Solve each equation with rational exponents in Exercises 31–40. Check all proposed solutions. (x - 4)^3/2 = 27

Verified step by step guidance

1. Recognize that the equation involves a rational exponent: \( (x - 4)^{\frac{3}{2}} = 27 \). The exponent \(\frac{3}{2}\) means we are dealing with a power and a root combined.
2. Rewrite the expression \( (x - 4)^{\frac{3}{2}} \) as \( \left( (x - 4)^{\frac{1}{2}} \right)^3 \) or equivalently \( \left( \sqrt{x - 4} \right)^3 \) to better understand the operation.
3. To isolate \( x - 4 \), raise both sides of the equation to the reciprocal power of \( \frac{3}{2} \), which is \( \frac{2}{3} \), so apply \( \left( \cdot \right)^{\frac{2}{3}} \) to both sides: \( \left( (x - 4)^{\frac{3}{2}} \right)^{\frac{2}{3}} = 27^{\frac{2}{3}} \).
4. Simplify the left side using the property \( (a^{m})^{n} = a^{mn} \), so \( (x - 4)^{\frac{3}{2} \times \frac{2}{3}} = (x - 4)^1 = x - 4 \). Then calculate the right side \( 27^{\frac{2}{3}} \) by first finding the cube root of 27 and then squaring the result.
5. After simplification, solve for \( x \) by adding 4 to both sides. Finally, check all proposed solutions by substituting them back into the original equation to ensure they satisfy it, especially because rational exponents can introduce extraneous solutions.

Key Concepts

Rational Exponents

Rational exponents represent roots and powers combined; for example, an exponent of m/n means taking the nth root of the base raised to the mth power. Understanding how to manipulate expressions with rational exponents is essential for solving equations like (x - 4)^(3/2) = 27.

Isolating the Variable

To solve equations, isolate the term containing the variable by undoing operations step-by-step. For (x - 4)^(3/2) = 27, this involves raising both sides to the reciprocal power (2/3) to eliminate the rational exponent and solve for x.

Checking Solutions

After finding potential solutions, substitute them back into the original equation to verify they satisfy it. This step is crucial because raising both sides to powers can introduce extraneous solutions, especially with rational exponents.