# Thread: Math Help Forum: My Last Hope

1. ## Math Help Forum: My Last Hope

I have been trying every which way to solve in terms of y and just can't seem to do it. I have posted on various help sources (vark.com and yahoo answers) but no one can do it. Someone pointed me here and you are my last hope! Thanks for your help.

z=x((y^t) -1)/(y -1)

I have been trying every which way to solve in terms of y and just can't seem to do it. I have posted on various help sources (vark.com and yahoo answers) but no one can do it. Someone pointed me here and you are my last hope! Thanks for your help.

z=x((y^t) -1)/(y -1)
Assuming t is an integer, $\frac{z}{x} = \frac{y^t - 1}{y - 1} = 1 + y + y^2 + y^3 + .... + y^{t-1}$ provided $y \neq 1$. And since there's no general formula for solving a polynomial of degree greater than 4, you're out of luck in finding a general solution for y. It's not that no-one's smart enough to do it, it's a mathematical theorem that such a solution does not exist.

If you're certain that you should be able to make y the subject, then you should re-visit where this equation has come from because it might be wrong.

3. ~extremely complexity~

I just played with it for a sec...

$z=\frac{x(y^t-1)}{y-1}$

$y-1=\frac{x(y^t-1)}{z}$

$y=\frac{x(y^t-1)}{z}+1$

$y=\frac{xy^t-x}{z}+1$

Thats the best I can do...

4. @Mr. Fantastic & integral
Thanks so much for giving it a shot. Actually, I started with 1+y+(y^2)+(y^3)+... part and simplified to the equation I posted about.

t (period of time), z (total sales after period of time, t), and x (the initial rate of sales per period when t=1) are all values that I know or can willingly change to find y (the rate at which 'sales per period' changes). Ideally I can plug my knowns (z, t and x) into an excel chart to figure out y, the rate at which sales/period must change to meet the goal, z, of total sales.

For example:
In 5 months (t=5) I have a goal to sell 31 computers (z=31). In my first month I sell 1 computer (x=1). IF my sales per period doubles (y=2), then I sell 1 computer in the first month, that rate doubles and I sell 2 the second month, it doubles again and I sell 4 the 3rd month, 8 the 5th month, and 16 computers the 5th month. Total sales, z, is 1+2+4+8+16= 31. Now this is a simple, unrealistic, example but aside from trial and error I am seeking a formula to find y.

All help is greatly appreciated!

@Mr. Fantastic & integral
Thanks so much for giving it a shot. Actually, I started with 1+y+(y^2)+(y^3)+... part and simplified to the equation I posted about.

t (period of time), z (total sales after period of time, t), and x (the initial rate of sales per period when t=1) are all values that I know or can willingly change to find y (the rate at which 'sales per period' changes). Ideally I can plug my knowns (z, t and x) into an excel chart to figure out y, the rate at which sales/period must change to meet the goal, z, of total sales.

For example:
In 5 months (t=5) I have a goal to sell 31 computers (z=31). In my first month I sell 1 computer (x=1). IF my sales per period doubles (y=2), then I sell 1 computer in the first month, that rate doubles and I sell 2 the second month, it doubles again and I sell 4 the 3rd month, 8 the 5th month, and 16 computers the 5th month. Total sales, z, is 1+2+4+8+16= 31. Now this is a simple, unrealistic, example but aside from trial and error I am seeking a formula to find y.

All help is greatly appreciated!
You would be better off getting an approximate solution for each particular case by using technology.

6. I feel the underlying root object is the exact solution:

$\text{myexp}=\text{Root}[\#{}^{\wedge}t-1-u \#+u\&,k]$

I know that's cryptic but that's 21'st century for ya'. Once I have that in Mathematica, I can access any of the solutions I want for any value of t, k, and u. For example, suppose the exponent is 10, the factor z/x=2.5, and I want all the solutions (roots) of the expression. I'd write:

Code:
myexp = Root[#^t - 1 - u # + u &, k]
Table[N[myexp /. {t -> 10, u -> 2.5, k -> kval}], {kval, 1, 10}]

{0.602522, 1., -1.0932 - 0.382235 I, -1.0932 +
0.382235 I, -0.608924 - 0.969707 I, -0.608924 + 0.969707 I,
0.130353 - 1.10888 I, 0.130353 + 1.10888 I, 0.770506 - 0.736209 I,
0.770506 + 0.736209 I}
. . . well, I guess it's only nine roots since y not equal to one. Suppose I should have coded it as:

$\text{Solve}\left[u==\frac{y^{10}-1}{y-1},y\right]$

in which case Mathematica would have returned only nine root objects. You work out all the kinks ok.