(16)} \). Use a calculator to find \( \log \left( \frac{1}{8} \right) \approx -0.903 \) and \( \log(16) \approx 1.204 \). Then \( \frac{-0.903}{1.204} \approx -0.75 \), which confirms our analytical result.

Step 5: Evaluate Using a Logarithm Calculator

Directly input \( \log_{16} \left( \frac{1}{8} \right) \) into a calculator that accepts any base. The calculator should return \( -0.75 \), confirming our earlier calculation.

Step 6: Verify Using the Graph of \( y = \log_{16}(x) \)

On the graph of \( y = \log_{16}(x) \), locate \( x = \frac{1}{8} \). The corresponding \( y \)-coordinate (which is \( -\frac{3}{4} \)) represents the value of \( \log_{16} \left( \frac{1}{8} \right) \), confirming the results obtained from the previous methods.

Key Concepts

Change-of-Base Formula

The change-of-base formula is a useful tool when evaluating logarithms, especially when working with bases other than 10 or when a given calculator does not support a specific base logarithm. The formula is given as:

• \( \log_b(a) = \frac{\log_c(a)}{\log_c(b)} \)

This equation allows you to convert a logarithm with any base \( b \) to a logarithm of a different base \( c \), which is commonly base 10 (\( \log \)) or base \( e \) (\( \ln \)), since these bases are available on most calculators.

In the given exercise, the change-of-base formula was used to transform \( \log_{16} \left( \frac{1}{8} \right) \) into a form that can easily be calculated using a typical calculator. We found that \( \frac{\log \left( \frac{1}{8} \right)}{\log(16)} \) yields \( -0.75 \), confirming the analytical result arrived at via direct calculation.

Graphing Logarithmic Functions

Graphing provides a visual method to confirm logarithmic values. A logarithmic function of the form \( y = \log_a(x) \) can be plotted on a graph, where \( y \) represents the logarithm and \( x \) is the value of which the logarithm is taken.

When graphing \( y = \log_{16}(x) \), each point on the graph has coordinates \( (x, y) \), where \( y = \log_{16}(x) \). For the exercise in question, locating \( x = \frac{1}{8} \) on this graph, the corresponding \( y \)-value should be \( -\frac{3}{4} \). This confirms that \( \log_{16} \left( \frac{1}{8} \right) = -\frac{3}{4} \). Such graphical interpretation not only verifies numerical answers but helps illustrate how logarithms behave with different bases and arguments.

Exponents

The concept of exponents is foundational to understanding logarithms, as a logarithm essentially asks the question: "To what power must the base \( b \) be raised, to yield the number \( a \)?" Therefore, the relationship between exponents and logarithms is interchangeable, described by the identity:

• If \( b^c = a \), then \( \log_b(a) = c \).

To tackle logarithmic problems, you might need to manipulate or rewrite numbers as their exponential forms. In our exercise, the numbers \( 16 \) and \( \frac{1}{8} \) were rewritten as \( 2^4 \) and \( 2^{-3} \) respectively, to use identical bases. Simplifying \( (2^4)^c = 2^{-3} \), we equate exponents to solve for \( c \) given \( 4c = -3 \), resulting in \( c = -\frac{3}{4} \). Mastering exponents is critical for effectively navigating and confirming outcomes in logarithmic calculations.