I've been asked to simplify this expressing it with positive indices/exponents. Could someone please show me in steps how to do this? The answer is shown next to it. Thanks a lot.
$$\dfrac{{(\large a^\frac{-5}{2})}^2 b^2}{\large {a}^3(b^\frac{1}{4})^6} = \dfrac{b^\frac{1}{2}}{a^8}$$
On
$$ \begin{eqnarray*} \left( a^{-\frac 52} \right)^2 = a^{-5} \\ \frac 1{a^3}=a^{-3} \\ \dfrac 1 { \left( b^{\frac 14} \right)^6}=b^{-\frac 32} \end{eqnarray*} $$
so
$$ \begin{eqnarray*} \dfrac {\left( a^{-\frac 52} \right)^2b^2 }{a^3\left( b^{\frac 14} \right)^6} &=a^{-5}a^{-3}b^2 b^{-\frac 32} \\&=a^{-8}b^\frac 12 \end{eqnarray*} $$
Power of $a$ in the numerator = $\dfrac{-5}{2} \cdot 2 = -5$
Power of $a$ in the denominator = $3$
Overall power of $a$ in the final equation = $-5 - 3 = -8$
Power of $b$ in the numerator = $2$
Power of $b$ in the denominator = $\dfrac{1}{4} \cdot 6 = \dfrac{3}{2}$
Overall power of $b$ in the final equation = $2 - \dfrac{3}{2} = \dfrac{1}{2}$
Hence the answer is $a^{-8} \cdot b^{\frac{1}{2}} = \boxed{\dfrac{\large{b}^{\frac{1}{2}}}{\large{a}^8}}$