Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \text { In Problems 17-24, solve for } x . \text { Hint: } \log _{a} b=c \Leftrightarrow a^{c}=b \text {. } $$ $$ \log _{4}\left(\frac{1}{2 x}\right)=3 $$

Step-by-Step Solution

Answer

\(x = \frac{1}{128}\)

Step 1: Understand the Given Equation

We are given the equation \( \log_{4}\left(\frac{1}{2x}\right) = 3 \). This implies that the logarithmic equation relates to how a number is expressed as a power of 4.

Step 2: Apply the Logarithm Rule

According to the hint, \( \log_{a} b = c \) means that \( a^{c} = b \). Applying this rule to our problem, we find that \( 4^{3} = \frac{1}{2x} \).

Step 3: Calculate the Exponentiation

Calculate \( 4^3 \). Since \( 4 \times 4 \times 4 = 64 \), we have \( 64 = \frac{1}{2x} \).

Step 4: Solve for \( x \) Using Algebra

To solve \( 64 = \frac{1}{2x} \), multiply both sides by \( 2x \) to get \( 64 \times 2x = 1 \). Simplify to find \( 128x = 1 \).

Step 5: Isolate \( x \)

Divide both sides of the equation by 128 to isolate \( x \). This results in \( x = \frac{1}{128} \).

Key Concepts

Exponentiation

Exponentiation is the mathematical operation involving two numbers, the base and the exponent. It is written in the form \( a^n \), where \( a \) is the base and \( n \) is the exponent. In the context of logarithmic equations, exponentiation helps in transforming logarithmic expressions into algebraic ones.
For example, in the given exercise, you are dealing with \( 4^3 \). Here, 4 is the base, and 3 is the exponent. The operation involves multiplying the base by itself the number of times indicated by the exponent, which is \( 4 \times 4 \times 4 = 64 \).

• Exponentiation gives us the power in which the base is raised.
• It is the inverse operation of logarithms.

In solving logarithmic equations, converting logarithmic form to exponential form allows you to work comfortably with numbers and speeds up calculations.

Algebraic Manipulation

Algebraic manipulation is a collection of strategies used to simplify, rearrange, or solve equations. In solving the exercise, algebraic manipulation was crucial to isolating the variable \( x \).
When you encounter the equation \( 64 = \frac{1}{2x} \), the goal is to express \( x \) in terms of known values. Multiply both sides by \( 2x \) to eliminate the fraction, which gives \( 64 \times 2x = 1 \). This step involves distributing to clear the fraction, a common tactic in algebra.

• Clearing fractions by multiplying is a key technique.
• Simplifying expressions makes it easier to isolate variables.

Finally, you divide both sides by 128, as you need to isolate \( x \). This leads to the solution \( x = \frac{1}{128} \). Breaking down steps carefully aids in understanding and solving algebra equations in systematic ways.

Logarithm Rules

Logarithm rules are essential for understanding and solving problems involving logarithmic equations. In this exercise, we utilized the conversion rule from logarithmic form to exponential form.
The given rule is \( \log_{a} b = c \Rightarrow a^c = b \). This means if \( a \) raised to the power of \( c \) equals \( b \), then \( \log_{a} b \) is \( c \). Applying this rule simplifies our logarithmic equation to an algebraic one: from \( \log_{4}(\frac{1}{2x}) = 3 \) to \( 4^3 = \frac{1}{2x} \).

• Conversion rules link logarithms to exponentiation.
• These rules simplify complex logarithmic expressions.

Understanding how to apply logarithm rules helps convert between forms, making it easier to find solutions. They are particularly useful in transforming equations to a more solvable form by switching between logarithmic and exponential expressions.