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# is there a formula to divde x by itself y times.

0
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is there a formula to divde x by itself y times.

eg.

x=3

y=4

3/3/3/3

P.S. This is not my homework, it's for a complex dice roller program

Guest Jun 23, 2017

#1
+5573
+3

Maybe this:     $$\frac{x}{x^y}\qquad\text{which is equal to}\qquad x^{1-y}$$

Here,   " 1 "   is the number of    x's    that you want in the numerator, and  " y "   is the number of    x's    that you want in the denominator.

So for example, if  y = 2   it becomes   x/(x*x)  =   ((x / x) / x)   =  x / x / x

..........

This doesn't work on your example though.

If you want your example to work, just replace   " y "  with  " (y - 1) "

hectictar  Jun 23, 2017
Sort:

#1
+5573
+3

Maybe this:     $$\frac{x}{x^y}\qquad\text{which is equal to}\qquad x^{1-y}$$

Here,   " 1 "   is the number of    x's    that you want in the numerator, and  " y "   is the number of    x's    that you want in the denominator.

So for example, if  y = 2   it becomes   x/(x*x)  =   ((x / x) / x)   =  x / x / x

..........

This doesn't work on your example though.

If you want your example to work, just replace   " y "  with  " (y - 1) "

hectictar  Jun 23, 2017
#2
0

I solved my question !!!!!

I thought it over in my head cuz' 3/3/3/3 = 1/9

Then I realized it's either 1/x^(sqrt(y)) or 1/x^(y/2) or 1/x^(y-2)

The 3rd one is correct. The calculator simplified it to x^(2-y)

Guest Jun 27, 2017
#3
+5573
+1

Also......1/9   =  1 / (32)

And.....x^(2 - y) is what you get if you replace  " y "  with  " y - 1 "  in:     $$x^{1-y}$$

$$x^{1-(y-1)}=x^{1-y+1}=x^{2-y}$$

hectictar  Jun 27, 2017

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