Hard time figuring out word problem Computers are beginning to lose power. The number of computers losing power doubles every minute. At 10 minutes half of the computers have lost power. After how many minutes will all computers be without power?

The only thing I could come up with is 

$$y- (2^9)x = .5y$$

Or a summation? 

$$y-x2^i \text{ from 1 to 9?}$$

I have no clue.

Is it even solvable?
The wording is slightly ambiguous, but the second sentence is probably intended to mean that the total number of computers that have lost power doubles every minute. If so, you don’t need to set up any sort of function: if half of the computers have lost power after ten minutes, and one minute later twice as many will have lost power, then in $11$ minutes all of the computers will have lost power. If you want to set up a formula, let $n_1$ be the number of computers losing power in the first minute. Then after $t$ minutes $n_12^{t-1}$ computers will have lost power. You’re told that $$n_12^{10-1}=\frac12n\;,$$ where $n$ is the total number of computers; multiplying by $2$ yields $n_12^{10}=n$, and $n_12^{10}$ is the number of computers that have lost power after $10+1=11$ minutes.

The wording could also be read as meaning that if $n_1$ computers lose power in the first minute, then $2n_1$ lose power in the second minute, $4n_1$ in the third minute, and so on. In that case the number that have lost power after $t$ minutes is

$$n_1+2n_1+2^2n_1+\ldots+2^{t-1}n_1=n_1\sum_{k=0}^{t-1}2^k=n_1(2^t-1)\;,$$

and we have $$n_1(2^{10}-1)=\frac12n\;.$$

Then $n=2046n_1$, so after $t$ minutes the number of computers that have lost power is

$$n_1(2^t-1)=\frac1{2046}n(2^t-1)\;.$$

This is equal to $n$, the total number of computers, when $t$ satisfies the equation $2^t-1=2046$, or $2^t=2047$. Since $2^{11}=2048$, this occurs just a hair before $11$ minutes have passed; numerically, $t\approx10.999295387$ minutes.
The answer is $t=10\sqrt{2}$.

The initial assumption is that $\frac{dy}{dt}$ is negative and doubles every minute, hence, $\frac{dy}{dt}=-2t(L)$ , where L is a positive integer that represents the initial number of computers that lost power.

Integrating gives $y=-L(t^2)+c$.

Solving with the initial condition [ $y(10)=.5y(0)$ ] gives:

$y=L(200-(t^2))$, which leads to the only reasonable solution: that $y=0$ when $t=10\sqrt{2}$.