Gage Potter

2022-04-30

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:

${\mathrm{log}}_{2}\left(\frac{256}{y}\right)=x$

The problem is that y can be zero. What should I do to solve for x?

Here is the original formula:

In order to solve for x, I've done this:

The problem is that y can be zero. What should I do to solve for x?

attefrimibeocx

Beginner2022-05-01Added 14 answers

If you rewrite $\frac{256}{{2}^{x}}=y$ 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$x={\mathrm{log}}_{2}\left(\frac{256}{y}\right)=8-{\mathrm{log}}_{2}\left(y\right)$ is fine.

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

So

On the other hand, it is possible to have

Jonas Dickerson

Beginner2022-05-02Added 22 answers

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