Answer
$b^{\frac{5}{4}}$
Work Step by Step
RECALL:
(i) $a^{-m} = \dfrac{1}{a^m}$
(ii) $\sqrt[n]{a^m} = a^{\frac{m}{n}}$
(iii) $a^m \cdot a^n = a^{m+n}$
Use rule (ii) above to obtain:
$=b^{\frac{3}{4}} \cdot b^{\frac{1}{2}}$
Use rule (iii) above to to obtain:
$=b^{\frac{3}{4} + \frac{1}{2}}
\\=b^{\frac{3}{4} + \frac{2}{4}}
\\=b^{\frac{5}{4}}$
