Simplify $\displaystyle \frac{(4a^{4}b)^{3}}{10a^{3}b^{7}}$, leaving your answer in positive index form. Answer ... Given that $2^{x} = 3$, $2^{y} = 7$ and $2^{z} = \dfrac{9}{49}$, use the laws of indices to find the value of $2^{3x - y + \tfrac{1}{2} - \tfrac{z}{2}}$. Answer ...

1. Expand the power in the numerator using the laws of indices.

$(4a^4b)^3=4^3(a^4)^3b^3=64a^{12}b^3$
So
$\frac{(4a^4b)^3}{10a^3b^7}=\frac{64a^{12}b^3}{10a^3b^7}.$

2. Simplify the coefficient and subtract indices for common bases, leaving positive indices only.

$\frac{64a^{12}b^3}{10a^3b^7}=\frac{64}{10}a^{12-3}b^{3-7}$
$=\frac{32}{5}a^9b^{-4}=\frac{32a^9}{5b^4}$

1. Split the power into products and quotients using the laws of indices.

$2^{3x-y+\tfrac12-\tfrac z2}=\dfrac{2^{3x}\cdot 2^{\tfrac12}}{2^y\cdot 2^{\tfrac z2}}=\dfrac{(2^x)^3\sqrt{2}}{2^y\sqrt{2^z}}$

2. Substitute the given values and simplify exactly.

$\dfrac{(2^x)^3\sqrt{2}}{2^y\sqrt{2^z}}=\dfrac{3^3\sqrt{2}}{7\sqrt{\tfrac{9}{49}}}=\dfrac{27\sqrt{2}}{7\cdot \tfrac37}=9\sqrt{2}$

Alternatively, since $2^z=\dfrac{9}{49}=\left(\dfrac37\right)^2$, $2^{z/2}=\dfrac37$.