## College Algebra 7th Edition

$\dfrac{1}{2\sqrt{2}}$
Note that $-1.5=-\frac{3}{2}$ so the given expression can be written as: $2^{-\frac{3}{2}}$ RECALL: (i) $\sqrt[n]{a^m} = a^{\frac{m}{n}}$ (ii) $a^{-m} = \dfrac{1}{a^m}, a\ne0$ Use rule (ii) above to obtain: $=\dfrac{1}{2^{\frac{3}{2}}}$ Use rule (i) above, where m=3 and n=2, to obtain: $=\dfrac{1}{\sqrt{2^3}}$ Simply the denominator to obtain: $=\dfrac{1}{\sqrt{2^2 \cdot 2}} \\=\dfrac{1}{2\sqrt{2}}$