For example:
$${\frac{{\mathtt{x}}}{{{\mathtt{2}}}^{-{\mathtt{3}}}}}$$
$${\frac{{\mathtt{x}}}{\left({\frac{{\mathtt{1}}}{{{\mathtt{2}}}^{{\mathtt{3}}}}}\right)}}$$
$$\left({\frac{{\mathtt{x}}}{{\mathtt{1}}}}\right){\mathtt{\,\times\,}}\left({\frac{\left({{\mathtt{2}}}^{{\mathtt{3}}}\right)}{{\mathtt{1}}}}\right)$$
$${{\mathtt{2}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\mathtt{x}}$$
For example:
$${\frac{{\mathtt{x}}}{{{\mathtt{2}}}^{-{\mathtt{3}}}}}$$
$${\frac{{\mathtt{x}}}{\left({\frac{{\mathtt{1}}}{{{\mathtt{2}}}^{{\mathtt{3}}}}}\right)}}$$
$$\left({\frac{{\mathtt{x}}}{{\mathtt{1}}}}\right){\mathtt{\,\times\,}}\left({\frac{\left({{\mathtt{2}}}^{{\mathtt{3}}}\right)}{{\mathtt{1}}}}\right)$$
$${{\mathtt{2}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\mathtt{x}}$$
Hi Zac,
you always get these correct but you do not need to make them looks so complicated.
$$\\2^{-3}=\frac{2^{-3}}{1}\\\\
$Now anything raised to a neg power gets swapped to the other side of the fraction line, and anything not raised to a negative power stays where it is$\\\\
$So the 1 stays , the $2^{-3}$ goes to the other side of the fraction line and the - becomes a +. $\\\\
$There is nothing on the top any more so just put a 1 there$\\\\
\frac{2^{-3}}{1}=\frac{?}{1*2^{+3}}=\frac{1}{2^{3}}\\\\$$
Once you do a couple you wil see that it is heaps easier.
Oh I did not do the correct question
$$\frac{x}{2^{-3}}=\frac{x*2^{+3}}{1}=8x$$
Well I like to show the in-between steps because I think it makes it easier to understand but once you know the rule you can just cut straight to the answer.