Experimentally finding order of a reaction For the reaction of bromocresol green and bleach, I was supposed to conduct an experiment to find the order of the reaction with respect to bleach. To do this, I used a fixed amount of bromocresol green and varied the amount of bleach in each trial. I recorded the amount of time it took for each reaction to finish. 

To find the order, I was told that if the order = x, $\frac{time_1}{time_2} = (\frac{[bleach]_2}{[bleach]_1})^x$. However, I'm having trouble understanding why this is true. For example, say we begin with $[bleach]_1=y$ and $[bleach]_2=2y$. If $\frac{time_1}{time_2} = 2^x$, this means that the rate of reaction 2 is always $2^x$ times faster than the rate of reaction 1. 

In the beginning, this is definitely true, since rate = $k[bleach]^x$. Initial $rate_1 = ky^x$ and initial $rate_2=k(2y)^x$. But at a time $dt$ later, $[bleach]_1=y-ky^x\cdot dt$ and $[bleach]_2=2y-k(2y)^x\cdot dt=2y-2^xky^x\cdot dt$. At this time, $\frac{rate_2}{rate_1} = (\frac{2y-2^xky^x\cdot dt}{y-ky^x\cdot dt})^x$, which does not equal $2^x$ unless $x=1$. 

Thus, I'm not understanding why $\frac{time_1}{time_2} = (\frac{[bleach]_2}{[bleach]_1})^x$ is true for x other than 1. Any clarifications would be much appreciated.
There are a couple of odd things here. 


the time for a reaction to be complete is usually infinite in theoretical kinetics, so I don't see how it could be useful and/or relevant. One usually looks at the initial rate of a reaction.
if bleach is in large excess, you may have a degenerate order. That means that the concentration of bleach barely varies during the reaction and can be considered a constant.
you use an expression with an infinitesimal amount of time ($\mathrm{d}t$), when $\mathrm{d}t$ is everything but infinitely small... 


Let me note bleach $\mathrm{ClO^-}$, and bromocresol green "BCG".

I assume that the reaction is monitored through the changes in absorbance, that is, the concentration of BCG. Its rate $r$ is the rate at which BCG is disappearing.

$$r = - \frac{\mathrm{d}[\mathrm{BCG}]}{\mathrm{d} t} = k \times [\mathrm{ClO^-}]^\alpha \times  [\mathrm{BCG}]^\beta $$ where $\alpha$ and $\beta$ are the partial orders.
If you use two different concentrations of bleach ($C_1$ and $C_2$), the ratio of rates will be :
$$\frac{r_1}{r_2} = \frac{k \times (C_1)^\alpha \times  [\mathrm{BCG}]^\beta}{k \times (C_2)^\alpha \times  [\mathrm{BCG}]^\beta} = \left(\frac{C_1}{C_2}\right)^\alpha$$

A simple way to get the initial rate is to measure the slope of the tangent to the graph of  ${[\mathrm{BCG}]}$ along time (dotted line).



This line crosses the time axis at $t=\Delta t$.

In this case, 
$$ r_i(t=0) = \frac{\mathrm{[BCG] (t=0)}}{\Delta t_i}$$
Therefore, one can write  :
$$\frac{r_1}{r_2} = \left(\frac{C_1}{C_2}\right)^\alpha =  \frac{\Delta t_2}{\Delta t_1}$$
which kind of looks like the formula you were given.