I'm looking at Eqn. 12 here

I tried the power rule $a^m \div a^n = a^{m-n}$ to get $2^1 / 2^s$. But probably this is wrong.

Looking at the equation this should probably be $1/2^s$.

Can you help?


I see in this question that the same expression is written in reverse as $2^{s-1}$ so this may be literally 2 raised to the power of $(1-s)$.

  • 1
    $\begingroup$ Your original understanding was right, $2^{1 - s} = \frac{2}{2^s}$. $\endgroup$
    – ConMan
    Commented Jan 31 at 23:12

1 Answer 1


$\frac{2^{1}}{2^{s}}$ is indeed $2^{1-s}$. There are many ways to prove this, with the easiest way of subtituting $s$ as an integer. Say $s=2$: $2^{1-(2)}=2^{-1}=0.5$. $\frac{2^{1}}{2^{(2)}}=0.5$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .