Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

Solve for \(x .\) Hint: \(\log _{a} b=c \Leftrightarrow a^{c}=b\). $$ \log _{4}\left(\frac{1}{2 x}\right)=3 $$

Step-by-Step Solution

Answer

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

Step 1: Understanding Logarithmic Form

We start by identifying the given logarithmic equation: \( \log_{4}\left(\frac{1}{2x}\right) = 3 \). This equation is in the form \( \log_{a}(b) = c \), where \( a = 4 \), \( b = \frac{1}{2x} \), and \( c = 3 \). The hint suggests using the property \( \log_{a}b=c \Leftrightarrow a^{c}=b \).

Step 2: Convert to Exponential Form

Using the property \( \log_{a}b=c \Leftrightarrow a^{c}=b \), we convert the logarithmic form to the exponential form: \( 4^{3} = \frac{1}{2x} \). This conversion helps us in solving for \( x \).

Step 3: Evaluate the Exponent

Now, calculate \( 4^{3} \). Since \( 4^{3} = 4 \times 4 \times 4 = 64 \), the equation becomes: \( 64 = \frac{1}{2x} \).

Step 4: Isolate the Variable

To solve for \( x \), isolate it by taking the reciprocal of both sides. \( 2x = \frac{1}{64} \). Now solve for \( x \) by multiplying by the reciprocal of 2: \( x = \frac{1}{2 \times 64} = \frac{1}{128} \).

Step 5: Verify the Solution

To confirm, substitute \( x = \frac{1}{128} \) back into the original logarithmic equation: \( \log_{4}\left(\frac{1}{2 \times \frac{1}{128}}\right) = 3 \). Simplify inside the logarithm: \( \frac{1}{2 \times \frac{1}{128}} = 64 \), so we check if \( \log_{4}(64) = 3 \) holds. Since \( 4^{3} = 64 \), the solution \( x = \frac{1}{128} \) is verified.

Key Concepts

Exponential Form

In mathematics, converting an equation from logarithmic form to exponential form can make it easier to solve. A logarithmic equation, like \( \log_a b = c \), can be rewritten using exponential form as \( a^c = b \). This transformation is crucial because it allows us to work with more straightforward numerical calculations and operations.

• Given \( \log_{4}\left(\frac{1}{2x}\right) = 3 \), we identify it in the form \( \log_a b = c \).
• The base \( a \) is \( 4 \), the power \( c \) is \( 3 \), and the result \( b \) is \( \frac{1}{2x} \).
• Thus, we can rewrite the equation exponentially as: \( 4^3 = \frac{1}{2x} \).

This transformation of the equation into exponential form allows us to simplify our approach when finding the solution, helping to isolate the variable. Remember that exponential form is about expressing how many times a number, the base, is multiplied by itself to reach another number, which in this context is the outcome of the transformation.

Solving Equations

After converting the logarithmic equation to an exponential form, the next step is solving for the unknown variable. Solving equations involves finding the value of the variable that makes the equation true.

• Starting from \( 4^3 = \frac{1}{2x} \), calculate \( 4^3 = 64 \).
• This simplifies the equation to \( 64 = \frac{1}{2x} \).

To isolate \( x \):

• First, take the reciprocal of both sides to begin solving for \( x \): \( 2x = \frac{1}{64} \).
• Then, multiply both sides by \( \frac{1}{2} \) to isolate \( x \), leading to \( x = \frac{1}{128} \).

By isolating \( x \), we are resolving the equation truly by determining the specific value of \( x \) that solves our original logarithmic equation. Keep in mind that checking your solution by substituting \( x \) back into the original problem can confirm the correctness of your answer.

Properties of Logarithms

Logarithms have several key properties that are useful in simplifying and solving equations. Understanding these properties can make it easier to manipulate logarithmic expressions.

• Conversion to Exponential Form: As seen, the key property \( \log_a b = c \leftrightarrow a^c = b \) is essential for solving equations.
• Change of Base: This property allows conversion of log bases, but is not needed for this exercise as direct conversion suffices.
• Product, Quotient, and Power Rules: Logarithms transform multiplication, division, and exponents into more manageable forms.

In our example, the direct linear translation of the log expression into exponential form is the primary technique used. However, familiarity with the properties can enable solutions to more complex logarithmic equations, providing a toolkit to simplify expressions and solve equations efficiently.