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Ch. 1 - Equations and Inequalities
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook
Chapter 2, Problem 143

Solve for x: xx3=9\(\sqrt\)[3]{x \(\sqrt{x}\)} = 9

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1
Rewrite the expression inside the cube root using exponents. Recall that \( \sqrt{x} = x^{\frac{1}{2}} \), so \( x \sqrt{x} = x \cdot x^{\frac{1}{2}} = x^{1 + \frac{1}{2}} = x^{\frac{3}{2}} \).
Express the cube root as an exponent: \( \sqrt[3]{x^{\frac{3}{2}}} = \left(x^{\frac{3}{2}}\right)^{\frac{1}{3}} \).
Use the power of a power property to simplify the exponent: \( \left(x^{\frac{3}{2}}\right)^{\frac{1}{3}} = x^{\frac{3}{2} \cdot \frac{1}{3}} = x^{\frac{1}{2}} \).
Set the simplified expression equal to 9: \( x^{\frac{1}{2}} = 9 \).
To solve for \( x \), square both sides to eliminate the fractional exponent: \( \left(x^{\frac{1}{2}}\right)^2 = 9^2 \), which simplifies to \( x = 81 \).

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