# Problems with the number $0$

Here is the original formula: $$\frac{256}{2^x}=y$$ In order to solve for $x$, I've done this: $$\log_{2}\left(\frac{256}{y}\right)=x$$ The problem is that $y$ can be zero. What should I do to solve for $x$?

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So $i$ is just a variable, and not the usual imaginary unit? – yunone Feb 26 '11 at 9:38
Correct, $s$ is also normal. (I probably should change them) – Mike Feb 26 '11 at 9:46
So $s$ has become $x$, and $i$ has become $y$ – Henry Feb 26 '11 at 10:08
If you are going to change the variables in the equations, you should change them in the text as well. It would make it easier to understand. – Ross Millikan Feb 26 '11 at 15:58
$y$ cannot equal zero, because $\frac{256}{2^x}$ can never be equal to $0$. – Arturo Magidin Feb 26 '11 at 20:15

If you rewrite $\dfrac{256}{2^x} = y$ as $256 = y \times 2^x$, then it is clearly impossible to have $y=0$ while remaining within the real numbers, as 256 is not a multiple of 0. Since $2^x$ is positive for all real $x$, you must have $y>0$.

So $x = \log_2 \left( \dfrac{256}{y} \right) = 8 - \log_2 \left( y \right)$ is fine.

On the other hand, it is possible to have $x=0$, in which case $2^x = 1$ and so $y = 256$ and $\log_2 \left( y \right) = 8$.

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Here let me change the equation to $\frac{256}{2^{((n+s) \mod 8)}}=256-m$ this proably makes more sense now. – Mike Feb 26 '11 at 10:31
I think that is a different question – Henry Feb 27 '11 at 12:32

HINT $\$ Rewrite it as $\rm\ Y = 2^{\:8 - X}$

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