## Algebra 2 Common Core

$\dfrac{1}{128}$
With $3.5=\frac{7}{2}$, the given expression can be written as: $=4^{-\frac{7}{2}}$ Recall the basic exponent property (pg. 360): $(a^m)^n=a^{mn}$ Using this property, we get: $4^{-\frac{7}{2}}=(4^{\frac{1}{2}})^{-7}$ Recall that $\sqrt[n]{x}=x^{\frac{1}{n}}$. Thus, the expression above can be written as: $=(\sqrt{4})^{-7}$ Since $2^2=4$, then the expression above simplifies to: $=(2)^{-7}$ Recall the basic exponent property (pg. 383): $a^{-m}=\frac{1}{a^m}$ Applying this, we get: $(2)^{-7}=\dfrac{1}{(2)^{7}}=\dfrac{1}{128}$