Question

Mark (✓) against the correct answer

${\left[{\left\{{\left(-\frac{1}{2}\right)}^2\right\}}^{-2}\right]}^{-1}=?$

(a) $\frac{1}{16}$
(b) 16
(c) $\frac{-1}{16}$
(d) −16

Answer 1

(a) $\frac{1}{16}$

${\left[{\left\{{\left(-\frac{1}{2}\right)}^2\right\}}^{-2}\right]}^{-1}={\left[{\left(-\frac{1}{2}\right)}^{2\times -2}\right]}^{-1}$ $\left[s\mathrm{ince}{\left\{{\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{\mathrm{m}}\right\}}^{\mathrm{n}}={\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{\mathrm{mn}}\right]$
$={\left[{\left(-\frac{1}{2}\right)}^{-4}\right]}^{-1}$
$$\begin{gathered} ={\left(-\frac{1}{2}\right)}^{\left(-4\right)\times \left(-1\right)} \\ ={\left(-\frac{1}{2}\right)}^4=\frac{{\left(-1\right)}^4}{{\left(2\right)}^4} \\ =\frac{1}{16} \end{gathered}$$