Problems with the number 0 Here is the original formula: \frac{256}{2^x}=y In order

Gage Potter

Gage Potter

Answered question

2022-04-30

Problems with the number 0
Here is the original formula:
2562x=y
In order to solve for x, I've done this:
log2(256y)=x
The problem is that y can be zero. What should I do to solve for x?

Answer & Explanation

attefrimibeocx

attefrimibeocx

Beginner2022-05-01Added 14 answers

If you rewrite 2562x=y as 256=y×2x , then it is clearly impossible to have y=0 while remaining within the real numbers, as 256 is not a multiple of 0. Since 2x is positive for all real x, you must have y>0.
So x=log2(256y)=8log2(y) is fine.
On the other hand, it is possible to have x=0, in which case 2x=1 and so y=256 and log2(y)=8.
Jonas Dickerson

Jonas Dickerson

Beginner2022-05-02Added 22 answers

HINT Rewrite it as Y=28X

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