Problem 24 - A Graphical Approach to College Algebra

Question

Solve each equation. Give the exact answer. $$\log _{4} x=-\frac{1}{6}$$

Step-by-Step Solution

Answer

\( x = \frac{1}{\sqrt[3]{2}} \).

Step 1: Understand the Equation

The given equation is \( \log_{4} x = -\frac{1}{6} \). This indicates that we are looking for the value of \( x \) such that the logarithm with base 4 equals \(-\frac{1}{6}\).

Step 2: Convert Logarithmic to Exponential Form

To solve the equation, convert the logarithmic equation to its exponential form. The equation \( \log_{b} a = c \) can be written as \( a = b^{c} \). Thus, \( x = 4^{-\frac{1}{6}} \).

Step 3: Simplify the Exponential Expression

Now simplify \( x = 4^{-\frac{1}{6}} \). Since 4 can be written as \( 2^2 \), we have \( x = (2^2)^{-\frac{1}{6}} = 2^{-\frac{1}{3}} \).

Step 4: Write Final Expression for x

The expression \( 2^{-\frac{1}{3}} \) implies the cube root of \( \, \frac{1}{2} \, \). Hence, the exact value for \( x \) is \( \frac{1}{\sqrt[3]{2}} \).

Key Concepts

Exponential Form

Converting a logarithmic equation into exponential form is a fundamental skill in solving logarithmic equations. The basic idea is rooted in the definition of a logarithm. When you have an equation like \( \log_b a = c \), it can be transformed into its corresponding exponential form: \( a = b^c \). This transformation allows us to express the problem in a form that might be easier to solve or interpret.
For example, in the exercise where \( \log _{4} x=-\frac{1}{6} \), we apply this conversion. We see that \( x \) equals \( 4^{-\frac{1}{6}} \) when rewritten in its exponential form. Why is this useful? Because it provides us with a way to calculate \( x \) by expressing it in terms of exponents, which are often easier to work with, especially with the help of calculators or further algebraic manipulation.

Logarithmic Properties

Logarithmic properties are essential tools when dealing with logarithmic equations. They help simplify and solve such equations effectively. Some key properties include:

• Logarithm of a power: \( \log_b (m^n) = n \cdot \log_b m \)
• Change of base formula: \( \log_b a = \frac{\log_k a}{\log_k b} \)
• Product rule: \( \log_b (mn) = \log_b m + \log_b n \)
• Quotient rule: \( \log_b \left( \frac{m}{n} \right) = \log_b m - \log_b n \)

The exercise makes implicit use of some properties when we simplify \( 4^{-\frac{1}{6}} \). Recognizing that \( 4 \) can be rewritten as \( 2^2 \), allows us to further simplify the expression by utilizing the power rule: \( (2^2)^{-\frac{1}{6}} = 2^{(-\frac{1}{6}) \times 2} = 2^{-\frac{1}{3}} \).
This step is crucial for finding the most reduced form of the solution. Understanding these properties makes it easier to manipulate logarithmic expressions and find answers efficiently.

Exact Solutions

Finding exact solutions is an important aspect of solving mathematical problems. In the context of logarithmic equations, an exact solution means expressing the answer in its simplest form, without approximations.
In the given exercise, we arrived at the expression \( 2^{-\frac{1}{3}} \) as the exact solution. This is further interpreted as \( \frac{1}{\sqrt[3]{2}} \), illustrating the importance of expressing solutions with clarity and precision.
Exact solutions are crucial as they provide a precise answer that doesn't rely on numerical approximations. This can be especially important when further calculations are required, or when an understanding of the mathematical subtleties is necessary. They ensure accuracy and convey a deeper insight into the mathematical relationships at play, which is invaluable in learning and applying mathematics properly. Always aim for clarity and precision in your solutions.