Problem 69 - Intermediate Algebra

Question

Solve for \(x .\) See Example 3. $$ \log _{x} \frac{1}{64}=-3 $$

Step-by-Step Solution

Answer

The solution is \( x = \frac{1}{4} \).

Step 1: Understand the Logarithmic Equation

The given equation is \( \log_{x} \frac{1}{64} = -3 \). This means that \( x \) raised to the power of \( -3 \) gives \( \frac{1}{64} \). In other words, \( x^{-3} = \frac{1}{64} \).

Step 2: Rewrite the Equation with Exponents

To solve for \( x \), we convert the logarithm to an exponential equation: \( x^{-3} = \frac{1}{64} \). This implies that we are dealing with a power of \( x \) that results in \( \frac{1}{64} \).

Step 3: Solve the Exponential Equation

Since \( x^{-3} = \left(\frac{1}{4}\right)^3 \), we can equate \( x \) to \( \frac{1}{4} \), because \( (\frac{1}{4})^3 = \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{64} \). So, \( x = \frac{1}{4} \).

Step 4: Verify the Solution

To ensure that \( x = \frac{1}{4} \) is correct, substitute back into the original equation: \( \log_{\frac{1}{4}} \frac{1}{64} = -3 \). Calculating, \( (\frac{1}{4})^{-3} = \left(\frac{1}{4}\right)^{-3} = \frac{1}{64} \), which matches both sides, confirming \( x = \frac{1}{4} \) is correct.

Key Concepts

Exponential Equations

Exponential equations are equations where variables appear in the exponent. They have the form \( a^x = b \), with \( a \) and \( b \) being constants. Solving exponential equations requires expressing both sides with a common base, if possible. For instance, if \( 2^x = 8 \), we can rewrite 8 as \( 2^3 \), leading to \( 2^x = 2^3 \). Therefore, \( x = 3 \).
In the provided exercise, the equation \( x^{-3} = \frac{1}{64} \) is an exponential equation because the variable \( x \) is in the exponent. By recognizing that \( \frac{1}{64} \) is \( (\frac{1}{4})^3 \), we can equate \( x \) to \( \frac{1}{4} \). This shows how solving exponential equations often involves comparing powers and bases.

• Look for patterns or common factors.
• Convert numbers into similar bases.
• Apply logarithms if needed to simplify.

Logarithmic Functions

Logarithmic functions, denoted by \( \log_b(a) \), are the inverse of exponential functions. This means if \( b^x = a \), then \( \log_b(a) = x \). It answers the question: To what power must the base \( b \) be raised to yield \( a \)?
In the original exercise, we used the logarithmic function \( \log_x\frac{1}{64} = -3 \) to lay the groundwork for solving the problem. This translates to \( x^{-3} = \frac{1}{64} \). Understanding how to switch between logarithmic and exponential forms is key to solving problems involving logs.
Key aspects of logarithmic functions include:

• The base \( b \) must be positive and not equal to 1.
• They can be used to solve equations involving upper or lower power limits.
• A change of base formula can help convert odd bases into more convenient ones.

Algebraic Problem Solving

Algebraic problem solving involves using a variety of methods to find unknown values. This is essential in solving equations and manipulating algebraic expressions to isolate variables. It often requires identifying the type of equation and choosing an appropriate solution strategy.
In the exercise you saw, algebraic problem solving was used to identify that \( x^{-3} = \frac{1}{64} \) needed exponential thinking to solve. By rewriting the equation to use similar bases, the solution became clear: \( x = \frac{1}{4} \).
Effective problem solving strategies include:

• Understanding the problem before trying to solve it.
• Translating word problems into mathematical equations.
• Checking solutions by substituting back into the original equation.
• Breaking down complex problems into simpler components.

By mastering these strategies, algebraic problem solving can become a straightforward process, making it easier to handle equations like the one presented.