Problem 5 - Algebra 2 and Trigonometry

Question

In \(3-17\) solve each equation and check. $$ x^{\frac{2}{3}}=9 $$

Step-by-Step Solution

Answer

The solution is \( x = 27 \).

Step 1: Raise Both Sides to the Reciprocal of the Fraction

To eliminate the fractional exponent, raise both sides of the equation to the power of the reciprocal of \( \frac{2}{3} \), which is \( \frac{3}{2} \). This gives us: \[ (x^{\frac{2}{3}})^{\frac{3}{2}} = 9^{\frac{3}{2}} \] Applying the property of exponents \( (a^{m})^{n} = a^{m \cdot n} \), we simplify the left side to \( x^{1} \) or just \( x \).

Step 2: Simplify the Right Side

Now, we calculate 9 raised to the power of \( \frac{3}{2} \). This can be separated into two parts: \[ 9^{\frac{1}{2}} = \sqrt{9} = 3 \]and then raise to the power of 3:\[ 3^3 = 27 \] Thus, \( 9^{\frac{3}{2}} = 27 \).

Step 3: Solution

After performing the operations, the equation simplifies to \( x = 27 \). Thus, the solution to the equation \( x^{\frac{2}{3}} = 9 \) is \( x = 27 \).

Step 4: Checking the Solution

To verify our solution, substitute \( x = 27 \) back into the original equation:\[ (27)^{\frac{2}{3}} = 9 \]Breaking it down:\[ 27^{\frac{1}{3}} = 3 \] (since \( 27^{\frac{1}{3}} = \sqrt[3]{27} = 3 \))Now raise 3 to the power of 2:\[ 3^2 = 9 \]The left side equals the right side of the original equation, confirming our solution is correct.

Key Concepts

Properties of Exponents

Understanding the properties of exponents is crucial when solving equations involving powers and roots. These properties allow us to manipulate and simplify expressions for easier solving. Here are some key exponent properties you need to know:

• Product of Powers: For any base \(a\), the expression \(a^m \cdot a^n = a^{m+n}\).
• Power of a Power: Applying a power to an exponentiated term means multiplying the exponents: \((a^m)^n = a^{m \cdot n}\).
• Power of a Product: The exponent applied to a product inside parentheses distributes to each factor: \((ab)^n = a^n b^n\).
• Reciprocal: For any non-zero number \(a\), the expression \(a^{-n} = \frac{1}{a^n}\).

These rules help simplify equations and solve problems like \(x^{\frac{2}{3}}=9\). By recognizing the need to transform both sides using reciprocal powers, you leverage the power of exponents to obtain solutions effectively.
When solving the equation \((x^{\frac{2}{3}})^{\frac{3}{2}} = 9^{\frac{3}{2}}\), apply the power of a power property to simplify to \(x^1 = 27\). This step essentially removes the fractional exponent from the variable x, making it straightforward to solve.

Rational Exponents

Rational exponents are an extension of integer exponents and indicate roots as well as powers. The general form \(a^{\frac{m}{n}}\) can be interpreted as the \(n\)th root of \(a\) raised to the \(m\)th power. This dual aspect allows us to handle roots and powers in a unified way.

• For example, \(x^{\frac{2}{3}}\) indicates the cube root of \(x\) squared: \((\sqrt[3]{x})^2\).
• It's equivalent to first finding the cube root of the number and then squaring the result.

In our exercise, you solve for \(x\) by manipulating these assessments, removing both the root and power through inverse operations.
Rational exponents provide a flexible way to express and solve equations that contain roots and powers, simplifying the process tremendously. Packing roots into a power equation allows a seamless transition to working with exponents alone.

Equation Verification

Equation verification is an essential final step to ensure the correctness of a solution. After solving an equation, you must verify that your obtained value truly satisfies the original equation. This approach not only confirms the answer is correct but also instills confidence in your problem-solving skills.
In our exercise of verifying \(x = 27\) for \(x^{\frac{2}{3}} = 9\), follow these back-checks:

• Substitute \(x = 27\) back into the equation, giving you \((27)^{\frac{2}{3}}\).
• Compute \((27^{\frac{1}{3}})^2 = 3^2\), since the cube root of 27 is 3.
• Calculate \(3^2 = 9\).
• Compare the solution to the original right side, ensuring they match.

The left and right balance confirms your solution. Taking this step in every problem validates not only the calculation but also aids understanding of the entire solving process.