Problem 30 - Precalculus with Calculus Previews

Question

Solve the given logarithmic equation. $$ \frac{\log _{2} 8^{x}}{\log _{2} \frac{1}{4}}=\frac{1}{2} $$

Step-by-Step Solution

Answer

The solution is \( x = -\frac{1}{3} \).

Step 1: Convert Logarithmic Terms

Start by converting the terms inside the logarithms to a simpler form. The term \(8^x\) can be rewritten using the base 2 as \((2^3)^x = 2^{3x}\). So, \(\log_{2} 8^x = \log_{2} (2^{3x})\). For the denominator \(\log_{2} \frac{1}{4} = \log_{2} 2^{-2} = -2\).

Step 2: Apply Logarithmic Power Rule

Apply the logarithmic power rule \(\log_{b} a^c = c \cdot \log_{b} a\). Thus, \(\log_{2} (2^{3x}) = 3x\cdot \log_{2} 2 = 3x\cdot 1 = 3x\). So the expression becomes \(\frac{3x}{-2}\).

Step 3: Solve for x

Set the expression \(\frac{3x}{-2}\) equal to \(\frac{1}{2}\), as given in the equation. Solve the equation: \[\frac{3x}{-2} = \frac{1}{2} \]Multiply both sides by -2 to get \[3x = -1\]Divide both sides by 3:\[ x = -\frac{1}{3} \]

Key Concepts

Logarithmic Power Rule

The logarithmic power rule is a handy tool when simplifying or solving logarithmic expressions. It lets you bring an exponent out of the logarithm as a multiplication factor, making equations easier to manage. This rule states that for any base \(b\) and positive number \(a\), the expression \(\log_b(a^c)\) simplifies to \(c \cdot \log_b(a)\).

For example, in the given exercise, we have \(8^x\), which can be rewritten as \( (2^3)^x = 2^{3x} \) due to the bases. Then, the logarithm \( \log_2 (2^{3x}) \) becomes \( 3x \cdot \log_2 (2) \). Since \( \log_2 (2) \) equals 1, this simplifies further to \( 3x \).

• This makes the problem much easier as we no longer have to deal with exponents within the logarithm.
• It reduces the complexity of the equation down to basic algebraic manipulation.

Using this rule effectively helps in breaking down complicated logarithmic expressions into simpler forms that can be easily solved.

Solving Equations

When solving equations that involve logarithms, we aim to isolate the variable of interest. After simplifying the equation, like in our original exercise, it becomes crucial to use algebraic operations effectively.

Once we express the problem using simpler terms, as seen with the \( \frac{3x}{-2} = \frac{1}{2} \) equation, the task is to solve for \(x\).

• The first step is to eliminate the fraction by multiplying both sides by the denominator's negative, -2, leading to \(3x = -1\).
• Then, divide both sides by 3, isolating \(x\) to give \(x = -\frac{1}{3}\).

At this stage, the process transitions smoothly from logarithmic manipulation to simple arithmetic, showcasing how logarithms and algebra work together seamlessly in equation solving.

Base Conversions

Understanding and applying base conversions is vital in dealing with logarithmic equations. Sometimes, it's necessary to convert numbers to a common base for easier manipulation.

In our exercise, we converted \( 8^x \) to \( (2^3)^x \) by recognizing that 8 can be expressed as \( 2^3 \). Similarly, \( \frac{1}{4} \) became \( 2^{-2} \), ensuring all logarithms had base 2.

• Converting to a common base simplifies calculations and aligns with the power rule.
• It also helps to uniformly handle different logarithmic terms, providing a consistent framework for calculation.

This conversion technique is practical not just for solving equations but also for understanding how numbers relate under different logarithm bases.