## Calculus with Applications (10th Edition)

$b^2\sqrt[4]{b}$
RECALL: (i) $a^m \cdot a^n = a^{m+n}$ (ii) $\sqrt[n]{a} = a^{\frac{1}{n}}$ (iii) $a^{\frac{m}{n}}=\sqrt[n]{a^m}$ Use rule (iii) to have: $=b^{\frac{3}{2}}\cdot b^{\frac{3}{4}} \\=b^{\frac{6}{4} + \frac{3}{4}} \\=b^{\frac{9}{4}}$ Use rule (i) to have: $\\=a^{\frac{1}{2} + \frac{1}{3}} \\=a^{\frac{3}{6} + \frac{2}{6}} \\=a^{\frac{5}{6}}$ Use rule (iii) to have: $=\sqrt[4]{b^9} \\=\sqrt[4]{b^8(b)} \\=b^2\sqrt[4]{b}$