Problem 17 - Intermediate Algebra

Question

For Problems \(1-34\), solve each equation. $$ \left(\frac{1}{2}\right)^{2 x}=64 $$

Step-by-Step Solution

Answer

The solution is \( x = -3 \).

Step 1: Understand Exponents and Rewrite Equation

We have the equation \( \left(\frac{1}{2}\right)^{2x} = 64 \). Notice that \( 64 \) can be rewritten as a power of 2: \( 64 = 2^6 \). Now, the goal is to express both sides of the equation with the same base.

Step 2: Rewrite the Equation with the Same Base

Rewrite \( \frac{1}{2} \) as \( 2^{-1} \). Thus, \( \left(\frac{1}{2}\right)^{2x} \) becomes \( (2^{-1})^{2x} = 2^{6} \). Using the power of a power property \((a^m)^n = a^{mn}\), we can rewrite the left side as \( 2^{-2x} \). The equation now is \( 2^{-2x} = 2^6 \).

Step 3: Set Exponents Equal

Since the bases are the same (both are base \(2\)), set the exponents equal to each other: \(-2x = 6\).

Step 4: Solve for x

To isolate \( x \), divide both sides of the equation by \(-2\): \(-2x = 6\) becomes \( x = -3 \).

Step 5: Verify Solution

Substitute \( x = -3 \) back into the original exponent equation to confirm: \( \left(\frac{1}{2}\right)^{2(-3)} = \left(\frac{1}{2}\right)^{-6} = 2^6 = 64\). Thus, the solution is verified.

Key Concepts

Solving Equations

Solving equations involves finding the value of the variable that makes the equation true. In the given exercise, the goal was to solve for \( x \) in the equation \( \left(\frac{1}{2}\right)^{2x} = 64 \). Here, equations with exponents were used, indicating that exponential values must be handled.
To tackle this type of equation, understanding the relationship between the bases and exponents is crucial. First, recognize that solving equations like these often requires both sides to have the same base. This allows you to set the exponents equal to each other, facilitating the solution more straightforwardly.
Approaching equations step-by-step, like in the provided solution, aids in clarity. Carefully translate the equation into an auxiliary form that makes solving for the variable simpler. Often, it involves rewriting one side of the equation, as seen when converting \( 64 \) into \( 2^6 \). The equation setup like \( 2^{-2x} = 2^6 \) allows us to use the property of equality for exponents effectively.

Exponents

Exponents are shorthand for expressing repeated multiplication of the same number. For instance, \( a^n \) implies that the number \( a \) is multiplied by itself \( n \) times. Understanding this concept is fundamental when working with exponential equations, such as the example \( \left(\frac{1}{2}\right)^{2x} = 64 \), where the concept of exponents was central.
In the exercise, understanding the principle

• \((a^m)^n = a^{mn}\)

was crucial because it allowed us to manipulate the equation into a solvable form by changing the base of the exponent. Recognizing equivalent exponents also means being comfortable with different powers of numbers, which we will address next.
By rewriting \( \frac{1}{2} \) as \( 2^{-1} \), we effectively change the structure while maintaining equivalence, showing how inverses work within exponents. This step reveals the depth and flexibility of exponents, much needed to solve these equations.

Powers of Numbers

The power of a number refers to the result you get when a number is multiplied by itself a certain number of times, as expressed by the exponent. This concept helps simplify equations by transforming complex expressions into manageable forms. For example, \( 64 \) expressed as \( 2^6 \) becomes manageable when solving our exponential equation.
To handle powers of numbers successfully:

• Recognize common powers: Knowing that 64 is equivalent to \( 2^6 \) quickly leads to the solution.
• Transform similar equations: Applying powers to both sides of an equation provides a balanced approach, ensuring equations stay true while simplifying their complexity.

This understanding was pivotal in converting the original equation \( \left(\frac{1}{2}\right)^{2x} = 64 \) into \( 2^{-2x} = 2^6 \). Such transformations allow you to apply properties of exponents effectively, leading directly to the solution by enabling a comparison of exponents on an identical base.